SQL query to find a list of city names that dont start with vowels
I'm trying to query the list of CITY names from table - STATION that do not start with vowels with results that cannot contain duplicates.
The table just has id, city, population
This is the query that I've written
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '[^bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';
This gives a wrong answer. What am I doing wrong here?
mysql
asked Apr 14 '16 at 21:59
ZeusZeus
7882830
add a comment |
I'm trying to query the list of CITY names from table - STATION that do not start with vowels with results that cannot contain duplicates.
The table just has id, city, population
This is the query that I've written
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '[^bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';
This gives a wrong answer. What am I doing wrong here?
mysql
asked Apr 14 '16 at 21:59
ZeusZeus
7882830
try theNOT RLIKE '^[aeiouAEIOU].*$'
– Tin Tran
Apr 14 '16 at 22:05
add a comment |
I'm trying to query the list of CITY names from table - STATION that do not start with vowels with results that cannot contain duplicates.
The table just has id, city, population
This is the query that I've written
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '[^bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';
This gives a wrong answer. What am I doing wrong here?
mysql
asked Apr 14 '16 at 21:59
ZeusZeus
7882830
I'm trying to query the list of CITY names from table - STATION that do not start with vowels with results that cannot contain duplicates.
The table just has id, city, population
This is the query that I've written
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '[^bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';
This gives a wrong answer. What am I doing wrong here?
mysql
mysql
asked Apr 14 '16 at 21:59
ZeusZeus
7882830
asked Apr 14 '16 at 21:59
ZeusZeus
7882830
asked Apr 14 '16 at 21:59
ZeusZeus
7882830
asked Apr 14 '16 at 21:59
ZeusZeus
7882830
asked Apr 14 '16 at 21:59
ZeusZeus
7882830
7882830
try theNOT RLIKE '^[aeiouAEIOU].*$'
– Tin Tran
Apr 14 '16 at 22:05
add a comment |
try theNOT RLIKE '^[aeiouAEIOU].*$'
– Tin Tran
Apr 14 '16 at 22:05
try the
NOT RLIKE '^[aeiouAEIOU].*$'– Tin Tran
Apr 14 '16 at 22:05
try the
NOT RLIKE '^[aeiouAEIOU].*$'– Tin Tran
Apr 14 '16 at 22:05
add a comment |
27 Answers
27
active
oldest
votes
try this.
SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '^[aeiouAEIOU].*$'
answered Apr 14 '16 at 22:06
Tin TranTin Tran
5,43021226
add a comment |
A ^ in regular expressions can have different meanings, depending on its location. When it's the first character in a regex, it refers to the start of the string. But when it's the first character in a set, like [^abc], it means not one of. And when it appears elsewhere, it just refers to the ^ itself.
So you would need something like:
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';
or, just exclude the letters you don't want:
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^[^aeiouAEIOU].*';
answered Apr 14 '16 at 22:07
ArjanArjan
8,39612132
add a comment |
Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. IN ORACLE
select distinct city
from station
where regexp_like(city, '^[^aeiouAEIOU]|*[^aeiouAEIOU]$');
In MySQL -
select distinct city
from station
where city RLIKE '^[^aeiouAEIOU].*' OR
city RLIKE '^.*[^aeiouAEIOU]$';
edited Aug 6 '18 at 18:00
Community♦
11
answered Feb 18 '18 at 18:45
Rahul YadavRahul Yadav
233
add a comment |
select distinct city from station
where city Not like 'A%' and city Not like 'E%' and city Not like 'I%' and
city Not like 'o%' and city not like 'U%';
edited Jul 14 '16 at 12:41
Bo Persson
78.2k17118184
answered Jul 14 '16 at 9:58
satishsatish
211
add a comment |
This worked on MS SQL SERVER:
select distinct city from station where city NOT like '[aeuio]%' Order by City;
edited Dec 24 '16 at 8:56
FelixSFD
4,29573086
answered Dec 24 '16 at 8:41
abhinay vijayabhinay vijay
311
it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41
add a comment |
SELECT DISTINCT CITY from STATION WHERE CITY RLIKE '^[^aeiouAEIOU].*$'
edited Dec 24 '16 at 8:59
Pang
6,9081664102
answered Sep 16 '16 at 0:47
SunnyShineColoradoSunnyShineColorado
111
This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48
add a comment |
select distinct city from station where city regexp
'^[^aeiou].*[^aeiou]$'
If your query doesn't starts and ends with a vowel.
answered May 21 '17 at 16:42
marwarimarwari
5818
add a comment |
Try the following:
SELECT DISTINCT CITY FROM STATION
WHERE CITY REGEXP '^[^aeiou].*';
MySQL Ref.: https://dev.mysql.com/doc/refman/5.5/en/regexp.html
edited Feb 21 '18 at 17:59
Syed Aslam
7,63953253
answered Feb 21 '18 at 16:42
simran kumarisimran kumari
111
add a comment |
Simple example without regular expressions
SELECT DISTINCT
City
FROM Station
WHERE LEFT(City, 1) NOT IN ("a", "e", "i", "o", "u");
edited Nov 25 '18 at 3:55
digital.aaron
3,2771331
answered Sep 3 '18 at 11:04
ed lovejoyed lovejoy
111
add a comment |
I tried this.
select distinct city
from station
where substring(city,1,1) not in('A','E','I','O','U');
edited Dec 5 '18 at 20:42
John Montgomery
2,64362237
answered Dec 5 '18 at 20:19
Makhmud GalyMakhmud Galy
112
add a comment |
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^.*[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ]$';
edited Dec 9 '16 at 17:25
Paul Roub
32.8k85874
answered Dec 9 '16 at 17:16
MANJUNATH KAMMARMANJUNATH KAMMAR
1
add a comment |
In MS SQL:
Select DISTINCT CITY
FROM
STATION
Where CITY NOT LIKE 'A%' AND CITY NOT LIKE 'E%' AND CITY NOT LIKE 'I%' AND CITY NOT LIKE 'O%' AND CITY NOT LIKE 'U%';
answered Jan 24 '17 at 23:59
KrishneilKrishneil
659816
add a comment |
you can use
select distinct city from station where city not RLIKE '^[aeiouAEIOU].*'
answered Mar 8 '17 at 4:33
Sajal SinghSajal Singh
146211
add a comment |
select distinct CITY from STATION where left(CITY,1) not in ('a','e','i','o','u') or right(CITY,1) not in ('a','e','i','o','u');
answered Mar 8 '17 at 17:42
MohitMohit
541312
add a comment |
select DISTINCT CITY from STATION where CITY NOT LIKE '[a,e,i,o,u]%'
answered May 26 '17 at 10:14
Aman VermaAman Verma
614
add a comment |
try this one
SELECT DISTINCT CITY FROM STATION WHERE
LOWER(SUBSTR(CITY,LENGTH(CITY),1)) IN ('a','e','i','o','u');
answered Oct 12 '17 at 11:12
Ganesh GiriGanesh Giri
30515
add a comment |
Select Distinct City from Station where City Like '[^aeiou]%';
MS SQL Server Tested.
edited Apr 5 '18 at 21:32
Jose Da Silva
2,09421026
answered Apr 5 '18 at 21:02
DarpanDarpan
1
add a comment |
SELECT DISTINCT city from STATION
WHERE city NOT LIKE "[aieuo]%[aieuo] [aieuo]%[aieuo]"
AND city NOT LIKE "[aieuo]%[aieuo]";
The first NOT LIKE means, that each word of the two-word city names doesn't start and end with vowels. The second NOT LIKE is for one-word city names.
answered Sep 25 '18 at 20:05
David OsipovDavid Osipov
12
add a comment |
select distinct city from station where (
left(city,1) not in ('a','e','i','o','u')
and
right(city,1) not in ('a','e','i','o','u')
);
edited Oct 21 '18 at 9:49
Mohammad
15.6k123562
answered Oct 21 '18 at 9:27
Rohit RawatRohit Rawat
1
1
PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34
Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05
add a comment |
In ORACLE:
select
distinct(city)
from station
where regexp_like (city, '^[^A|E|I|O|U](*)');
There is no need to specify the ending letter in this solution. It's not called out for in the problem presented.
edited Nov 19 '18 at 20:50
Hossein Golshani
2,09051123
answered Nov 19 '18 at 20:40
Aaron W.Aaron W.
1
Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41
add a comment |
for oracle
select distinct
city
from station
where regexp_like(city, '^[^aeiouAEIOU].*');
edited Nov 25 '18 at 3:56
digital.aaron
3,2771331
answered Apr 28 '18 at 19:47
Rasim SENRasim SEN
11
add a comment |
select distinct city from station where city regexp '^[^aeiou].*[^aeiou]$'
distinct: to avoid the duplication on the following column
regexp: MySql function (regexp) to obtain Regular Expressions. expression starts with ^ and ends with $. In Oracle its regexp_like (city,'RegEx') Regular Expressions With Oracle Database.
[^...] means any character NOT contain any of ...[^...]. means a single character NOT contain any of ...[^...].* means first character NOT contain any of ...[^...].*[^...] means first character NOT contains any of ... AND not ends with ...[^aeiou].*[^aeiou] mean the first character NOT starts with vowels AND not ends with vowels
answered Nov 25 '18 at 3:11
JayJay
32226
While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28
add a comment |
Mysql we can use this.
select distinct city from station where city regexp '^[^aeiou].*';
To Know More About MySQL Regular Expression
For Oracle we can use
select distinct(city) from station where regexp_like (city, '^[^AEIOU](*)');
To Know More About Oracle Regular Expression
For MS SQL Server
select distinct city from station where city not like '[aeuio]%';
answered Nov 30 '18 at 4:22
Arun SolomonArun Solomon
72
add a comment |
I worked with this
SELECT DISTINCT CITY FROM STATION
WHERE CITY NOT LIKE 'a%'
AND CITY NOT LIKE 'e%'
AND CITY NOT LIKE 'i%'
AND CITY NOT LIKE 'o%'
AND CITY NOT LIKE 'u%'
answered Jan 15 at 16:54
Kashif Faraz ShamsiKashif Faraz Shamsi
4431619
add a comment |
select distinct <COLUMN NAME> from <table name>
WHERE SUBSTRING(<COLUMN NAME>,1,1) not in ('a','e','i','o','u','A','E','I','O','U')
order by <COLUMN NAME>
Distinct words do not start with a vowel.
edited Jan 22 at 7:59
Prakash Pazhanisamy
8681923
answered Jan 22 at 6:13
srinidhi sundarrajansrinidhi sundarrajan
11
SUBSTRING(<COLUMN NAME>,1,1)
– srinidhi sundarrajan
Jan 22 at 6:14
Could you please add a bit more of context to your answer?
– toti08
Jan 22 at 6:38
add a comment |
select distinct city from station where
city not in
(select distinct city from station where
(city like "A%" or
city like "E%" or
city like "I%" or
city like "O%" or
city like "U%"
));
This will work i guess.
edited Oct 18 '16 at 4:14
noufalcep
2,25682236
answered Oct 18 '16 at 3:04
Saikrishna SyamalaSaikrishna Syamala
1
please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27
add a comment |
SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '[aeiouAEIOU]$'
This works for MYSQL
answered Dec 9 '17 at 7:33
Vivek GhanchiVivek Ghanchi
328
You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11
Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19
add a comment |
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27 Answers
27
active
oldest
votes
27 Answers
27
active
oldest
votes
active
oldest
votes
active
oldest
votes
try this.
SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '^[aeiouAEIOU].*$'
answered Apr 14 '16 at 22:06
Tin TranTin Tran
5,43021226
add a comment |
try this.
SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '^[aeiouAEIOU].*$'
answered Apr 14 '16 at 22:06
Tin TranTin Tran
5,43021226
add a comment |
try this.
SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '^[aeiouAEIOU].*$'
answered Apr 14 '16 at 22:06
Tin TranTin Tran
5,43021226
try this.
SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '^[aeiouAEIOU].*$'
answered Apr 14 '16 at 22:06
Tin TranTin Tran
5,43021226
answered Apr 14 '16 at 22:06
Tin TranTin Tran
5,43021226
answered Apr 14 '16 at 22:06
Tin TranTin Tran
5,43021226
answered Apr 14 '16 at 22:06
Tin TranTin Tran
5,43021226
5,43021226
add a comment |
add a comment |
A ^ in regular expressions can have different meanings, depending on its location. When it's the first character in a regex, it refers to the start of the string. But when it's the first character in a set, like [^abc], it means not one of. And when it appears elsewhere, it just refers to the ^ itself.
So you would need something like:
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';
or, just exclude the letters you don't want:
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^[^aeiouAEIOU].*';
answered Apr 14 '16 at 22:07
ArjanArjan
8,39612132
add a comment |
A ^ in regular expressions can have different meanings, depending on its location. When it's the first character in a regex, it refers to the start of the string. But when it's the first character in a set, like [^abc], it means not one of. And when it appears elsewhere, it just refers to the ^ itself.
So you would need something like:
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';
or, just exclude the letters you don't want:
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^[^aeiouAEIOU].*';
answered Apr 14 '16 at 22:07
ArjanArjan
8,39612132
add a comment |
A ^ in regular expressions can have different meanings, depending on its location. When it's the first character in a regex, it refers to the start of the string. But when it's the first character in a set, like [^abc], it means not one of. And when it appears elsewhere, it just refers to the ^ itself.
So you would need something like:
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';
or, just exclude the letters you don't want:
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^[^aeiouAEIOU].*';
answered Apr 14 '16 at 22:07
ArjanArjan
8,39612132
A ^ in regular expressions can have different meanings, depending on its location. When it's the first character in a regex, it refers to the start of the string. But when it's the first character in a set, like [^abc], it means not one of. And when it appears elsewhere, it just refers to the ^ itself.
So you would need something like:
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';
or, just exclude the letters you don't want:
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^[^aeiouAEIOU].*';
answered Apr 14 '16 at 22:07
ArjanArjan
8,39612132
answered Apr 14 '16 at 22:07
ArjanArjan
8,39612132
answered Apr 14 '16 at 22:07
ArjanArjan
8,39612132
answered Apr 14 '16 at 22:07
ArjanArjan
8,39612132
8,39612132
add a comment |
add a comment |
Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. IN ORACLE
select distinct city
from station
where regexp_like(city, '^[^aeiouAEIOU]|*[^aeiouAEIOU]$');
In MySQL -
select distinct city
from station
where city RLIKE '^[^aeiouAEIOU].*' OR
city RLIKE '^.*[^aeiouAEIOU]$';
edited Aug 6 '18 at 18:00
Community♦
11
answered Feb 18 '18 at 18:45
Rahul YadavRahul Yadav
233
add a comment |
Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. IN ORACLE
select distinct city
from station
where regexp_like(city, '^[^aeiouAEIOU]|*[^aeiouAEIOU]$');
In MySQL -
select distinct city
from station
where city RLIKE '^[^aeiouAEIOU].*' OR
city RLIKE '^.*[^aeiouAEIOU]$';
edited Aug 6 '18 at 18:00
Community♦
11
answered Feb 18 '18 at 18:45
Rahul YadavRahul Yadav
233
add a comment |
Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. IN ORACLE
select distinct city
from station
where regexp_like(city, '^[^aeiouAEIOU]|*[^aeiouAEIOU]$');
In MySQL -
select distinct city
from station
where city RLIKE '^[^aeiouAEIOU].*' OR
city RLIKE '^.*[^aeiouAEIOU]$';
edited Aug 6 '18 at 18:00
Community♦
11
answered Feb 18 '18 at 18:45
Rahul YadavRahul Yadav
233
Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. IN ORACLE
select distinct city
from station
where regexp_like(city, '^[^aeiouAEIOU]|*[^aeiouAEIOU]$');
In MySQL -
select distinct city
from station
where city RLIKE '^[^aeiouAEIOU].*' OR
city RLIKE '^.*[^aeiouAEIOU]$';
edited Aug 6 '18 at 18:00
Community♦
11
answered Feb 18 '18 at 18:45
Rahul YadavRahul Yadav
233
edited Aug 6 '18 at 18:00
Community♦
11
edited Aug 6 '18 at 18:00
Community♦
11
edited Aug 6 '18 at 18:00
Community♦
11
11
answered Feb 18 '18 at 18:45
Rahul YadavRahul Yadav
233
answered Feb 18 '18 at 18:45
Rahul YadavRahul Yadav
233
answered Feb 18 '18 at 18:45
Rahul YadavRahul Yadav
233
233
add a comment |
add a comment |
select distinct city from station
where city Not like 'A%' and city Not like 'E%' and city Not like 'I%' and
city Not like 'o%' and city not like 'U%';
edited Jul 14 '16 at 12:41
Bo Persson
78.2k17118184
answered Jul 14 '16 at 9:58
satishsatish
211
add a comment |
select distinct city from station
where city Not like 'A%' and city Not like 'E%' and city Not like 'I%' and
city Not like 'o%' and city not like 'U%';
edited Jul 14 '16 at 12:41
Bo Persson
78.2k17118184
answered Jul 14 '16 at 9:58
satishsatish
211
add a comment |
select distinct city from station
where city Not like 'A%' and city Not like 'E%' and city Not like 'I%' and
city Not like 'o%' and city not like 'U%';
edited Jul 14 '16 at 12:41
Bo Persson
78.2k17118184
answered Jul 14 '16 at 9:58
satishsatish
211
select distinct city from station
where city Not like 'A%' and city Not like 'E%' and city Not like 'I%' and
city Not like 'o%' and city not like 'U%';
edited Jul 14 '16 at 12:41
Bo Persson
78.2k17118184
answered Jul 14 '16 at 9:58
satishsatish
211
edited Jul 14 '16 at 12:41
Bo Persson
78.2k17118184
edited Jul 14 '16 at 12:41
Bo Persson
78.2k17118184
edited Jul 14 '16 at 12:41
Bo Persson
78.2k17118184
78.2k17118184
answered Jul 14 '16 at 9:58
satishsatish
211
answered Jul 14 '16 at 9:58
satishsatish
211
answered Jul 14 '16 at 9:58
satishsatish
211
211
add a comment |
add a comment |
This worked on MS SQL SERVER:
select distinct city from station where city NOT like '[aeuio]%' Order by City;
edited Dec 24 '16 at 8:56
FelixSFD
4,29573086
answered Dec 24 '16 at 8:41
abhinay vijayabhinay vijay
311
it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41
add a comment |
This worked on MS SQL SERVER:
select distinct city from station where city NOT like '[aeuio]%' Order by City;
edited Dec 24 '16 at 8:56
FelixSFD
4,29573086
answered Dec 24 '16 at 8:41
abhinay vijayabhinay vijay
311
it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41
add a comment |
This worked on MS SQL SERVER:
select distinct city from station where city NOT like '[aeuio]%' Order by City;
edited Dec 24 '16 at 8:56
FelixSFD
4,29573086
answered Dec 24 '16 at 8:41
abhinay vijayabhinay vijay
311
This worked on MS SQL SERVER:
select distinct city from station where city NOT like '[aeuio]%' Order by City;
edited Dec 24 '16 at 8:56
FelixSFD
4,29573086
answered Dec 24 '16 at 8:41
abhinay vijayabhinay vijay
311
edited Dec 24 '16 at 8:56
FelixSFD
4,29573086
edited Dec 24 '16 at 8:56
FelixSFD
4,29573086
edited Dec 24 '16 at 8:56
FelixSFD
4,29573086
4,29573086
answered Dec 24 '16 at 8:41
abhinay vijayabhinay vijay
311
answered Dec 24 '16 at 8:41
abhinay vijayabhinay vijay
311
answered Dec 24 '16 at 8:41
abhinay vijayabhinay vijay
311
311
it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41
add a comment |
it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41
it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41
it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41
add a comment |
SELECT DISTINCT CITY from STATION WHERE CITY RLIKE '^[^aeiouAEIOU].*$'
edited Dec 24 '16 at 8:59
Pang
6,9081664102
answered Sep 16 '16 at 0:47
SunnyShineColoradoSunnyShineColorado
111
This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48
add a comment |
SELECT DISTINCT CITY from STATION WHERE CITY RLIKE '^[^aeiouAEIOU].*$'
edited Dec 24 '16 at 8:59
Pang
6,9081664102
answered Sep 16 '16 at 0:47
SunnyShineColoradoSunnyShineColorado
111
This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48
add a comment |
SELECT DISTINCT CITY from STATION WHERE CITY RLIKE '^[^aeiouAEIOU].*$'
edited Dec 24 '16 at 8:59
Pang
6,9081664102
answered Sep 16 '16 at 0:47
SunnyShineColoradoSunnyShineColorado
111
SELECT DISTINCT CITY from STATION WHERE CITY RLIKE '^[^aeiouAEIOU].*$'
edited Dec 24 '16 at 8:59
Pang
6,9081664102
answered Sep 16 '16 at 0:47
SunnyShineColoradoSunnyShineColorado
111
edited Dec 24 '16 at 8:59
Pang
6,9081664102
edited Dec 24 '16 at 8:59
Pang
6,9081664102
edited Dec 24 '16 at 8:59
Pang
6,9081664102
6,9081664102
answered Sep 16 '16 at 0:47
SunnyShineColoradoSunnyShineColorado
111
answered Sep 16 '16 at 0:47
SunnyShineColoradoSunnyShineColorado
111
answered Sep 16 '16 at 0:47
SunnyShineColoradoSunnyShineColorado
111
111
This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48
add a comment |
This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48
This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48
This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48
add a comment |
select distinct city from station where city regexp
'^[^aeiou].*[^aeiou]$'
If your query doesn't starts and ends with a vowel.
answered May 21 '17 at 16:42
marwarimarwari
5818
add a comment |
select distinct city from station where city regexp
'^[^aeiou].*[^aeiou]$'
If your query doesn't starts and ends with a vowel.
answered May 21 '17 at 16:42
marwarimarwari
5818
add a comment |
select distinct city from station where city regexp
'^[^aeiou].*[^aeiou]$'
If your query doesn't starts and ends with a vowel.
answered May 21 '17 at 16:42
marwarimarwari
5818
select distinct city from station where city regexp
'^[^aeiou].*[^aeiou]$'
If your query doesn't starts and ends with a vowel.
answered May 21 '17 at 16:42
marwarimarwari
5818
answered May 21 '17 at 16:42
marwarimarwari
5818
answered May 21 '17 at 16:42
marwarimarwari
5818
answered May 21 '17 at 16:42
marwarimarwari
5818
5818
add a comment |
add a comment |
Try the following:
SELECT DISTINCT CITY FROM STATION
WHERE CITY REGEXP '^[^aeiou].*';
MySQL Ref.: https://dev.mysql.com/doc/refman/5.5/en/regexp.html
edited Feb 21 '18 at 17:59
Syed Aslam
7,63953253
answered Feb 21 '18 at 16:42
simran kumarisimran kumari
111
add a comment |
Try the following:
SELECT DISTINCT CITY FROM STATION
WHERE CITY REGEXP '^[^aeiou].*';
MySQL Ref.: https://dev.mysql.com/doc/refman/5.5/en/regexp.html
edited Feb 21 '18 at 17:59
Syed Aslam
7,63953253
answered Feb 21 '18 at 16:42
simran kumarisimran kumari
111
add a comment |
Try the following:
SELECT DISTINCT CITY FROM STATION
WHERE CITY REGEXP '^[^aeiou].*';
MySQL Ref.: https://dev.mysql.com/doc/refman/5.5/en/regexp.html
edited Feb 21 '18 at 17:59
Syed Aslam
7,63953253
answered Feb 21 '18 at 16:42
simran kumarisimran kumari
111
Try the following:
SELECT DISTINCT CITY FROM STATION
WHERE CITY REGEXP '^[^aeiou].*';
MySQL Ref.: https://dev.mysql.com/doc/refman/5.5/en/regexp.html
edited Feb 21 '18 at 17:59
Syed Aslam
7,63953253
answered Feb 21 '18 at 16:42
simran kumarisimran kumari
111
edited Feb 21 '18 at 17:59
Syed Aslam
7,63953253
edited Feb 21 '18 at 17:59
Syed Aslam
7,63953253
edited Feb 21 '18 at 17:59
Syed Aslam
7,63953253
7,63953253
answered Feb 21 '18 at 16:42
simran kumarisimran kumari
111
answered Feb 21 '18 at 16:42
simran kumarisimran kumari
111
answered Feb 21 '18 at 16:42
simran kumarisimran kumari
111
111
add a comment |
add a comment |
Simple example without regular expressions
SELECT DISTINCT
City
FROM Station
WHERE LEFT(City, 1) NOT IN ("a", "e", "i", "o", "u");
edited Nov 25 '18 at 3:55
digital.aaron
3,2771331
answered Sep 3 '18 at 11:04
ed lovejoyed lovejoy
111
add a comment |
Simple example without regular expressions
SELECT DISTINCT
City
FROM Station
WHERE LEFT(City, 1) NOT IN ("a", "e", "i", "o", "u");
edited Nov 25 '18 at 3:55
digital.aaron
3,2771331
answered Sep 3 '18 at 11:04
ed lovejoyed lovejoy
111
add a comment |
Simple example without regular expressions
SELECT DISTINCT
City
FROM Station
WHERE LEFT(City, 1) NOT IN ("a", "e", "i", "o", "u");
edited Nov 25 '18 at 3:55
digital.aaron
3,2771331
answered Sep 3 '18 at 11:04
ed lovejoyed lovejoy
111
Simple example without regular expressions
SELECT DISTINCT
City
FROM Station
WHERE LEFT(City, 1) NOT IN ("a", "e", "i", "o", "u");
edited Nov 25 '18 at 3:55
digital.aaron
3,2771331
answered Sep 3 '18 at 11:04
ed lovejoyed lovejoy
111
edited Nov 25 '18 at 3:55
digital.aaron
3,2771331
edited Nov 25 '18 at 3:55
digital.aaron
3,2771331
edited Nov 25 '18 at 3:55
digital.aaron
3,2771331
3,2771331
answered Sep 3 '18 at 11:04
ed lovejoyed lovejoy
111
answered Sep 3 '18 at 11:04
ed lovejoyed lovejoy
111
answered Sep 3 '18 at 11:04
ed lovejoyed lovejoy
111
111
add a comment |
add a comment |
I tried this.
select distinct city
from station
where substring(city,1,1) not in('A','E','I','O','U');
edited Dec 5 '18 at 20:42
John Montgomery
2,64362237
answered Dec 5 '18 at 20:19
Makhmud GalyMakhmud Galy
112
add a comment |
I tried this.
select distinct city
from station
where substring(city,1,1) not in('A','E','I','O','U');
edited Dec 5 '18 at 20:42
John Montgomery
2,64362237
answered Dec 5 '18 at 20:19
Makhmud GalyMakhmud Galy
112
add a comment |
I tried this.
select distinct city
from station
where substring(city,1,1) not in('A','E','I','O','U');
edited Dec 5 '18 at 20:42
John Montgomery
2,64362237
answered Dec 5 '18 at 20:19
Makhmud GalyMakhmud Galy
112
I tried this.
select distinct city
from station
where substring(city,1,1) not in('A','E','I','O','U');
edited Dec 5 '18 at 20:42
John Montgomery
2,64362237
answered Dec 5 '18 at 20:19
Makhmud GalyMakhmud Galy
112
edited Dec 5 '18 at 20:42
John Montgomery
2,64362237
edited Dec 5 '18 at 20:42
John Montgomery
2,64362237
edited Dec 5 '18 at 20:42
John Montgomery
2,64362237
2,64362237
answered Dec 5 '18 at 20:19
Makhmud GalyMakhmud Galy
112
answered Dec 5 '18 at 20:19
Makhmud GalyMakhmud Galy
112
answered Dec 5 '18 at 20:19
Makhmud GalyMakhmud Galy
112
112
add a comment |
add a comment |
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^.*[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ]$';
edited Dec 9 '16 at 17:25
Paul Roub
32.8k85874
answered Dec 9 '16 at 17:16
MANJUNATH KAMMARMANJUNATH KAMMAR
1
add a comment |
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^.*[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ]$';
edited Dec 9 '16 at 17:25
Paul Roub
32.8k85874
answered Dec 9 '16 at 17:16
MANJUNATH KAMMARMANJUNATH KAMMAR
1
add a comment |
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^.*[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ]$';
edited Dec 9 '16 at 17:25
Paul Roub
32.8k85874
answered Dec 9 '16 at 17:16
MANJUNATH KAMMARMANJUNATH KAMMAR
1
SELECT DISTINCT CITY FROM STATION
WHERE CITY RLIKE '^.*[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ]$';
edited Dec 9 '16 at 17:25
Paul Roub
32.8k85874
answered Dec 9 '16 at 17:16
MANJUNATH KAMMARMANJUNATH KAMMAR
1
edited Dec 9 '16 at 17:25
Paul Roub
32.8k85874
edited Dec 9 '16 at 17:25
Paul Roub
32.8k85874
edited Dec 9 '16 at 17:25
Paul Roub
32.8k85874
32.8k85874
answered Dec 9 '16 at 17:16
MANJUNATH KAMMARMANJUNATH KAMMAR
1
answered Dec 9 '16 at 17:16
MANJUNATH KAMMARMANJUNATH KAMMAR
1
answered Dec 9 '16 at 17:16
MANJUNATH KAMMARMANJUNATH KAMMAR
1
1
add a comment |
add a comment |
In MS SQL:
Select DISTINCT CITY
FROM
STATION
Where CITY NOT LIKE 'A%' AND CITY NOT LIKE 'E%' AND CITY NOT LIKE 'I%' AND CITY NOT LIKE 'O%' AND CITY NOT LIKE 'U%';
answered Jan 24 '17 at 23:59
KrishneilKrishneil
659816
add a comment |
In MS SQL:
Select DISTINCT CITY
FROM
STATION
Where CITY NOT LIKE 'A%' AND CITY NOT LIKE 'E%' AND CITY NOT LIKE 'I%' AND CITY NOT LIKE 'O%' AND CITY NOT LIKE 'U%';
answered Jan 24 '17 at 23:59
KrishneilKrishneil
659816
add a comment |
In MS SQL:
Select DISTINCT CITY
FROM
STATION
Where CITY NOT LIKE 'A%' AND CITY NOT LIKE 'E%' AND CITY NOT LIKE 'I%' AND CITY NOT LIKE 'O%' AND CITY NOT LIKE 'U%';
answered Jan 24 '17 at 23:59
KrishneilKrishneil
659816
In MS SQL:
Select DISTINCT CITY
FROM
STATION
Where CITY NOT LIKE 'A%' AND CITY NOT LIKE 'E%' AND CITY NOT LIKE 'I%' AND CITY NOT LIKE 'O%' AND CITY NOT LIKE 'U%';
answered Jan 24 '17 at 23:59
KrishneilKrishneil
659816
answered Jan 24 '17 at 23:59
KrishneilKrishneil
659816
answered Jan 24 '17 at 23:59
KrishneilKrishneil
659816
answered Jan 24 '17 at 23:59
KrishneilKrishneil
659816
659816
add a comment |
add a comment |
you can use
select distinct city from station where city not RLIKE '^[aeiouAEIOU].*'
answered Mar 8 '17 at 4:33
Sajal SinghSajal Singh
146211
add a comment |
you can use
select distinct city from station where city not RLIKE '^[aeiouAEIOU].*'
answered Mar 8 '17 at 4:33
Sajal SinghSajal Singh
146211
add a comment |
you can use
select distinct city from station where city not RLIKE '^[aeiouAEIOU].*'
answered Mar 8 '17 at 4:33
Sajal SinghSajal Singh
146211
you can use
select distinct city from station where city not RLIKE '^[aeiouAEIOU].*'
answered Mar 8 '17 at 4:33
Sajal SinghSajal Singh
146211
answered Mar 8 '17 at 4:33
Sajal SinghSajal Singh
146211
answered Mar 8 '17 at 4:33
Sajal SinghSajal Singh
146211
answered Mar 8 '17 at 4:33
Sajal SinghSajal Singh
146211
146211
add a comment |
add a comment |
select distinct CITY from STATION where left(CITY,1) not in ('a','e','i','o','u') or right(CITY,1) not in ('a','e','i','o','u');
answered Mar 8 '17 at 17:42
MohitMohit
541312
add a comment |
select distinct CITY from STATION where left(CITY,1) not in ('a','e','i','o','u') or right(CITY,1) not in ('a','e','i','o','u');
answered Mar 8 '17 at 17:42
MohitMohit
541312
add a comment |
select distinct CITY from STATION where left(CITY,1) not in ('a','e','i','o','u') or right(CITY,1) not in ('a','e','i','o','u');
answered Mar 8 '17 at 17:42
MohitMohit
541312
select distinct CITY from STATION where left(CITY,1) not in ('a','e','i','o','u') or right(CITY,1) not in ('a','e','i','o','u');
answered Mar 8 '17 at 17:42
MohitMohit
541312
answered Mar 8 '17 at 17:42
MohitMohit
541312
answered Mar 8 '17 at 17:42
MohitMohit
541312
answered Mar 8 '17 at 17:42
MohitMohit
541312
541312
add a comment |
add a comment |
select DISTINCT CITY from STATION where CITY NOT LIKE '[a,e,i,o,u]%'
answered May 26 '17 at 10:14
Aman VermaAman Verma
614
add a comment |
select DISTINCT CITY from STATION where CITY NOT LIKE '[a,e,i,o,u]%'
answered May 26 '17 at 10:14
Aman VermaAman Verma
614
add a comment |
select DISTINCT CITY from STATION where CITY NOT LIKE '[a,e,i,o,u]%'
answered May 26 '17 at 10:14
Aman VermaAman Verma
614
select DISTINCT CITY from STATION where CITY NOT LIKE '[a,e,i,o,u]%'
answered May 26 '17 at 10:14
Aman VermaAman Verma
614
answered May 26 '17 at 10:14
Aman VermaAman Verma
614
answered May 26 '17 at 10:14
Aman VermaAman Verma
614
answered May 26 '17 at 10:14
Aman VermaAman Verma
614
614
add a comment |
add a comment |
try this one
SELECT DISTINCT CITY FROM STATION WHERE
LOWER(SUBSTR(CITY,LENGTH(CITY),1)) IN ('a','e','i','o','u');
answered Oct 12 '17 at 11:12
Ganesh GiriGanesh Giri
30515
add a comment |
try this one
SELECT DISTINCT CITY FROM STATION WHERE
LOWER(SUBSTR(CITY,LENGTH(CITY),1)) IN ('a','e','i','o','u');
answered Oct 12 '17 at 11:12
Ganesh GiriGanesh Giri
30515
add a comment |
try this one
SELECT DISTINCT CITY FROM STATION WHERE
LOWER(SUBSTR(CITY,LENGTH(CITY),1)) IN ('a','e','i','o','u');
answered Oct 12 '17 at 11:12
Ganesh GiriGanesh Giri
30515
try this one
SELECT DISTINCT CITY FROM STATION WHERE
LOWER(SUBSTR(CITY,LENGTH(CITY),1)) IN ('a','e','i','o','u');
answered Oct 12 '17 at 11:12
Ganesh GiriGanesh Giri
30515
answered Oct 12 '17 at 11:12
Ganesh GiriGanesh Giri
30515
answered Oct 12 '17 at 11:12
Ganesh GiriGanesh Giri
30515
answered Oct 12 '17 at 11:12
Ganesh GiriGanesh Giri
30515
30515
add a comment |
add a comment |
Select Distinct City from Station where City Like '[^aeiou]%';
MS SQL Server Tested.
edited Apr 5 '18 at 21:32
Jose Da Silva
2,09421026
answered Apr 5 '18 at 21:02
DarpanDarpan
1
add a comment |
Select Distinct City from Station where City Like '[^aeiou]%';
MS SQL Server Tested.
edited Apr 5 '18 at 21:32
Jose Da Silva
2,09421026
answered Apr 5 '18 at 21:02
DarpanDarpan
1
add a comment |
Select Distinct City from Station where City Like '[^aeiou]%';
MS SQL Server Tested.
edited Apr 5 '18 at 21:32
Jose Da Silva
2,09421026
answered Apr 5 '18 at 21:02
DarpanDarpan
1
Select Distinct City from Station where City Like '[^aeiou]%';
MS SQL Server Tested.
edited Apr 5 '18 at 21:32
Jose Da Silva
2,09421026
answered Apr 5 '18 at 21:02
DarpanDarpan
1
edited Apr 5 '18 at 21:32
Jose Da Silva
2,09421026
edited Apr 5 '18 at 21:32
Jose Da Silva
2,09421026
edited Apr 5 '18 at 21:32
Jose Da Silva
2,09421026
2,09421026
answered Apr 5 '18 at 21:02
DarpanDarpan
1
answered Apr 5 '18 at 21:02
DarpanDarpan
1
answered Apr 5 '18 at 21:02
DarpanDarpan
1
1
add a comment |
add a comment |
SELECT DISTINCT city from STATION
WHERE city NOT LIKE "[aieuo]%[aieuo] [aieuo]%[aieuo]"
AND city NOT LIKE "[aieuo]%[aieuo]";
The first NOT LIKE means, that each word of the two-word city names doesn't start and end with vowels. The second NOT LIKE is for one-word city names.
answered Sep 25 '18 at 20:05
David OsipovDavid Osipov
12
add a comment |
SELECT DISTINCT city from STATION
WHERE city NOT LIKE "[aieuo]%[aieuo] [aieuo]%[aieuo]"
AND city NOT LIKE "[aieuo]%[aieuo]";
The first NOT LIKE means, that each word of the two-word city names doesn't start and end with vowels. The second NOT LIKE is for one-word city names.
answered Sep 25 '18 at 20:05
David OsipovDavid Osipov
12
add a comment |
SELECT DISTINCT city from STATION
WHERE city NOT LIKE "[aieuo]%[aieuo] [aieuo]%[aieuo]"
AND city NOT LIKE "[aieuo]%[aieuo]";
The first NOT LIKE means, that each word of the two-word city names doesn't start and end with vowels. The second NOT LIKE is for one-word city names.
answered Sep 25 '18 at 20:05
David OsipovDavid Osipov
12
SELECT DISTINCT city from STATION
WHERE city NOT LIKE "[aieuo]%[aieuo] [aieuo]%[aieuo]"
AND city NOT LIKE "[aieuo]%[aieuo]";
The first NOT LIKE means, that each word of the two-word city names doesn't start and end with vowels. The second NOT LIKE is for one-word city names.
answered Sep 25 '18 at 20:05
David OsipovDavid Osipov
12
answered Sep 25 '18 at 20:05
David OsipovDavid Osipov
12
answered Sep 25 '18 at 20:05
David OsipovDavid Osipov
12
answered Sep 25 '18 at 20:05
David OsipovDavid Osipov
12
12
add a comment |
add a comment |
select distinct city from station where (
left(city,1) not in ('a','e','i','o','u')
and
right(city,1) not in ('a','e','i','o','u')
);
edited Oct 21 '18 at 9:49
Mohammad
15.6k123562
answered Oct 21 '18 at 9:27
Rohit RawatRohit Rawat
1
1
PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34
Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05
add a comment |
select distinct city from station where (
left(city,1) not in ('a','e','i','o','u')
and
right(city,1) not in ('a','e','i','o','u')
);
edited Oct 21 '18 at 9:49
Mohammad
15.6k123562
answered Oct 21 '18 at 9:27
Rohit RawatRohit Rawat
1
1
PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34
Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05
add a comment |
select distinct city from station where (
left(city,1) not in ('a','e','i','o','u')
and
right(city,1) not in ('a','e','i','o','u')
);
edited Oct 21 '18 at 9:49
Mohammad
15.6k123562
answered Oct 21 '18 at 9:27
Rohit RawatRohit Rawat
1
select distinct city from station where (
left(city,1) not in ('a','e','i','o','u')
and
right(city,1) not in ('a','e','i','o','u')
);
edited Oct 21 '18 at 9:49
Mohammad
15.6k123562
answered Oct 21 '18 at 9:27
Rohit RawatRohit Rawat
1
edited Oct 21 '18 at 9:49
Mohammad
15.6k123562
edited Oct 21 '18 at 9:49
Mohammad
15.6k123562
edited Oct 21 '18 at 9:49
Mohammad
15.6k123562
15.6k123562
answered Oct 21 '18 at 9:27
Rohit RawatRohit Rawat
1
answered Oct 21 '18 at 9:27
Rohit RawatRohit Rawat
1
answered Oct 21 '18 at 9:27
Rohit RawatRohit Rawat
1
1
1
PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34
Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05
add a comment |
1
PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34
Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05
1
1
PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34
PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34
Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05
Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05
add a comment |
In ORACLE:
select
distinct(city)
from station
where regexp_like (city, '^[^A|E|I|O|U](*)');
There is no need to specify the ending letter in this solution. It's not called out for in the problem presented.
edited Nov 19 '18 at 20:50
Hossein Golshani
2,09051123
answered Nov 19 '18 at 20:40
Aaron W.Aaron W.
1
Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41
add a comment |
In ORACLE:
select
distinct(city)
from station
where regexp_like (city, '^[^A|E|I|O|U](*)');
There is no need to specify the ending letter in this solution. It's not called out for in the problem presented.
edited Nov 19 '18 at 20:50
Hossein Golshani
2,09051123
answered Nov 19 '18 at 20:40
Aaron W.Aaron W.
1
Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41
add a comment |
In ORACLE:
select
distinct(city)
from station
where regexp_like (city, '^[^A|E|I|O|U](*)');
There is no need to specify the ending letter in this solution. It's not called out for in the problem presented.
edited Nov 19 '18 at 20:50
Hossein Golshani
2,09051123
answered Nov 19 '18 at 20:40
Aaron W.Aaron W.
1
In ORACLE:
select
distinct(city)
from station
where regexp_like (city, '^[^A|E|I|O|U](*)');
There is no need to specify the ending letter in this solution. It's not called out for in the problem presented.
edited Nov 19 '18 at 20:50
Hossein Golshani
2,09051123
answered Nov 19 '18 at 20:40
Aaron W.Aaron W.
1
edited Nov 19 '18 at 20:50
Hossein Golshani
2,09051123
edited Nov 19 '18 at 20:50
Hossein Golshani
2,09051123
edited Nov 19 '18 at 20:50
Hossein Golshani
2,09051123
2,09051123
answered Nov 19 '18 at 20:40
Aaron W.Aaron W.
1
answered Nov 19 '18 at 20:40
Aaron W.Aaron W.
1
answered Nov 19 '18 at 20:40
Aaron W.Aaron W.
1
1
Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41
add a comment |
Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41
Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41
Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41
add a comment |
for oracle
select distinct
city
from station
where regexp_like(city, '^[^aeiouAEIOU].*');
edited Nov 25 '18 at 3:56
digital.aaron
3,2771331
answered Apr 28 '18 at 19:47
Rasim SENRasim SEN
11
add a comment |
for oracle
select distinct
city
from station
where regexp_like(city, '^[^aeiouAEIOU].*');
edited Nov 25 '18 at 3:56
digital.aaron
3,2771331
answered Apr 28 '18 at 19:47
Rasim SENRasim SEN
11
add a comment |
for oracle
select distinct
city
from station
where regexp_like(city, '^[^aeiouAEIOU].*');
edited Nov 25 '18 at 3:56
digital.aaron
3,2771331
answered Apr 28 '18 at 19:47
Rasim SENRasim SEN
11
for oracle
select distinct
city
from station
where regexp_like(city, '^[^aeiouAEIOU].*');
edited Nov 25 '18 at 3:56
digital.aaron
3,2771331
answered Apr 28 '18 at 19:47
Rasim SENRasim SEN
11
edited Nov 25 '18 at 3:56
digital.aaron
3,2771331
edited Nov 25 '18 at 3:56
digital.aaron
3,2771331
edited Nov 25 '18 at 3:56
digital.aaron
3,2771331
3,2771331
answered Apr 28 '18 at 19:47
Rasim SENRasim SEN
11
answered Apr 28 '18 at 19:47
Rasim SENRasim SEN
11
answered Apr 28 '18 at 19:47
Rasim SENRasim SEN
11
11
add a comment |
add a comment |
select distinct city from station where city regexp '^[^aeiou].*[^aeiou]$'
distinct: to avoid the duplication on the following column
regexp: MySql function (regexp) to obtain Regular Expressions. expression starts with ^ and ends with $. In Oracle its regexp_like (city,'RegEx') Regular Expressions With Oracle Database.
[^...] means any character NOT contain any of ...[^...]. means a single character NOT contain any of ...[^...].* means first character NOT contain any of ...[^...].*[^...] means first character NOT contains any of ... AND not ends with ...[^aeiou].*[^aeiou] mean the first character NOT starts with vowels AND not ends with vowels
answered Nov 25 '18 at 3:11
JayJay
32226
While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28
add a comment |
select distinct city from station where city regexp '^[^aeiou].*[^aeiou]$'
distinct: to avoid the duplication on the following column
regexp: MySql function (regexp) to obtain Regular Expressions. expression starts with ^ and ends with $. In Oracle its regexp_like (city,'RegEx') Regular Expressions With Oracle Database.
[^...] means any character NOT contain any of ...[^...]. means a single character NOT contain any of ...[^...].* means first character NOT contain any of ...[^...].*[^...] means first character NOT contains any of ... AND not ends with ...[^aeiou].*[^aeiou] mean the first character NOT starts with vowels AND not ends with vowels
answered Nov 25 '18 at 3:11
JayJay
32226
While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28
add a comment |
select distinct city from station where city regexp '^[^aeiou].*[^aeiou]$'
distinct: to avoid the duplication on the following column
regexp: MySql function (regexp) to obtain Regular Expressions. expression starts with ^ and ends with $. In Oracle its regexp_like (city,'RegEx') Regular Expressions With Oracle Database.
[^...] means any character NOT contain any of ...[^...]. means a single character NOT contain any of ...[^...].* means first character NOT contain any of ...[^...].*[^...] means first character NOT contains any of ... AND not ends with ...[^aeiou].*[^aeiou] mean the first character NOT starts with vowels AND not ends with vowels
answered Nov 25 '18 at 3:11
JayJay
32226
select distinct city from station where city regexp '^[^aeiou].*[^aeiou]$'
distinct: to avoid the duplication on the following column
regexp: MySql function (regexp) to obtain Regular Expressions. expression starts with ^ and ends with $. In Oracle its regexp_like (city,'RegEx') Regular Expressions With Oracle Database.
[^...] means any character NOT contain any of ...[^...]. means a single character NOT contain any of ...[^...].* means first character NOT contain any of ...[^...].*[^...] means first character NOT contains any of ... AND not ends with ...[^aeiou].*[^aeiou] mean the first character NOT starts with vowels AND not ends with vowels
answered Nov 25 '18 at 3:11
JayJay
32226
edited Nov 26 '18 at 3:09
answered Nov 25 '18 at 3:11
JayJay
32226
answered Nov 25 '18 at 3:11
JayJay
32226
answered Nov 25 '18 at 3:11
JayJay
32226
32226
While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28
add a comment |
While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28
While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28
While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28
add a comment |
Mysql we can use this.
select distinct city from station where city regexp '^[^aeiou].*';
To Know More About MySQL Regular Expression
For Oracle we can use
select distinct(city) from station where regexp_like (city, '^[^AEIOU](*)');
To Know More About Oracle Regular Expression
For MS SQL Server
select distinct city from station where city not like '[aeuio]%';
answered Nov 30 '18 at 4:22
Arun SolomonArun Solomon
72
add a comment |
Mysql we can use this.
select distinct city from station where city regexp '^[^aeiou].*';
To Know More About MySQL Regular Expression
For Oracle we can use
select distinct(city) from station where regexp_like (city, '^[^AEIOU](*)');
To Know More About Oracle Regular Expression
For MS SQL Server
select distinct city from station where city not like '[aeuio]%';
answered Nov 30 '18 at 4:22
Arun SolomonArun Solomon
72
add a comment |
Mysql we can use this.
select distinct city from station where city regexp '^[^aeiou].*';
To Know More About MySQL Regular Expression
For Oracle we can use
select distinct(city) from station where regexp_like (city, '^[^AEIOU](*)');
To Know More About Oracle Regular Expression
For MS SQL Server
select distinct city from station where city not like '[aeuio]%';
answered Nov 30 '18 at 4:22
Arun SolomonArun Solomon
72
Mysql we can use this.
select distinct city from station where city regexp '^[^aeiou].*';
To Know More About MySQL Regular Expression
For Oracle we can use
select distinct(city) from station where regexp_like (city, '^[^AEIOU](*)');
To Know More About Oracle Regular Expression
For MS SQL Server
select distinct city from station where city not like '[aeuio]%';
answered Nov 30 '18 at 4:22
Arun SolomonArun Solomon
72
answered Nov 30 '18 at 4:22
Arun SolomonArun Solomon
72
answered Nov 30 '18 at 4:22
Arun SolomonArun Solomon
72
answered Nov 30 '18 at 4:22
Arun SolomonArun Solomon
72
72
add a comment |
add a comment |
I worked with this
SELECT DISTINCT CITY FROM STATION
WHERE CITY NOT LIKE 'a%'
AND CITY NOT LIKE 'e%'
AND CITY NOT LIKE 'i%'
AND CITY NOT LIKE 'o%'
AND CITY NOT LIKE 'u%'
answered Jan 15 at 16:54
Kashif Faraz ShamsiKashif Faraz Shamsi
4431619
add a comment |
I worked with this
SELECT DISTINCT CITY FROM STATION
WHERE CITY NOT LIKE 'a%'
AND CITY NOT LIKE 'e%'
AND CITY NOT LIKE 'i%'
AND CITY NOT LIKE 'o%'
AND CITY NOT LIKE 'u%'
answered Jan 15 at 16:54
Kashif Faraz ShamsiKashif Faraz Shamsi
4431619
add a comment |
I worked with this
SELECT DISTINCT CITY FROM STATION
WHERE CITY NOT LIKE 'a%'
AND CITY NOT LIKE 'e%'
AND CITY NOT LIKE 'i%'
AND CITY NOT LIKE 'o%'
AND CITY NOT LIKE 'u%'
answered Jan 15 at 16:54
Kashif Faraz ShamsiKashif Faraz Shamsi
4431619
I worked with this
SELECT DISTINCT CITY FROM STATION
WHERE CITY NOT LIKE 'a%'
AND CITY NOT LIKE 'e%'
AND CITY NOT LIKE 'i%'
AND CITY NOT LIKE 'o%'
AND CITY NOT LIKE 'u%'
answered Jan 15 at 16:54
Kashif Faraz ShamsiKashif Faraz Shamsi
4431619
answered Jan 15 at 16:54
Kashif Faraz ShamsiKashif Faraz Shamsi
4431619
answered Jan 15 at 16:54
Kashif Faraz ShamsiKashif Faraz Shamsi
4431619
answered Jan 15 at 16:54
Kashif Faraz ShamsiKashif Faraz Shamsi
4431619
4431619
add a comment |
add a comment |
select distinct <COLUMN NAME> from <table name>
WHERE SUBSTRING(<COLUMN NAME>,1,1) not in ('a','e','i','o','u','A','E','I','O','U')
order by <COLUMN NAME>
Distinct words do not start with a vowel.
edited Jan 22 at 7:59
Prakash Pazhanisamy
8681923
answered Jan 22 at 6:13
srinidhi sundarrajansrinidhi sundarrajan
11
SUBSTRING(<COLUMN NAME>,1,1)
– srinidhi sundarrajan
Jan 22 at 6:14
Could you please add a bit more of context to your answer?
– toti08
Jan 22 at 6:38
add a comment |
select distinct <COLUMN NAME> from <table name>
WHERE SUBSTRING(<COLUMN NAME>,1,1) not in ('a','e','i','o','u','A','E','I','O','U')
order by <COLUMN NAME>
Distinct words do not start with a vowel.
edited Jan 22 at 7:59
Prakash Pazhanisamy
8681923
answered Jan 22 at 6:13
srinidhi sundarrajansrinidhi sundarrajan
11
SUBSTRING(<COLUMN NAME>,1,1)
– srinidhi sundarrajan
Jan 22 at 6:14
Could you please add a bit more of context to your answer?
– toti08
Jan 22 at 6:38
add a comment |
select distinct <COLUMN NAME> from <table name>
WHERE SUBSTRING(<COLUMN NAME>,1,1) not in ('a','e','i','o','u','A','E','I','O','U')
order by <COLUMN NAME>
Distinct words do not start with a vowel.
edited Jan 22 at 7:59
Prakash Pazhanisamy
8681923
answered Jan 22 at 6:13
srinidhi sundarrajansrinidhi sundarrajan
11
select distinct <COLUMN NAME> from <table name>
WHERE SUBSTRING(<COLUMN NAME>,1,1) not in ('a','e','i','o','u','A','E','I','O','U')
order by <COLUMN NAME>
Distinct words do not start with a vowel.
edited Jan 22 at 7:59
Prakash Pazhanisamy
8681923
answered Jan 22 at 6:13
srinidhi sundarrajansrinidhi sundarrajan
11
edited Jan 22 at 7:59
Prakash Pazhanisamy
8681923
edited Jan 22 at 7:59
Prakash Pazhanisamy
8681923
edited Jan 22 at 7:59
Prakash Pazhanisamy
8681923
8681923
answered Jan 22 at 6:13
srinidhi sundarrajansrinidhi sundarrajan
11
answered Jan 22 at 6:13
srinidhi sundarrajansrinidhi sundarrajan
11
answered Jan 22 at 6:13
srinidhi sundarrajansrinidhi sundarrajan
11
11
SUBSTRING(<COLUMN NAME>,1,1)
– srinidhi sundarrajan
Jan 22 at 6:14
Could you please add a bit more of context to your answer?
– toti08
Jan 22 at 6:38
add a comment |
SUBSTRING(<COLUMN NAME>,1,1)
– srinidhi sundarrajan
Jan 22 at 6:14
Could you please add a bit more of context to your answer?
– toti08
Jan 22 at 6:38
SUBSTRING(<COLUMN NAME>,1,1)
– srinidhi sundarrajan
Jan 22 at 6:14
SUBSTRING(<COLUMN NAME>,1,1)
– srinidhi sundarrajan
Jan 22 at 6:14
Could you please add a bit more of context to your answer?
– toti08
Jan 22 at 6:38
Could you please add a bit more of context to your answer?
– toti08
Jan 22 at 6:38
add a comment |
select distinct city from station where
city not in
(select distinct city from station where
(city like "A%" or
city like "E%" or
city like "I%" or
city like "O%" or
city like "U%"
));
This will work i guess.
edited Oct 18 '16 at 4:14
noufalcep
2,25682236
answered Oct 18 '16 at 3:04
Saikrishna SyamalaSaikrishna Syamala
1
please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27
add a comment |
select distinct city from station where
city not in
(select distinct city from station where
(city like "A%" or
city like "E%" or
city like "I%" or
city like "O%" or
city like "U%"
));
This will work i guess.
edited Oct 18 '16 at 4:14
noufalcep
2,25682236
answered Oct 18 '16 at 3:04
Saikrishna SyamalaSaikrishna Syamala
1
please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27
add a comment |
select distinct city from station where
city not in
(select distinct city from station where
(city like "A%" or
city like "E%" or
city like "I%" or
city like "O%" or
city like "U%"
));
This will work i guess.
edited Oct 18 '16 at 4:14
noufalcep
2,25682236
answered Oct 18 '16 at 3:04
Saikrishna SyamalaSaikrishna Syamala
1
select distinct city from station where
city not in
(select distinct city from station where
(city like "A%" or
city like "E%" or
city like "I%" or
city like "O%" or
city like "U%"
));
This will work i guess.
edited Oct 18 '16 at 4:14
noufalcep
2,25682236
answered Oct 18 '16 at 3:04
Saikrishna SyamalaSaikrishna Syamala
1
edited Oct 18 '16 at 4:14
noufalcep
2,25682236
edited Oct 18 '16 at 4:14
noufalcep
2,25682236
edited Oct 18 '16 at 4:14
noufalcep
2,25682236
2,25682236
answered Oct 18 '16 at 3:04
Saikrishna SyamalaSaikrishna Syamala
1
answered Oct 18 '16 at 3:04
Saikrishna SyamalaSaikrishna Syamala
1
answered Oct 18 '16 at 3:04
Saikrishna SyamalaSaikrishna Syamala
1
1
please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27
add a comment |
please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27
please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27
please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27
add a comment |
SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '[aeiouAEIOU]$'
This works for MYSQL
answered Dec 9 '17 at 7:33
Vivek GhanchiVivek Ghanchi
328
You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11
Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19
add a comment |
SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '[aeiouAEIOU]$'
This works for MYSQL
answered Dec 9 '17 at 7:33
Vivek GhanchiVivek Ghanchi
328
You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11
Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19
add a comment |
SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '[aeiouAEIOU]$'
This works for MYSQL
answered Dec 9 '17 at 7:33
Vivek GhanchiVivek Ghanchi
328
SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '[aeiouAEIOU]$'
This works for MYSQL
answered Dec 9 '17 at 7:33
Vivek GhanchiVivek Ghanchi
328
answered Dec 9 '17 at 7:33
Vivek GhanchiVivek Ghanchi
328
answered Dec 9 '17 at 7:33
Vivek GhanchiVivek Ghanchi
328
answered Dec 9 '17 at 7:33
Vivek GhanchiVivek Ghanchi
328
328
You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11
Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19
add a comment |
You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11
Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19
You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11
You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11
Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19
Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19
add a comment |
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try the
NOT RLIKE '^[aeiouAEIOU].*$'– Tin Tran
Apr 14 '16 at 22:05