awk how to find first available date field?
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Fields 1,2,3,4 are date fields yyyy-mm-dd.
Delimited by ";"
"-" if no date.
Field 4 will always have a date
Examples;
-; 2016-08-19; 2016-08-19; 2018-07-17; Beach-Rangiroa.jpg
-; -; -; 2018-09-12; MV3_0034-copy.webp
2016-12-10; 2016-12-10; 2016-12-20; 2018-07-18; Sukhothai-61.jpg
-; -; -; 2018-07-19; Gdu9Rwhu6W3Q5W6q_1Qag.jpg
Objective: Use awk to print the 1st available date in order fields 1,2,3,4
I've tried this;
awk -F";" '{if ($1!="-") print $1; else if ($2!="-") print $2; else if ($3!="-") prin$3; else if ($4!="-") print $4}'
Results...
2016-08-19
-
-
bash version 4.3.48
I am trying to achieve this: e.g. line 1 in example...
2016-08-19; Beach-Rangiroa.jpg
echo '-; -; -; 2018-07-15; Stock-Photo-114398301.webp; WEBP; image/webp; 2000; 1333' |
awk -F';' 'OFS=";" {for(i=1; i<5; ++i) { if ($i ~ /[0-9]{4}-[0-9]{,2}-[0-9]{,2}/) { print $i,$5,$6,$7,$8,$9; next; }}}'
Result;
2018-07-15; Stock-Photo-114398301.webp; WEBP; image/webp; 2000; 1333
This works nicely, except the 1st space on the date, also is there a method available to verify the date, e.g. date -d "%Y-%m-%d" ?
Thank you.
bash awk
asked Nov 25 '18 at 8:59
DooUKnowDooUKnow
62
add a comment |
Fields 1,2,3,4 are date fields yyyy-mm-dd.
Delimited by ";"
"-" if no date.
Field 4 will always have a date
Examples;
-; 2016-08-19; 2016-08-19; 2018-07-17; Beach-Rangiroa.jpg
-; -; -; 2018-09-12; MV3_0034-copy.webp
2016-12-10; 2016-12-10; 2016-12-20; 2018-07-18; Sukhothai-61.jpg
-; -; -; 2018-07-19; Gdu9Rwhu6W3Q5W6q_1Qag.jpg
Objective: Use awk to print the 1st available date in order fields 1,2,3,4
I've tried this;
awk -F";" '{if ($1!="-") print $1; else if ($2!="-") print $2; else if ($3!="-") prin$3; else if ($4!="-") print $4}'
Results...
2016-08-19
-
-
bash version 4.3.48
I am trying to achieve this: e.g. line 1 in example...
2016-08-19; Beach-Rangiroa.jpg
echo '-; -; -; 2018-07-15; Stock-Photo-114398301.webp; WEBP; image/webp; 2000; 1333' |
awk -F';' 'OFS=";" {for(i=1; i<5; ++i) { if ($i ~ /[0-9]{4}-[0-9]{,2}-[0-9]{,2}/) { print $i,$5,$6,$7,$8,$9; next; }}}'
Result;
2018-07-15; Stock-Photo-114398301.webp; WEBP; image/webp; 2000; 1333
This works nicely, except the 1st space on the date, also is there a method available to verify the date, e.g. date -d "%Y-%m-%d" ?
Thank you.
bash awk
asked Nov 25 '18 at 8:59
DooUKnowDooUKnow
62
If you clean up your question to get rid of all the edits and just tell us the question with the sample input/output it'll be much easier for us to help you.
– Ed Morton
Nov 26 '18 at 1:54
add a comment |
Fields 1,2,3,4 are date fields yyyy-mm-dd.
Delimited by ";"
"-" if no date.
Field 4 will always have a date
Examples;
-; 2016-08-19; 2016-08-19; 2018-07-17; Beach-Rangiroa.jpg
-; -; -; 2018-09-12; MV3_0034-copy.webp
2016-12-10; 2016-12-10; 2016-12-20; 2018-07-18; Sukhothai-61.jpg
-; -; -; 2018-07-19; Gdu9Rwhu6W3Q5W6q_1Qag.jpg
Objective: Use awk to print the 1st available date in order fields 1,2,3,4
I've tried this;
awk -F";" '{if ($1!="-") print $1; else if ($2!="-") print $2; else if ($3!="-") prin$3; else if ($4!="-") print $4}'
Results...
2016-08-19
-
-
bash version 4.3.48
I am trying to achieve this: e.g. line 1 in example...
2016-08-19; Beach-Rangiroa.jpg
echo '-; -; -; 2018-07-15; Stock-Photo-114398301.webp; WEBP; image/webp; 2000; 1333' |
awk -F';' 'OFS=";" {for(i=1; i<5; ++i) { if ($i ~ /[0-9]{4}-[0-9]{,2}-[0-9]{,2}/) { print $i,$5,$6,$7,$8,$9; next; }}}'
Result;
2018-07-15; Stock-Photo-114398301.webp; WEBP; image/webp; 2000; 1333
This works nicely, except the 1st space on the date, also is there a method available to verify the date, e.g. date -d "%Y-%m-%d" ?
Thank you.
bash awk
asked Nov 25 '18 at 8:59
DooUKnowDooUKnow
62
Fields 1,2,3,4 are date fields yyyy-mm-dd.
Delimited by ";"
"-" if no date.
Field 4 will always have a date
Examples;
-; 2016-08-19; 2016-08-19; 2018-07-17; Beach-Rangiroa.jpg
-; -; -; 2018-09-12; MV3_0034-copy.webp
2016-12-10; 2016-12-10; 2016-12-20; 2018-07-18; Sukhothai-61.jpg
-; -; -; 2018-07-19; Gdu9Rwhu6W3Q5W6q_1Qag.jpg
Objective: Use awk to print the 1st available date in order fields 1,2,3,4
I've tried this;
awk -F";" '{if ($1!="-") print $1; else if ($2!="-") print $2; else if ($3!="-") prin$3; else if ($4!="-") print $4}'
Results...
2016-08-19
-
-
bash version 4.3.48
I am trying to achieve this: e.g. line 1 in example...
2016-08-19; Beach-Rangiroa.jpg
echo '-; -; -; 2018-07-15; Stock-Photo-114398301.webp; WEBP; image/webp; 2000; 1333' |
awk -F';' 'OFS=";" {for(i=1; i<5; ++i) { if ($i ~ /[0-9]{4}-[0-9]{,2}-[0-9]{,2}/) { print $i,$5,$6,$7,$8,$9; next; }}}'
Result;
2018-07-15; Stock-Photo-114398301.webp; WEBP; image/webp; 2000; 1333
This works nicely, except the 1st space on the date, also is there a method available to verify the date, e.g. date -d "%Y-%m-%d" ?
Thank you.
bash awk
bash awk
asked Nov 25 '18 at 8:59
DooUKnowDooUKnow
62
asked Nov 25 '18 at 8:59
DooUKnowDooUKnow
62
edited Nov 26 '18 at 4:01
DooUKnow
asked Nov 25 '18 at 8:59
DooUKnowDooUKnow
62
asked Nov 25 '18 at 8:59
DooUKnowDooUKnow
62
asked Nov 25 '18 at 8:59
DooUKnowDooUKnow
62
62
If you clean up your question to get rid of all the edits and just tell us the question with the sample input/output it'll be much easier for us to help you.
– Ed Morton
Nov 26 '18 at 1:54
add a comment |
If you clean up your question to get rid of all the edits and just tell us the question with the sample input/output it'll be much easier for us to help you.
– Ed Morton
Nov 26 '18 at 1:54
If you clean up your question to get rid of all the edits and just tell us the question with the sample input/output it'll be much easier for us to help you.
– Ed Morton
Nov 26 '18 at 1:54
If you clean up your question to get rid of all the edits and just tell us the question with the sample input/output it'll be much easier for us to help you.
– Ed Morton
Nov 26 '18 at 1:54
add a comment |
4 Answers
4
active
oldest
votes
This is a gnu only gawk solution using FPAT:
awk 'BEGIN{FPAT="[0-9]{4}-[0-9]{,2}-[0-9]{,2}"}{print $1}' file1
2016-08-19
2018-09-12
2018-07-19
With FPAT you actually instruct gawk what to consider as a field, a whole regex here. If the input line has also a second date this will appear as $2, $NF will return the last date field of each line,NF will return the total date fields,and so on.
answered Nov 25 '18 at 9:29
George VasiliouGeorge Vasiliou
4,27221019
There's no FPAT in awk. Perhaps you're using gawk?
– ghoti
Nov 25 '18 at 15:09
doesn't work in macos El Capitan.
– Graham
Nov 25 '18 at 16:13
add a comment |
You can use a variable for field numbers:
awk -F; '{for(i=1; i<5; ++i) { if ($i ~ /[0-9]/) { print $i; next; }}}' in
answered Nov 25 '18 at 9:05
perrealperreal
73.1k10113142
Should be-F' *; *'to trim the date value.FSis a full regular expression, according to the man page. You can also set thei < NFfor the condition.
– Amessihel
Nov 25 '18 at 9:42
add a comment |
Solution without awk:
You said you wanted the 1st available date. When you only want 1 line output, you can use
grep -Eo "[0-9]{4}-[0-9]{2}-[0-9]{2}" inputfile| head -1
When you want to have the first date for each line, change the grep or use sed:
grep -Eo "[0-9]{4}-[0-9]{2}-[0-9]{2}.*" inputfile| cut -d';' -f1
# or
sed -r 's/([0-9]{4}-[0-9]{2}-[0-9]{2}).*/1/; s/.*([0-9]{4}-[0-9]{2}-[0-9]{2})/1/' inputfile
answered Nov 25 '18 at 9:27
Walter AWalter A
11.2k21132
add a comment |
Thank you for all for your help.
I think this acomplishes the objective;
echo '-; -; -; 2018-07-25; Redwood-Forest-Sequoia-4.jpg; JPEG; image/jpeg; 1280; 720' |
awk -F'; ' 'OFS="; " {for(i=1; i<5; ++i) { if ($i ~ /[0-9]{4}-[0-9]{,2}-[0-9]{,2}/) { print $i,$5,$6,$7,$8,$9; next; }}}'
Result;
2018-07-25; Redwood-Forest-Sequoia-4.jpg; JPEG; image/jpeg; 1280; 720
Best regards.
answered Nov 25 '18 at 17:31
DooUKnowDooUKnow
62
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is a gnu only gawk solution using FPAT:
awk 'BEGIN{FPAT="[0-9]{4}-[0-9]{,2}-[0-9]{,2}"}{print $1}' file1
2016-08-19
2018-09-12
2018-07-19
With FPAT you actually instruct gawk what to consider as a field, a whole regex here. If the input line has also a second date this will appear as $2, $NF will return the last date field of each line,NF will return the total date fields,and so on.
answered Nov 25 '18 at 9:29
George VasiliouGeorge Vasiliou
4,27221019
There's no FPAT in awk. Perhaps you're using gawk?
– ghoti
Nov 25 '18 at 15:09
doesn't work in macos El Capitan.
– Graham
Nov 25 '18 at 16:13
add a comment |
This is a gnu only gawk solution using FPAT:
awk 'BEGIN{FPAT="[0-9]{4}-[0-9]{,2}-[0-9]{,2}"}{print $1}' file1
2016-08-19
2018-09-12
2018-07-19
With FPAT you actually instruct gawk what to consider as a field, a whole regex here. If the input line has also a second date this will appear as $2, $NF will return the last date field of each line,NF will return the total date fields,and so on.
answered Nov 25 '18 at 9:29
George VasiliouGeorge Vasiliou
4,27221019
There's no FPAT in awk. Perhaps you're using gawk?
– ghoti
Nov 25 '18 at 15:09
doesn't work in macos El Capitan.
– Graham
Nov 25 '18 at 16:13
add a comment |
This is a gnu only gawk solution using FPAT:
awk 'BEGIN{FPAT="[0-9]{4}-[0-9]{,2}-[0-9]{,2}"}{print $1}' file1
2016-08-19
2018-09-12
2018-07-19
With FPAT you actually instruct gawk what to consider as a field, a whole regex here. If the input line has also a second date this will appear as $2, $NF will return the last date field of each line,NF will return the total date fields,and so on.
answered Nov 25 '18 at 9:29
George VasiliouGeorge Vasiliou
4,27221019
This is a gnu only gawk solution using FPAT:
awk 'BEGIN{FPAT="[0-9]{4}-[0-9]{,2}-[0-9]{,2}"}{print $1}' file1
2016-08-19
2018-09-12
2018-07-19
With FPAT you actually instruct gawk what to consider as a field, a whole regex here. If the input line has also a second date this will appear as $2, $NF will return the last date field of each line,NF will return the total date fields,and so on.
answered Nov 25 '18 at 9:29
George VasiliouGeorge Vasiliou
4,27221019
edited Nov 25 '18 at 16:52
answered Nov 25 '18 at 9:29
George VasiliouGeorge Vasiliou
4,27221019
answered Nov 25 '18 at 9:29
George VasiliouGeorge Vasiliou
4,27221019
answered Nov 25 '18 at 9:29
George VasiliouGeorge Vasiliou
4,27221019
4,27221019
There's no FPAT in awk. Perhaps you're using gawk?
– ghoti
Nov 25 '18 at 15:09
doesn't work in macos El Capitan.
– Graham
Nov 25 '18 at 16:13
add a comment |
There's no FPAT in awk. Perhaps you're using gawk?
– ghoti
Nov 25 '18 at 15:09
doesn't work in macos El Capitan.
– Graham
Nov 25 '18 at 16:13
There's no FPAT in awk. Perhaps you're using gawk?
– ghoti
Nov 25 '18 at 15:09
There's no FPAT in awk. Perhaps you're using gawk?
– ghoti
Nov 25 '18 at 15:09
doesn't work in macos El Capitan.
– Graham
Nov 25 '18 at 16:13
doesn't work in macos El Capitan.
– Graham
Nov 25 '18 at 16:13
add a comment |
You can use a variable for field numbers:
awk -F; '{for(i=1; i<5; ++i) { if ($i ~ /[0-9]/) { print $i; next; }}}' in
answered Nov 25 '18 at 9:05
perrealperreal
73.1k10113142
Should be-F' *; *'to trim the date value.FSis a full regular expression, according to the man page. You can also set thei < NFfor the condition.
– Amessihel
Nov 25 '18 at 9:42
add a comment |
You can use a variable for field numbers:
awk -F; '{for(i=1; i<5; ++i) { if ($i ~ /[0-9]/) { print $i; next; }}}' in
answered Nov 25 '18 at 9:05
perrealperreal
73.1k10113142
Should be-F' *; *'to trim the date value.FSis a full regular expression, according to the man page. You can also set thei < NFfor the condition.
– Amessihel
Nov 25 '18 at 9:42
add a comment |
You can use a variable for field numbers:
awk -F; '{for(i=1; i<5; ++i) { if ($i ~ /[0-9]/) { print $i; next; }}}' in
answered Nov 25 '18 at 9:05
perrealperreal
73.1k10113142
You can use a variable for field numbers:
awk -F; '{for(i=1; i<5; ++i) { if ($i ~ /[0-9]/) { print $i; next; }}}' in
answered Nov 25 '18 at 9:05
perrealperreal
73.1k10113142
answered Nov 25 '18 at 9:05
perrealperreal
73.1k10113142
answered Nov 25 '18 at 9:05
perrealperreal
73.1k10113142
answered Nov 25 '18 at 9:05
perrealperreal
73.1k10113142
73.1k10113142
Should be-F' *; *'to trim the date value.FSis a full regular expression, according to the man page. You can also set thei < NFfor the condition.
– Amessihel
Nov 25 '18 at 9:42
add a comment |
Should be-F' *; *'to trim the date value.FSis a full regular expression, according to the man page. You can also set thei < NFfor the condition.
– Amessihel
Nov 25 '18 at 9:42
Should be
-F' *; *' to trim the date value. FS is a full regular expression, according to the man page. You can also set the i < NF for the condition.– Amessihel
Nov 25 '18 at 9:42
Should be
-F' *; *' to trim the date value. FS is a full regular expression, according to the man page. You can also set the i < NF for the condition.– Amessihel
Nov 25 '18 at 9:42
add a comment |
Solution without awk:
You said you wanted the 1st available date. When you only want 1 line output, you can use
grep -Eo "[0-9]{4}-[0-9]{2}-[0-9]{2}" inputfile| head -1
When you want to have the first date for each line, change the grep or use sed:
grep -Eo "[0-9]{4}-[0-9]{2}-[0-9]{2}.*" inputfile| cut -d';' -f1
# or
sed -r 's/([0-9]{4}-[0-9]{2}-[0-9]{2}).*/1/; s/.*([0-9]{4}-[0-9]{2}-[0-9]{2})/1/' inputfile
answered Nov 25 '18 at 9:27
Walter AWalter A
11.2k21132
add a comment |
Solution without awk:
You said you wanted the 1st available date. When you only want 1 line output, you can use
grep -Eo "[0-9]{4}-[0-9]{2}-[0-9]{2}" inputfile| head -1
When you want to have the first date for each line, change the grep or use sed:
grep -Eo "[0-9]{4}-[0-9]{2}-[0-9]{2}.*" inputfile| cut -d';' -f1
# or
sed -r 's/([0-9]{4}-[0-9]{2}-[0-9]{2}).*/1/; s/.*([0-9]{4}-[0-9]{2}-[0-9]{2})/1/' inputfile
answered Nov 25 '18 at 9:27
Walter AWalter A
11.2k21132
add a comment |
Solution without awk:
You said you wanted the 1st available date. When you only want 1 line output, you can use
grep -Eo "[0-9]{4}-[0-9]{2}-[0-9]{2}" inputfile| head -1
When you want to have the first date for each line, change the grep or use sed:
grep -Eo "[0-9]{4}-[0-9]{2}-[0-9]{2}.*" inputfile| cut -d';' -f1
# or
sed -r 's/([0-9]{4}-[0-9]{2}-[0-9]{2}).*/1/; s/.*([0-9]{4}-[0-9]{2}-[0-9]{2})/1/' inputfile
answered Nov 25 '18 at 9:27
Walter AWalter A
11.2k21132
Solution without awk:
You said you wanted the 1st available date. When you only want 1 line output, you can use
grep -Eo "[0-9]{4}-[0-9]{2}-[0-9]{2}" inputfile| head -1
When you want to have the first date for each line, change the grep or use sed:
grep -Eo "[0-9]{4}-[0-9]{2}-[0-9]{2}.*" inputfile| cut -d';' -f1
# or
sed -r 's/([0-9]{4}-[0-9]{2}-[0-9]{2}).*/1/; s/.*([0-9]{4}-[0-9]{2}-[0-9]{2})/1/' inputfile
answered Nov 25 '18 at 9:27
Walter AWalter A
11.2k21132
answered Nov 25 '18 at 9:27
Walter AWalter A
11.2k21132
answered Nov 25 '18 at 9:27
Walter AWalter A
11.2k21132
answered Nov 25 '18 at 9:27
Walter AWalter A
11.2k21132
11.2k21132
add a comment |
add a comment |
Thank you for all for your help.
I think this acomplishes the objective;
echo '-; -; -; 2018-07-25; Redwood-Forest-Sequoia-4.jpg; JPEG; image/jpeg; 1280; 720' |
awk -F'; ' 'OFS="; " {for(i=1; i<5; ++i) { if ($i ~ /[0-9]{4}-[0-9]{,2}-[0-9]{,2}/) { print $i,$5,$6,$7,$8,$9; next; }}}'
Result;
2018-07-25; Redwood-Forest-Sequoia-4.jpg; JPEG; image/jpeg; 1280; 720
Best regards.
answered Nov 25 '18 at 17:31
DooUKnowDooUKnow
62
add a comment |
Thank you for all for your help.
I think this acomplishes the objective;
echo '-; -; -; 2018-07-25; Redwood-Forest-Sequoia-4.jpg; JPEG; image/jpeg; 1280; 720' |
awk -F'; ' 'OFS="; " {for(i=1; i<5; ++i) { if ($i ~ /[0-9]{4}-[0-9]{,2}-[0-9]{,2}/) { print $i,$5,$6,$7,$8,$9; next; }}}'
Result;
2018-07-25; Redwood-Forest-Sequoia-4.jpg; JPEG; image/jpeg; 1280; 720
Best regards.
answered Nov 25 '18 at 17:31
DooUKnowDooUKnow
62
add a comment |
Thank you for all for your help.
I think this acomplishes the objective;
echo '-; -; -; 2018-07-25; Redwood-Forest-Sequoia-4.jpg; JPEG; image/jpeg; 1280; 720' |
awk -F'; ' 'OFS="; " {for(i=1; i<5; ++i) { if ($i ~ /[0-9]{4}-[0-9]{,2}-[0-9]{,2}/) { print $i,$5,$6,$7,$8,$9; next; }}}'
Result;
2018-07-25; Redwood-Forest-Sequoia-4.jpg; JPEG; image/jpeg; 1280; 720
Best regards.
answered Nov 25 '18 at 17:31
DooUKnowDooUKnow
62
Thank you for all for your help.
I think this acomplishes the objective;
echo '-; -; -; 2018-07-25; Redwood-Forest-Sequoia-4.jpg; JPEG; image/jpeg; 1280; 720' |
awk -F'; ' 'OFS="; " {for(i=1; i<5; ++i) { if ($i ~ /[0-9]{4}-[0-9]{,2}-[0-9]{,2}/) { print $i,$5,$6,$7,$8,$9; next; }}}'
Result;
2018-07-25; Redwood-Forest-Sequoia-4.jpg; JPEG; image/jpeg; 1280; 720
Best regards.
answered Nov 25 '18 at 17:31
DooUKnowDooUKnow
62
answered Nov 25 '18 at 17:31
DooUKnowDooUKnow
62
answered Nov 25 '18 at 17:31
DooUKnowDooUKnow
62
answered Nov 25 '18 at 17:31
DooUKnowDooUKnow
62
62
add a comment |
add a comment |
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If you clean up your question to get rid of all the edits and just tell us the question with the sample input/output it'll be much easier for us to help you.
– Ed Morton
Nov 26 '18 at 1:54