How to match a string in this way?
I need to check if a String matches this specific pattern.
The pattern is:
(Numbers)(all characters allowed)(numbers)
and the numbers may have a comma ("." or ",")!
For instance the input could be 500+400 or 400,021+213.443.
I tried Pattern.matches("[0-9],?.?+[0-9],?.?+", theequation2), but it didn't work!
I know that I have to use the method Pattern.match(regex, String), but I am not being able to find the correct regex.
java regex
edited Nov 10 at 20:11
Wiktor Stribiżew
307k16126202
asked Nov 10 at 18:19
AyhamSYR
53
add a comment |
I need to check if a String matches this specific pattern.
The pattern is:
(Numbers)(all characters allowed)(numbers)
and the numbers may have a comma ("." or ",")!
For instance the input could be 500+400 or 400,021+213.443.
I tried Pattern.matches("[0-9],?.?+[0-9],?.?+", theequation2), but it didn't work!
I know that I have to use the method Pattern.match(regex, String), but I am not being able to find the correct regex.
java regex
edited Nov 10 at 20:11
Wiktor Stribiżew
307k16126202
asked Nov 10 at 18:19
AyhamSYR
53
2
please, give possible input and output, and show your attempt.
– The Scientific Method
Nov 10 at 18:21
Done! I have added them"
– AyhamSYR
Nov 10 at 18:26
If "all characters are allowed", then there is no way to know the start of the 2nd numbers group (as numbers are a subset of "all"). So there must be a better differentiation here.
– KevinO
Nov 10 at 18:39
Could you give an example please?
– AyhamSYR
Nov 10 at 18:41
Your requirement is too vague. You might as well uses.matches("\d.*\d")to match a string starting and ending with a digit and having any 0+ chars in between.
– Wiktor Stribiżew
Nov 10 at 20:13
add a comment |
I need to check if a String matches this specific pattern.
The pattern is:
(Numbers)(all characters allowed)(numbers)
and the numbers may have a comma ("." or ",")!
For instance the input could be 500+400 or 400,021+213.443.
I tried Pattern.matches("[0-9],?.?+[0-9],?.?+", theequation2), but it didn't work!
I know that I have to use the method Pattern.match(regex, String), but I am not being able to find the correct regex.
java regex
edited Nov 10 at 20:11
Wiktor Stribiżew
307k16126202
asked Nov 10 at 18:19
AyhamSYR
53
I need to check if a String matches this specific pattern.
The pattern is:
(Numbers)(all characters allowed)(numbers)
and the numbers may have a comma ("." or ",")!
For instance the input could be 500+400 or 400,021+213.443.
I tried Pattern.matches("[0-9],?.?+[0-9],?.?+", theequation2), but it didn't work!
I know that I have to use the method Pattern.match(regex, String), but I am not being able to find the correct regex.
java regex
java regex
edited Nov 10 at 20:11
Wiktor Stribiżew
307k16126202
asked Nov 10 at 18:19
AyhamSYR
53
edited Nov 10 at 20:11
Wiktor Stribiżew
307k16126202
asked Nov 10 at 18:19
AyhamSYR
53
edited Nov 10 at 20:11
Wiktor Stribiżew
307k16126202
edited Nov 10 at 20:11
Wiktor Stribiżew
307k16126202
edited Nov 10 at 20:11
Wiktor Stribiżew
307k16126202
307k16126202
asked Nov 10 at 18:19
AyhamSYR
53
asked Nov 10 at 18:19
AyhamSYR
53
asked Nov 10 at 18:19
AyhamSYR
53
53
2
please, give possible input and output, and show your attempt.
– The Scientific Method
Nov 10 at 18:21
Done! I have added them"
– AyhamSYR
Nov 10 at 18:26
If "all characters are allowed", then there is no way to know the start of the 2nd numbers group (as numbers are a subset of "all"). So there must be a better differentiation here.
– KevinO
Nov 10 at 18:39
Could you give an example please?
– AyhamSYR
Nov 10 at 18:41
Your requirement is too vague. You might as well uses.matches("\d.*\d")to match a string starting and ending with a digit and having any 0+ chars in between.
– Wiktor Stribiżew
Nov 10 at 20:13
add a comment |
2
please, give possible input and output, and show your attempt.
– The Scientific Method
Nov 10 at 18:21
Done! I have added them"
– AyhamSYR
Nov 10 at 18:26
If "all characters are allowed", then there is no way to know the start of the 2nd numbers group (as numbers are a subset of "all"). So there must be a better differentiation here.
– KevinO
Nov 10 at 18:39
Could you give an example please?
– AyhamSYR
Nov 10 at 18:41
Your requirement is too vague. You might as well uses.matches("\d.*\d")to match a string starting and ending with a digit and having any 0+ chars in between.
– Wiktor Stribiżew
Nov 10 at 20:13
2
2
please, give possible input and output, and show your attempt.
– The Scientific Method
Nov 10 at 18:21
please, give possible input and output, and show your attempt.
– The Scientific Method
Nov 10 at 18:21
Done! I have added them"
– AyhamSYR
Nov 10 at 18:26
Done! I have added them"
– AyhamSYR
Nov 10 at 18:26
If "all characters are allowed", then there is no way to know the start of the 2nd numbers group (as numbers are a subset of "all"). So there must be a better differentiation here.
– KevinO
Nov 10 at 18:39
If "all characters are allowed", then there is no way to know the start of the 2nd numbers group (as numbers are a subset of "all"). So there must be a better differentiation here.
– KevinO
Nov 10 at 18:39
Could you give an example please?
– AyhamSYR
Nov 10 at 18:41
Could you give an example please?
– AyhamSYR
Nov 10 at 18:41
Your requirement is too vague. You might as well use
s.matches("\d.*\d") to match a string starting and ending with a digit and having any 0+ chars in between.– Wiktor Stribiżew
Nov 10 at 20:13
Your requirement is too vague. You might as well use
s.matches("\d.*\d") to match a string starting and ending with a digit and having any 0+ chars in between.– Wiktor Stribiżew
Nov 10 at 20:13
add a comment |
2 Answers
2
active
oldest
votes
Dealing with numbers can be difficult. This approach will deal with your examples, but check carefully. I also didn't do "all characters" in the middle grouping, as "all" would include numbers, so instead I assumed that finding the next non-number would be appropriate.
This Java regex handles the requirements:
"((-?)[\d,.]+)([^\d-]+)((-?)[\d,.]+)"
However, there is a potential issue in the above. Consider the following:
300 - -200. The foregoing won't match that case.
Now, based upon the examples, I think the point is that one should have a valid operator. The number of math operations is likely limited, so I would whitelist the operators in the middle. Thus, something like:
"((-?)[\d,.]+)([\s]*[*/+-]+[\s]*)((-?)[\d,.]+)"
Would, I think, be more appropriate. The [*/+-] can be expanded for the power operator ^ or whatever. Now, if one is going to start adding words (such as mod) in the equation, then the expression will need to be modified.
You can see this regular expression here
answered Nov 10 at 18:42
KevinO
3,07131628
add a comment |
In your regex you have to escape the dot . to match it literally and escape the + or else it would make the ? a possessive quantifier. To match 1+ digits you have to use a quantifier [0-9]+
For your example data, you could match 1+ digits followed by an optional part which matches either a dot or a comma at the start and at the end. If you want to match 1 time any character you could use a dot.
Instead of using a dot, you could also use for example a character class [-+*] to list some operators or list what you would allow to match. If this should be the only match, you could use anchors to assert the start ^ and the end $ of the string.
d+(?:[.,]d+)?.d+(?:[.,]d+)?
In Java:
String regex = "\d+(?:[.,]\d+)?.\d+(?:[.,]\d+)?";
Regex demo
That would match:
d+(?:[.,]d+)?1+ digits followed by an optional part that matches.or,followed by 1+ digits
.Match any character (Use .+) to repeat 1+ times- Same as the first pattern
answered Nov 11 at 12:22
The fourth bird
20.4k71326
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Dealing with numbers can be difficult. This approach will deal with your examples, but check carefully. I also didn't do "all characters" in the middle grouping, as "all" would include numbers, so instead I assumed that finding the next non-number would be appropriate.
This Java regex handles the requirements:
"((-?)[\d,.]+)([^\d-]+)((-?)[\d,.]+)"
However, there is a potential issue in the above. Consider the following:
300 - -200. The foregoing won't match that case.
Now, based upon the examples, I think the point is that one should have a valid operator. The number of math operations is likely limited, so I would whitelist the operators in the middle. Thus, something like:
"((-?)[\d,.]+)([\s]*[*/+-]+[\s]*)((-?)[\d,.]+)"
Would, I think, be more appropriate. The [*/+-] can be expanded for the power operator ^ or whatever. Now, if one is going to start adding words (such as mod) in the equation, then the expression will need to be modified.
You can see this regular expression here
answered Nov 10 at 18:42
KevinO
3,07131628
add a comment |
Dealing with numbers can be difficult. This approach will deal with your examples, but check carefully. I also didn't do "all characters" in the middle grouping, as "all" would include numbers, so instead I assumed that finding the next non-number would be appropriate.
This Java regex handles the requirements:
"((-?)[\d,.]+)([^\d-]+)((-?)[\d,.]+)"
However, there is a potential issue in the above. Consider the following:
300 - -200. The foregoing won't match that case.
Now, based upon the examples, I think the point is that one should have a valid operator. The number of math operations is likely limited, so I would whitelist the operators in the middle. Thus, something like:
"((-?)[\d,.]+)([\s]*[*/+-]+[\s]*)((-?)[\d,.]+)"
Would, I think, be more appropriate. The [*/+-] can be expanded for the power operator ^ or whatever. Now, if one is going to start adding words (such as mod) in the equation, then the expression will need to be modified.
You can see this regular expression here
answered Nov 10 at 18:42
KevinO
3,07131628
add a comment |
Dealing with numbers can be difficult. This approach will deal with your examples, but check carefully. I also didn't do "all characters" in the middle grouping, as "all" would include numbers, so instead I assumed that finding the next non-number would be appropriate.
This Java regex handles the requirements:
"((-?)[\d,.]+)([^\d-]+)((-?)[\d,.]+)"
However, there is a potential issue in the above. Consider the following:
300 - -200. The foregoing won't match that case.
Now, based upon the examples, I think the point is that one should have a valid operator. The number of math operations is likely limited, so I would whitelist the operators in the middle. Thus, something like:
"((-?)[\d,.]+)([\s]*[*/+-]+[\s]*)((-?)[\d,.]+)"
Would, I think, be more appropriate. The [*/+-] can be expanded for the power operator ^ or whatever. Now, if one is going to start adding words (such as mod) in the equation, then the expression will need to be modified.
You can see this regular expression here
answered Nov 10 at 18:42
KevinO
3,07131628
Dealing with numbers can be difficult. This approach will deal with your examples, but check carefully. I also didn't do "all characters" in the middle grouping, as "all" would include numbers, so instead I assumed that finding the next non-number would be appropriate.
This Java regex handles the requirements:
"((-?)[\d,.]+)([^\d-]+)((-?)[\d,.]+)"
However, there is a potential issue in the above. Consider the following:
300 - -200. The foregoing won't match that case.
Now, based upon the examples, I think the point is that one should have a valid operator. The number of math operations is likely limited, so I would whitelist the operators in the middle. Thus, something like:
"((-?)[\d,.]+)([\s]*[*/+-]+[\s]*)((-?)[\d,.]+)"
Would, I think, be more appropriate. The [*/+-] can be expanded for the power operator ^ or whatever. Now, if one is going to start adding words (such as mod) in the equation, then the expression will need to be modified.
You can see this regular expression here
answered Nov 10 at 18:42
KevinO
3,07131628
edited Nov 10 at 19:00
answered Nov 10 at 18:42
KevinO
3,07131628
answered Nov 10 at 18:42
KevinO
3,07131628
answered Nov 10 at 18:42
KevinO
3,07131628
3,07131628
add a comment |
add a comment |
In your regex you have to escape the dot . to match it literally and escape the + or else it would make the ? a possessive quantifier. To match 1+ digits you have to use a quantifier [0-9]+
For your example data, you could match 1+ digits followed by an optional part which matches either a dot or a comma at the start and at the end. If you want to match 1 time any character you could use a dot.
Instead of using a dot, you could also use for example a character class [-+*] to list some operators or list what you would allow to match. If this should be the only match, you could use anchors to assert the start ^ and the end $ of the string.
d+(?:[.,]d+)?.d+(?:[.,]d+)?
In Java:
String regex = "\d+(?:[.,]\d+)?.\d+(?:[.,]\d+)?";
Regex demo
That would match:
d+(?:[.,]d+)?1+ digits followed by an optional part that matches.or,followed by 1+ digits
.Match any character (Use .+) to repeat 1+ times- Same as the first pattern
answered Nov 11 at 12:22
The fourth bird
20.4k71326
add a comment |
In your regex you have to escape the dot . to match it literally and escape the + or else it would make the ? a possessive quantifier. To match 1+ digits you have to use a quantifier [0-9]+
For your example data, you could match 1+ digits followed by an optional part which matches either a dot or a comma at the start and at the end. If you want to match 1 time any character you could use a dot.
Instead of using a dot, you could also use for example a character class [-+*] to list some operators or list what you would allow to match. If this should be the only match, you could use anchors to assert the start ^ and the end $ of the string.
d+(?:[.,]d+)?.d+(?:[.,]d+)?
In Java:
String regex = "\d+(?:[.,]\d+)?.\d+(?:[.,]\d+)?";
Regex demo
That would match:
d+(?:[.,]d+)?1+ digits followed by an optional part that matches.or,followed by 1+ digits
.Match any character (Use .+) to repeat 1+ times- Same as the first pattern
answered Nov 11 at 12:22
The fourth bird
20.4k71326
add a comment |
In your regex you have to escape the dot . to match it literally and escape the + or else it would make the ? a possessive quantifier. To match 1+ digits you have to use a quantifier [0-9]+
For your example data, you could match 1+ digits followed by an optional part which matches either a dot or a comma at the start and at the end. If you want to match 1 time any character you could use a dot.
Instead of using a dot, you could also use for example a character class [-+*] to list some operators or list what you would allow to match. If this should be the only match, you could use anchors to assert the start ^ and the end $ of the string.
d+(?:[.,]d+)?.d+(?:[.,]d+)?
In Java:
String regex = "\d+(?:[.,]\d+)?.\d+(?:[.,]\d+)?";
Regex demo
That would match:
d+(?:[.,]d+)?1+ digits followed by an optional part that matches.or,followed by 1+ digits
.Match any character (Use .+) to repeat 1+ times- Same as the first pattern
answered Nov 11 at 12:22
The fourth bird
20.4k71326
In your regex you have to escape the dot . to match it literally and escape the + or else it would make the ? a possessive quantifier. To match 1+ digits you have to use a quantifier [0-9]+
For your example data, you could match 1+ digits followed by an optional part which matches either a dot or a comma at the start and at the end. If you want to match 1 time any character you could use a dot.
Instead of using a dot, you could also use for example a character class [-+*] to list some operators or list what you would allow to match. If this should be the only match, you could use anchors to assert the start ^ and the end $ of the string.
d+(?:[.,]d+)?.d+(?:[.,]d+)?
In Java:
String regex = "\d+(?:[.,]\d+)?.\d+(?:[.,]\d+)?";
Regex demo
That would match:
d+(?:[.,]d+)?1+ digits followed by an optional part that matches.or,followed by 1+ digits
.Match any character (Use .+) to repeat 1+ times- Same as the first pattern
answered Nov 11 at 12:22
The fourth bird
20.4k71326
edited Nov 11 at 13:35
answered Nov 11 at 12:22
The fourth bird
20.4k71326
answered Nov 11 at 12:22
The fourth bird
20.4k71326
answered Nov 11 at 12:22
The fourth bird
20.4k71326
20.4k71326
add a comment |
add a comment |
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2
please, give possible input and output, and show your attempt.
– The Scientific Method
Nov 10 at 18:21
Done! I have added them"
– AyhamSYR
Nov 10 at 18:26
If "all characters are allowed", then there is no way to know the start of the 2nd numbers group (as numbers are a subset of "all"). So there must be a better differentiation here.
– KevinO
Nov 10 at 18:39
Could you give an example please?
– AyhamSYR
Nov 10 at 18:41
Your requirement is too vague. You might as well use
s.matches("\d.*\d")to match a string starting and ending with a digit and having any 0+ chars in between.– Wiktor Stribiżew
Nov 10 at 20:13