Match the alphanumeric words(NOT NUMERIC-ONLY words) which have unique digits
7
Using regular expression, I want to select only the words which:
- are alphanumeric
- do not contain only numbers
- do not contain only alphabets
- have unique numbers(1 or more)
I am not really good with the regex but so far, I have tried [^ds]*(d+)(?!.*1) which takes me nowhere close to the desired output :(
Here are the input strings:
I would like abc123 to match but not 123.
ab12s should also match
Only number-words like 1234 should not match
Words containing same numbers like ab22s should not match
234 should not match
hel1lo2haha3hoho4
hel1lo2haha3hoho3
Expected Matches:
abc123
ab12s
hel1lo2haha3hoho4
javascript regex
edited Feb 2 at 6:58
CertainPerformance
89.5k165177
asked Feb 2 at 6:44
ManJoeyManJoey
11616
add a comment |
7
Using regular expression, I want to select only the words which:
- are alphanumeric
- do not contain only numbers
- do not contain only alphabets
- have unique numbers(1 or more)
I am not really good with the regex but so far, I have tried [^ds]*(d+)(?!.*1) which takes me nowhere close to the desired output :(
Here are the input strings:
I would like abc123 to match but not 123.
ab12s should also match
Only number-words like 1234 should not match
Words containing same numbers like ab22s should not match
234 should not match
hel1lo2haha3hoho4
hel1lo2haha3hoho3
Expected Matches:
abc123
ab12s
hel1lo2haha3hoho4
javascript regex
edited Feb 2 at 6:58
CertainPerformance
89.5k165177
asked Feb 2 at 6:44
ManJoeyManJoey
11616
add a comment |
7
7
7
1
Using regular expression, I want to select only the words which:
- are alphanumeric
- do not contain only numbers
- do not contain only alphabets
- have unique numbers(1 or more)
I am not really good with the regex but so far, I have tried [^ds]*(d+)(?!.*1) which takes me nowhere close to the desired output :(
Here are the input strings:
I would like abc123 to match but not 123.
ab12s should also match
Only number-words like 1234 should not match
Words containing same numbers like ab22s should not match
234 should not match
hel1lo2haha3hoho4
hel1lo2haha3hoho3
Expected Matches:
abc123
ab12s
hel1lo2haha3hoho4
javascript regex
edited Feb 2 at 6:58
CertainPerformance
89.5k165177
asked Feb 2 at 6:44
ManJoeyManJoey
11616
Using regular expression, I want to select only the words which:
- are alphanumeric
- do not contain only numbers
- do not contain only alphabets
- have unique numbers(1 or more)
I am not really good with the regex but so far, I have tried [^ds]*(d+)(?!.*1) which takes me nowhere close to the desired output :(
Here are the input strings:
I would like abc123 to match but not 123.
ab12s should also match
Only number-words like 1234 should not match
Words containing same numbers like ab22s should not match
234 should not match
hel1lo2haha3hoho4
hel1lo2haha3hoho3
Expected Matches:
abc123
ab12s
hel1lo2haha3hoho4
javascript regex
javascript regex
edited Feb 2 at 6:58
CertainPerformance
89.5k165177
asked Feb 2 at 6:44
ManJoeyManJoey
11616
edited Feb 2 at 6:58
CertainPerformance
89.5k165177
asked Feb 2 at 6:44
ManJoeyManJoey
11616
edited Feb 2 at 6:58
CertainPerformance
89.5k165177
edited Feb 2 at 6:58
CertainPerformance
89.5k165177
edited Feb 2 at 6:58
CertainPerformance
89.5k165177
89.5k165177
asked Feb 2 at 6:44
ManJoeyManJoey
11616
asked Feb 2 at 6:44
ManJoeyManJoey
11616
asked Feb 2 at 6:44
ManJoeyManJoey
11616
11616
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
8
You can use
b(?=d*[a-z])(?=[a-z]*d)(?:[a-z]|(d)(?!w*1))+b
https://regex101.com/r/TimjdW/3
Anchor the start and end of the pattern at word boundaries with b, then:
(?=d*[a-z])- Lookahead for an alphabetical character somewhere in the word
(?=[a-z]*d)- Lookahead for a digit somewhere in the word
(?:[a-z]|(d)(?!w*1))+Repeatedly match either:
[a-z]- Any alphabetical character, or
(d)(?!w*1)- A digit which does not occur again in the same word
answered Feb 2 at 6:50
CertainPerformanceCertainPerformance
89.5k165177
add a comment |
3
Here is a bit shorter & faster regex to make it happen since it doesn't assert negative lookahead for each character:
/b(?=[a-z]*d)(?=d*[a-z])(?!w*(d)w*1)[a-zd]+b/ig
RegEx Demo
RegEx Details:
b: Word boundary
(?=[a-z]*d): Make sure we have at least a digit
(?=d*[a-z]): Make sure we have at least a letter
(?!w*(d)w*1): Make sure digits are not repeated anywhere in the word
[a-zd]+: Match 1+ alphanumericals
b: Word boundary
answered Feb 2 at 7:00
anubhavaanubhava
529k46327404
@ManJoey: I believe the that selected regex will be slower for bigger strings
– anubhava
Feb 2 at 7:37
@anubhava,1in(?!w*(d)w*1)means(d)saved in memory's first buffer(likesed) kind of thing, sorry I am learning regex so thought to check with you on same once.
– RavinderSingh13
Feb 2 at 7:39
1
Yes that's correct.1is back-reference for captured group #1 i.e.(d)
– anubhava
Feb 2 at 7:41
add a comment |
2
You could assert all the conditions using one negative lookahead:
b(?![a-z]+b|d+b|w*(d)w*1)[a-zd]+b
See live demo here
The important parts are starting match from b and immediately looking for the conditions:
[a-z]+bOnly alphabeticd+bOnly numericw*(d)w*1Has a repeating digit
answered Feb 2 at 7:00
revorevo
33.3k135085
Thanks, good fix though not sure why one single negative lookahead with alternations shows more steps (as in my answer) than multiple assertions on regex101
– anubhava
Feb 2 at 7:07
1
@anubhava Yes, it depends on the input string. For example consider switching the first two lookaheads in your solution. You'll see it increases.
– revo
Feb 2 at 7:10
add a comment |
1
You can use this
b(?!w*(d)w*1)(?=(?:[a-z]+d+)|(?:d+[a-z]+))[a-z0-9]+b
b- Word boundary.
(?!w*(d)w*1)- Condition to check unique digits.
(?=(?:[a-z]+d+)|(?:d+[a-z]+))- Condition to check alphanumeric words.
[a-z0-9]+- Matchesa to zand0 to 9
Demo
answered Feb 2 at 7:04
Code ManiacCode Maniac
7,6001526
Looking at all the answers here reminds me why I think regexes aren't the best way to solve this sort of problem… They're highly ingenious, but I'd hate to have to debug or maintain any of them!
– gidds
Feb 2 at 8:36
1
@gidds You can approach them the same way you approach any programming problem - break the problem down into logical groups, (re?)write and verify each group, and put it together into a single pattern. REs are a great concise way to match strings - they're extremely flexible and mostly language agnostic, which is a huge plus. As long as the pattern to debug has descriptive comments (like in the answers here), it shouldn't be hard at all for someone with a bit of experience with REs, IMO
– CertainPerformance
Feb 3 at 21:02
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
8
You can use
b(?=d*[a-z])(?=[a-z]*d)(?:[a-z]|(d)(?!w*1))+b
https://regex101.com/r/TimjdW/3
Anchor the start and end of the pattern at word boundaries with b, then:
(?=d*[a-z])- Lookahead for an alphabetical character somewhere in the word
(?=[a-z]*d)- Lookahead for a digit somewhere in the word
(?:[a-z]|(d)(?!w*1))+Repeatedly match either:
[a-z]- Any alphabetical character, or
(d)(?!w*1)- A digit which does not occur again in the same word
answered Feb 2 at 6:50
CertainPerformanceCertainPerformance
89.5k165177
add a comment |
8
You can use
b(?=d*[a-z])(?=[a-z]*d)(?:[a-z]|(d)(?!w*1))+b
https://regex101.com/r/TimjdW/3
Anchor the start and end of the pattern at word boundaries with b, then:
(?=d*[a-z])- Lookahead for an alphabetical character somewhere in the word
(?=[a-z]*d)- Lookahead for a digit somewhere in the word
(?:[a-z]|(d)(?!w*1))+Repeatedly match either:
[a-z]- Any alphabetical character, or
(d)(?!w*1)- A digit which does not occur again in the same word
answered Feb 2 at 6:50
CertainPerformanceCertainPerformance
89.5k165177
add a comment |
8
8
8
You can use
b(?=d*[a-z])(?=[a-z]*d)(?:[a-z]|(d)(?!w*1))+b
https://regex101.com/r/TimjdW/3
Anchor the start and end of the pattern at word boundaries with b, then:
(?=d*[a-z])- Lookahead for an alphabetical character somewhere in the word
(?=[a-z]*d)- Lookahead for a digit somewhere in the word
(?:[a-z]|(d)(?!w*1))+Repeatedly match either:
[a-z]- Any alphabetical character, or
(d)(?!w*1)- A digit which does not occur again in the same word
answered Feb 2 at 6:50
CertainPerformanceCertainPerformance
89.5k165177
You can use
b(?=d*[a-z])(?=[a-z]*d)(?:[a-z]|(d)(?!w*1))+b
https://regex101.com/r/TimjdW/3
Anchor the start and end of the pattern at word boundaries with b, then:
(?=d*[a-z])- Lookahead for an alphabetical character somewhere in the word
(?=[a-z]*d)- Lookahead for a digit somewhere in the word
(?:[a-z]|(d)(?!w*1))+Repeatedly match either:
[a-z]- Any alphabetical character, or
(d)(?!w*1)- A digit which does not occur again in the same word
answered Feb 2 at 6:50
CertainPerformanceCertainPerformance
89.5k165177
edited Feb 2 at 6:56
answered Feb 2 at 6:50
CertainPerformanceCertainPerformance
89.5k165177
answered Feb 2 at 6:50
CertainPerformanceCertainPerformance
89.5k165177
answered Feb 2 at 6:50
CertainPerformanceCertainPerformance
89.5k165177
89.5k165177
add a comment |
add a comment |
3
Here is a bit shorter & faster regex to make it happen since it doesn't assert negative lookahead for each character:
/b(?=[a-z]*d)(?=d*[a-z])(?!w*(d)w*1)[a-zd]+b/ig
RegEx Demo
RegEx Details:
b: Word boundary
(?=[a-z]*d): Make sure we have at least a digit
(?=d*[a-z]): Make sure we have at least a letter
(?!w*(d)w*1): Make sure digits are not repeated anywhere in the word
[a-zd]+: Match 1+ alphanumericals
b: Word boundary
answered Feb 2 at 7:00
anubhavaanubhava
529k46327404
@ManJoey: I believe the that selected regex will be slower for bigger strings
– anubhava
Feb 2 at 7:37
@anubhava,1in(?!w*(d)w*1)means(d)saved in memory's first buffer(likesed) kind of thing, sorry I am learning regex so thought to check with you on same once.
– RavinderSingh13
Feb 2 at 7:39
1
Yes that's correct.1is back-reference for captured group #1 i.e.(d)
– anubhava
Feb 2 at 7:41
add a comment |
3
Here is a bit shorter & faster regex to make it happen since it doesn't assert negative lookahead for each character:
/b(?=[a-z]*d)(?=d*[a-z])(?!w*(d)w*1)[a-zd]+b/ig
RegEx Demo
RegEx Details:
b: Word boundary
(?=[a-z]*d): Make sure we have at least a digit
(?=d*[a-z]): Make sure we have at least a letter
(?!w*(d)w*1): Make sure digits are not repeated anywhere in the word
[a-zd]+: Match 1+ alphanumericals
b: Word boundary
answered Feb 2 at 7:00
anubhavaanubhava
529k46327404
@ManJoey: I believe the that selected regex will be slower for bigger strings
– anubhava
Feb 2 at 7:37
@anubhava,1in(?!w*(d)w*1)means(d)saved in memory's first buffer(likesed) kind of thing, sorry I am learning regex so thought to check with you on same once.
– RavinderSingh13
Feb 2 at 7:39
1
Yes that's correct.1is back-reference for captured group #1 i.e.(d)
– anubhava
Feb 2 at 7:41
add a comment |
3
3
3
Here is a bit shorter & faster regex to make it happen since it doesn't assert negative lookahead for each character:
/b(?=[a-z]*d)(?=d*[a-z])(?!w*(d)w*1)[a-zd]+b/ig
RegEx Demo
RegEx Details:
b: Word boundary
(?=[a-z]*d): Make sure we have at least a digit
(?=d*[a-z]): Make sure we have at least a letter
(?!w*(d)w*1): Make sure digits are not repeated anywhere in the word
[a-zd]+: Match 1+ alphanumericals
b: Word boundary
answered Feb 2 at 7:00
anubhavaanubhava
529k46327404
Here is a bit shorter & faster regex to make it happen since it doesn't assert negative lookahead for each character:
/b(?=[a-z]*d)(?=d*[a-z])(?!w*(d)w*1)[a-zd]+b/ig
RegEx Demo
RegEx Details:
b: Word boundary
(?=[a-z]*d): Make sure we have at least a digit
(?=d*[a-z]): Make sure we have at least a letter
(?!w*(d)w*1): Make sure digits are not repeated anywhere in the word
[a-zd]+: Match 1+ alphanumericals
b: Word boundary
answered Feb 2 at 7:00
anubhavaanubhava
529k46327404
answered Feb 2 at 7:00
anubhavaanubhava
529k46327404
answered Feb 2 at 7:00
anubhavaanubhava
529k46327404
answered Feb 2 at 7:00
anubhavaanubhava
529k46327404
529k46327404
@ManJoey: I believe the that selected regex will be slower for bigger strings
– anubhava
Feb 2 at 7:37
@anubhava,1in(?!w*(d)w*1)means(d)saved in memory's first buffer(likesed) kind of thing, sorry I am learning regex so thought to check with you on same once.
– RavinderSingh13
Feb 2 at 7:39
1
Yes that's correct.1is back-reference for captured group #1 i.e.(d)
– anubhava
Feb 2 at 7:41
add a comment |
@ManJoey: I believe the that selected regex will be slower for bigger strings
– anubhava
Feb 2 at 7:37
@anubhava,1in(?!w*(d)w*1)means(d)saved in memory's first buffer(likesed) kind of thing, sorry I am learning regex so thought to check with you on same once.
– RavinderSingh13
Feb 2 at 7:39
1
Yes that's correct.1is back-reference for captured group #1 i.e.(d)
– anubhava
Feb 2 at 7:41
@ManJoey: I believe the that selected regex will be slower for bigger strings
– anubhava
Feb 2 at 7:37
@ManJoey: I believe the that selected regex will be slower for bigger strings
– anubhava
Feb 2 at 7:37
@anubhava,
1 in (?!w*(d)w*1) means (d) saved in memory's first buffer(like sed) kind of thing, sorry I am learning regex so thought to check with you on same once.– RavinderSingh13
Feb 2 at 7:39
@anubhava,
1 in (?!w*(d)w*1) means (d) saved in memory's first buffer(like sed) kind of thing, sorry I am learning regex so thought to check with you on same once.– RavinderSingh13
Feb 2 at 7:39
1
1
Yes that's correct.
1 is back-reference for captured group #1 i.e. (d)– anubhava
Feb 2 at 7:41
Yes that's correct.
1 is back-reference for captured group #1 i.e. (d)– anubhava
Feb 2 at 7:41
add a comment |
2
You could assert all the conditions using one negative lookahead:
b(?![a-z]+b|d+b|w*(d)w*1)[a-zd]+b
See live demo here
The important parts are starting match from b and immediately looking for the conditions:
[a-z]+bOnly alphabeticd+bOnly numericw*(d)w*1Has a repeating digit
answered Feb 2 at 7:00
revorevo
33.3k135085
Thanks, good fix though not sure why one single negative lookahead with alternations shows more steps (as in my answer) than multiple assertions on regex101
– anubhava
Feb 2 at 7:07
1
@anubhava Yes, it depends on the input string. For example consider switching the first two lookaheads in your solution. You'll see it increases.
– revo
Feb 2 at 7:10
add a comment |
2
You could assert all the conditions using one negative lookahead:
b(?![a-z]+b|d+b|w*(d)w*1)[a-zd]+b
See live demo here
The important parts are starting match from b and immediately looking for the conditions:
[a-z]+bOnly alphabeticd+bOnly numericw*(d)w*1Has a repeating digit
answered Feb 2 at 7:00
revorevo
33.3k135085
Thanks, good fix though not sure why one single negative lookahead with alternations shows more steps (as in my answer) than multiple assertions on regex101
– anubhava
Feb 2 at 7:07
1
@anubhava Yes, it depends on the input string. For example consider switching the first two lookaheads in your solution. You'll see it increases.
– revo
Feb 2 at 7:10
add a comment |
2
2
2
You could assert all the conditions using one negative lookahead:
b(?![a-z]+b|d+b|w*(d)w*1)[a-zd]+b
See live demo here
The important parts are starting match from b and immediately looking for the conditions:
[a-z]+bOnly alphabeticd+bOnly numericw*(d)w*1Has a repeating digit
answered Feb 2 at 7:00
revorevo
33.3k135085
You could assert all the conditions using one negative lookahead:
b(?![a-z]+b|d+b|w*(d)w*1)[a-zd]+b
See live demo here
The important parts are starting match from b and immediately looking for the conditions:
[a-z]+bOnly alphabeticd+bOnly numericw*(d)w*1Has a repeating digit
answered Feb 2 at 7:00
revorevo
33.3k135085
edited Feb 2 at 7:04
answered Feb 2 at 7:00
revorevo
33.3k135085
answered Feb 2 at 7:00
revorevo
33.3k135085
answered Feb 2 at 7:00
revorevo
33.3k135085
33.3k135085
Thanks, good fix though not sure why one single negative lookahead with alternations shows more steps (as in my answer) than multiple assertions on regex101
– anubhava
Feb 2 at 7:07
1
@anubhava Yes, it depends on the input string. For example consider switching the first two lookaheads in your solution. You'll see it increases.
– revo
Feb 2 at 7:10
add a comment |
Thanks, good fix though not sure why one single negative lookahead with alternations shows more steps (as in my answer) than multiple assertions on regex101
– anubhava
Feb 2 at 7:07
1
@anubhava Yes, it depends on the input string. For example consider switching the first two lookaheads in your solution. You'll see it increases.
– revo
Feb 2 at 7:10
Thanks, good fix though not sure why one single negative lookahead with alternations shows more steps (as in my answer) than multiple assertions on regex101
– anubhava
Feb 2 at 7:07
Thanks, good fix though not sure why one single negative lookahead with alternations shows more steps (as in my answer) than multiple assertions on regex101
– anubhava
Feb 2 at 7:07
1
1
@anubhava Yes, it depends on the input string. For example consider switching the first two lookaheads in your solution. You'll see it increases.
– revo
Feb 2 at 7:10
@anubhava Yes, it depends on the input string. For example consider switching the first two lookaheads in your solution. You'll see it increases.
– revo
Feb 2 at 7:10
add a comment |
1
You can use this
b(?!w*(d)w*1)(?=(?:[a-z]+d+)|(?:d+[a-z]+))[a-z0-9]+b
b- Word boundary.
(?!w*(d)w*1)- Condition to check unique digits.
(?=(?:[a-z]+d+)|(?:d+[a-z]+))- Condition to check alphanumeric words.
[a-z0-9]+- Matchesa to zand0 to 9
Demo
answered Feb 2 at 7:04
Code ManiacCode Maniac
7,6001526
Looking at all the answers here reminds me why I think regexes aren't the best way to solve this sort of problem… They're highly ingenious, but I'd hate to have to debug or maintain any of them!
– gidds
Feb 2 at 8:36
1
@gidds You can approach them the same way you approach any programming problem - break the problem down into logical groups, (re?)write and verify each group, and put it together into a single pattern. REs are a great concise way to match strings - they're extremely flexible and mostly language agnostic, which is a huge plus. As long as the pattern to debug has descriptive comments (like in the answers here), it shouldn't be hard at all for someone with a bit of experience with REs, IMO
– CertainPerformance
Feb 3 at 21:02
add a comment |
1
You can use this
b(?!w*(d)w*1)(?=(?:[a-z]+d+)|(?:d+[a-z]+))[a-z0-9]+b
b- Word boundary.
(?!w*(d)w*1)- Condition to check unique digits.
(?=(?:[a-z]+d+)|(?:d+[a-z]+))- Condition to check alphanumeric words.
[a-z0-9]+- Matchesa to zand0 to 9
Demo
answered Feb 2 at 7:04
Code ManiacCode Maniac
7,6001526
Looking at all the answers here reminds me why I think regexes aren't the best way to solve this sort of problem… They're highly ingenious, but I'd hate to have to debug or maintain any of them!
– gidds
Feb 2 at 8:36
1
@gidds You can approach them the same way you approach any programming problem - break the problem down into logical groups, (re?)write and verify each group, and put it together into a single pattern. REs are a great concise way to match strings - they're extremely flexible and mostly language agnostic, which is a huge plus. As long as the pattern to debug has descriptive comments (like in the answers here), it shouldn't be hard at all for someone with a bit of experience with REs, IMO
– CertainPerformance
Feb 3 at 21:02
add a comment |
1
1
1
You can use this
b(?!w*(d)w*1)(?=(?:[a-z]+d+)|(?:d+[a-z]+))[a-z0-9]+b
b- Word boundary.
(?!w*(d)w*1)- Condition to check unique digits.
(?=(?:[a-z]+d+)|(?:d+[a-z]+))- Condition to check alphanumeric words.
[a-z0-9]+- Matchesa to zand0 to 9
Demo
answered Feb 2 at 7:04
Code ManiacCode Maniac
7,6001526
You can use this
b(?!w*(d)w*1)(?=(?:[a-z]+d+)|(?:d+[a-z]+))[a-z0-9]+b
b- Word boundary.
(?!w*(d)w*1)- Condition to check unique digits.
(?=(?:[a-z]+d+)|(?:d+[a-z]+))- Condition to check alphanumeric words.
[a-z0-9]+- Matchesa to zand0 to 9
Demo
answered Feb 2 at 7:04
Code ManiacCode Maniac
7,6001526
answered Feb 2 at 7:04
Code ManiacCode Maniac
7,6001526
answered Feb 2 at 7:04
Code ManiacCode Maniac
7,6001526
answered Feb 2 at 7:04
Code ManiacCode Maniac
7,6001526
7,6001526
Looking at all the answers here reminds me why I think regexes aren't the best way to solve this sort of problem… They're highly ingenious, but I'd hate to have to debug or maintain any of them!
– gidds
Feb 2 at 8:36
1
@gidds You can approach them the same way you approach any programming problem - break the problem down into logical groups, (re?)write and verify each group, and put it together into a single pattern. REs are a great concise way to match strings - they're extremely flexible and mostly language agnostic, which is a huge plus. As long as the pattern to debug has descriptive comments (like in the answers here), it shouldn't be hard at all for someone with a bit of experience with REs, IMO
– CertainPerformance
Feb 3 at 21:02
add a comment |
Looking at all the answers here reminds me why I think regexes aren't the best way to solve this sort of problem… They're highly ingenious, but I'd hate to have to debug or maintain any of them!
– gidds
Feb 2 at 8:36
1
@gidds You can approach them the same way you approach any programming problem - break the problem down into logical groups, (re?)write and verify each group, and put it together into a single pattern. REs are a great concise way to match strings - they're extremely flexible and mostly language agnostic, which is a huge plus. As long as the pattern to debug has descriptive comments (like in the answers here), it shouldn't be hard at all for someone with a bit of experience with REs, IMO
– CertainPerformance
Feb 3 at 21:02
Looking at all the answers here reminds me why I think regexes aren't the best way to solve this sort of problem… They're highly ingenious, but I'd hate to have to debug or maintain any of them!
– gidds
Feb 2 at 8:36
Looking at all the answers here reminds me why I think regexes aren't the best way to solve this sort of problem… They're highly ingenious, but I'd hate to have to debug or maintain any of them!
– gidds
Feb 2 at 8:36
1
1
@gidds You can approach them the same way you approach any programming problem - break the problem down into logical groups, (re?)write and verify each group, and put it together into a single pattern. REs are a great concise way to match strings - they're extremely flexible and mostly language agnostic, which is a huge plus. As long as the pattern to debug has descriptive comments (like in the answers here), it shouldn't be hard at all for someone with a bit of experience with REs, IMO
– CertainPerformance
Feb 3 at 21:02
@gidds You can approach them the same way you approach any programming problem - break the problem down into logical groups, (re?)write and verify each group, and put it together into a single pattern. REs are a great concise way to match strings - they're extremely flexible and mostly language agnostic, which is a huge plus. As long as the pattern to debug has descriptive comments (like in the answers here), it shouldn't be hard at all for someone with a bit of experience with REs, IMO
– CertainPerformance
Feb 3 at 21:02
add a comment |
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