Regex for strings containing substring that contains substring
Let's say I have this string:
[section][module_text id="123"]text[/module][module_number id=""]000[/module][/section]
I can find each entire [module...[/module...] substring with
([module)(.*?)([/module.*?])
But I would like to only match the [module] string that has a digit in the id="" parameter. In this case the first [module].
I've been trying a bunch of stuff but i cant seem to get any closer.
php regex
asked Nov 20 at 7:20
kylmark
98112
add a comment |
Let's say I have this string:
[section][module_text id="123"]text[/module][module_number id=""]000[/module][/section]
I can find each entire [module...[/module...] substring with
([module)(.*?)([/module.*?])
But I would like to only match the [module] string that has a digit in the id="" parameter. In this case the first [module].
I've been trying a bunch of stuff but i cant seem to get any closer.
php regex
asked Nov 20 at 7:20
kylmark
98112
Are you sure you don't want tu use XML with an XML parser ?
– Blusky
Nov 20 at 7:21
add a comment |
Let's say I have this string:
[section][module_text id="123"]text[/module][module_number id=""]000[/module][/section]
I can find each entire [module...[/module...] substring with
([module)(.*?)([/module.*?])
But I would like to only match the [module] string that has a digit in the id="" parameter. In this case the first [module].
I've been trying a bunch of stuff but i cant seem to get any closer.
php regex
asked Nov 20 at 7:20
kylmark
98112
Let's say I have this string:
[section][module_text id="123"]text[/module][module_number id=""]000[/module][/section]
I can find each entire [module...[/module...] substring with
([module)(.*?)([/module.*?])
But I would like to only match the [module] string that has a digit in the id="" parameter. In this case the first [module].
I've been trying a bunch of stuff but i cant seem to get any closer.
php regex
php regex
asked Nov 20 at 7:20
kylmark
98112
asked Nov 20 at 7:20
kylmark
98112
asked Nov 20 at 7:20
kylmark
98112
asked Nov 20 at 7:20
kylmark
98112
asked Nov 20 at 7:20
kylmark
98112
98112
Are you sure you don't want tu use XML with an XML parser ?
– Blusky
Nov 20 at 7:21
add a comment |
Are you sure you don't want tu use XML with an XML parser ?
– Blusky
Nov 20 at 7:21
Are you sure you don't want tu use XML with an XML parser ?
– Blusky
Nov 20 at 7:21
Are you sure you don't want tu use XML with an XML parser ?
– Blusky
Nov 20 at 7:21
add a comment |
1 Answer
1
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votes
After matching module, you can match non-] characters until you get to the id part, and after the id's ", you can match [^"]*d to ensure that the id has a digit in it:
([module[^]]*id="[^"]*d)(.*?)([/module[^]]*])
https://regex101.com/r/25JcEP/1
If the modules you want to match have ids which always start with digits, then you can simplify it a bit to
([module[^]]*id="d)(.*?)([/module[^]]*])
Note the use of a negative character class rather than .*? for the final ([/module[^]]*]) - negative character classes are a bit more efficient than lazy repetition when possible.
answered Nov 20 at 7:23
CertainPerformance
75.3k143659
@kylmark Same regex, just usepreg_match_all
– CertainPerformance
Nov 20 at 7:39
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
After matching module, you can match non-] characters until you get to the id part, and after the id's ", you can match [^"]*d to ensure that the id has a digit in it:
([module[^]]*id="[^"]*d)(.*?)([/module[^]]*])
https://regex101.com/r/25JcEP/1
If the modules you want to match have ids which always start with digits, then you can simplify it a bit to
([module[^]]*id="d)(.*?)([/module[^]]*])
Note the use of a negative character class rather than .*? for the final ([/module[^]]*]) - negative character classes are a bit more efficient than lazy repetition when possible.
answered Nov 20 at 7:23
CertainPerformance
75.3k143659
@kylmark Same regex, just usepreg_match_all
– CertainPerformance
Nov 20 at 7:39
add a comment |
After matching module, you can match non-] characters until you get to the id part, and after the id's ", you can match [^"]*d to ensure that the id has a digit in it:
([module[^]]*id="[^"]*d)(.*?)([/module[^]]*])
https://regex101.com/r/25JcEP/1
If the modules you want to match have ids which always start with digits, then you can simplify it a bit to
([module[^]]*id="d)(.*?)([/module[^]]*])
Note the use of a negative character class rather than .*? for the final ([/module[^]]*]) - negative character classes are a bit more efficient than lazy repetition when possible.
answered Nov 20 at 7:23
CertainPerformance
75.3k143659
@kylmark Same regex, just usepreg_match_all
– CertainPerformance
Nov 20 at 7:39
add a comment |
After matching module, you can match non-] characters until you get to the id part, and after the id's ", you can match [^"]*d to ensure that the id has a digit in it:
([module[^]]*id="[^"]*d)(.*?)([/module[^]]*])
https://regex101.com/r/25JcEP/1
If the modules you want to match have ids which always start with digits, then you can simplify it a bit to
([module[^]]*id="d)(.*?)([/module[^]]*])
Note the use of a negative character class rather than .*? for the final ([/module[^]]*]) - negative character classes are a bit more efficient than lazy repetition when possible.
answered Nov 20 at 7:23
CertainPerformance
75.3k143659
After matching module, you can match non-] characters until you get to the id part, and after the id's ", you can match [^"]*d to ensure that the id has a digit in it:
([module[^]]*id="[^"]*d)(.*?)([/module[^]]*])
https://regex101.com/r/25JcEP/1
If the modules you want to match have ids which always start with digits, then you can simplify it a bit to
([module[^]]*id="d)(.*?)([/module[^]]*])
Note the use of a negative character class rather than .*? for the final ([/module[^]]*]) - negative character classes are a bit more efficient than lazy repetition when possible.
answered Nov 20 at 7:23
CertainPerformance
75.3k143659
edited Nov 20 at 7:32
answered Nov 20 at 7:23
CertainPerformance
75.3k143659
answered Nov 20 at 7:23
CertainPerformance
75.3k143659
answered Nov 20 at 7:23
CertainPerformance
75.3k143659
75.3k143659
@kylmark Same regex, just usepreg_match_all
– CertainPerformance
Nov 20 at 7:39
add a comment |
@kylmark Same regex, just usepreg_match_all
– CertainPerformance
Nov 20 at 7:39
@kylmark Same regex, just use
preg_match_all– CertainPerformance
Nov 20 at 7:39
@kylmark Same regex, just use
preg_match_all– CertainPerformance
Nov 20 at 7:39
add a comment |
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Are you sure you don't want tu use XML with an XML parser ?
– Blusky
Nov 20 at 7:21