How to invert the regular expression group capture logic?
To create a capturing group in a regex you use (match) and you prefix it with ?: to make it non-capturing, like (?:match). The thing is, in any kind of complicated regular expression I find myself wanting to create far more non-capturing groups than capturing ones, so I'd like to reverse this logic and only capture groups beginning with ?: (or whatever). How can I do this? I mainly use regular expressions with .NET, but I wouldn't mind answers for other languages with regular expressions like Perl, PHP, Python, JavaScript, etc.
c# python .net regex perl
asked Nov 12 '18 at 21:37
JezJez
11.8k1881152
add a comment |
To create a capturing group in a regex you use (match) and you prefix it with ?: to make it non-capturing, like (?:match). The thing is, in any kind of complicated regular expression I find myself wanting to create far more non-capturing groups than capturing ones, so I'd like to reverse this logic and only capture groups beginning with ?: (or whatever). How can I do this? I mainly use regular expressions with .NET, but I wouldn't mind answers for other languages with regular expressions like Perl, PHP, Python, JavaScript, etc.
c# python .net regex perl
asked Nov 12 '18 at 21:37
JezJez
11.8k1881152
You could perhaps instantiate your RegEx object withRegexOptions.ExplicitCapture(docs.microsoft.com/en-gb/dotnet/api/…)
– elgonzo
Nov 12 '18 at 21:40
@elgonzo That looks good, if you put it as an answer I can accept.
– Jez
Nov 12 '18 at 21:41
add a comment |
To create a capturing group in a regex you use (match) and you prefix it with ?: to make it non-capturing, like (?:match). The thing is, in any kind of complicated regular expression I find myself wanting to create far more non-capturing groups than capturing ones, so I'd like to reverse this logic and only capture groups beginning with ?: (or whatever). How can I do this? I mainly use regular expressions with .NET, but I wouldn't mind answers for other languages with regular expressions like Perl, PHP, Python, JavaScript, etc.
c# python .net regex perl
asked Nov 12 '18 at 21:37
JezJez
11.8k1881152
To create a capturing group in a regex you use (match) and you prefix it with ?: to make it non-capturing, like (?:match). The thing is, in any kind of complicated regular expression I find myself wanting to create far more non-capturing groups than capturing ones, so I'd like to reverse this logic and only capture groups beginning with ?: (or whatever). How can I do this? I mainly use regular expressions with .NET, but I wouldn't mind answers for other languages with regular expressions like Perl, PHP, Python, JavaScript, etc.
c# python .net regex perl
c# python .net regex perl
asked Nov 12 '18 at 21:37
JezJez
11.8k1881152
asked Nov 12 '18 at 21:37
JezJez
11.8k1881152
asked Nov 12 '18 at 21:37
JezJez
11.8k1881152
asked Nov 12 '18 at 21:37
JezJez
11.8k1881152
asked Nov 12 '18 at 21:37
JezJez
11.8k1881152
11.8k1881152
You could perhaps instantiate your RegEx object withRegexOptions.ExplicitCapture(docs.microsoft.com/en-gb/dotnet/api/…)
– elgonzo
Nov 12 '18 at 21:40
@elgonzo That looks good, if you put it as an answer I can accept.
– Jez
Nov 12 '18 at 21:41
add a comment |
You could perhaps instantiate your RegEx object withRegexOptions.ExplicitCapture(docs.microsoft.com/en-gb/dotnet/api/…)
– elgonzo
Nov 12 '18 at 21:40
@elgonzo That looks good, if you put it as an answer I can accept.
– Jez
Nov 12 '18 at 21:41
You could perhaps instantiate your RegEx object with
RegexOptions.ExplicitCapture (docs.microsoft.com/en-gb/dotnet/api/…)– elgonzo
Nov 12 '18 at 21:40
You could perhaps instantiate your RegEx object with
RegexOptions.ExplicitCapture (docs.microsoft.com/en-gb/dotnet/api/…)– elgonzo
Nov 12 '18 at 21:40
@elgonzo That looks good, if you put it as an answer I can accept.
– Jez
Nov 12 '18 at 21:41
@elgonzo That looks good, if you put it as an answer I can accept.
– Jez
Nov 12 '18 at 21:41
add a comment |
2 Answers
2
active
oldest
votes
If you want to avoid the clumsiness of (?: ) and turn ( ) groups into non-capturing groups, use the RegexOptions.ExplicitCapture option. Only named groups ((?<name>subexpression)) will be captured if this option is being used.
However, you cannot turn non-capturing groups (?: ) into capturing groups, unfortunately.
The RegEx constructor as well as other methods from the RegEx class accept RegexOptions flags.
For example:
Regex.Matches(input, pattern, RegexOptions.ExplicitCapture)
answered Nov 12 '18 at 21:44
elgonzoelgonzo
4,49921323
add a comment |
In any language that supports named capture groups you can simply use them for what you want captured, and ignore the numbered ones.
my $string = q(Available from v5.10 in Perl.);
$string =~ /([A-Z].+?)(?<v>[0-9.]+)s+(.*?)./;
say "Version: $+{v}";
After the regex the capture is available in %+ hash, inside the regex in k<name> or g{name}.
The downside is that you still capture all that other stuff (what hurts efficiency a little), while the upside is that you still capture all that other stuff (what helps flexibility, if some of it turns needed).
answered Nov 13 '18 at 1:05
zdimzdim
32.3k32041
add a comment |
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2 Answers
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active
oldest
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2 Answers
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active
oldest
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If you want to avoid the clumsiness of (?: ) and turn ( ) groups into non-capturing groups, use the RegexOptions.ExplicitCapture option. Only named groups ((?<name>subexpression)) will be captured if this option is being used.
However, you cannot turn non-capturing groups (?: ) into capturing groups, unfortunately.
The RegEx constructor as well as other methods from the RegEx class accept RegexOptions flags.
For example:
Regex.Matches(input, pattern, RegexOptions.ExplicitCapture)
answered Nov 12 '18 at 21:44
elgonzoelgonzo
4,49921323
add a comment |
If you want to avoid the clumsiness of (?: ) and turn ( ) groups into non-capturing groups, use the RegexOptions.ExplicitCapture option. Only named groups ((?<name>subexpression)) will be captured if this option is being used.
However, you cannot turn non-capturing groups (?: ) into capturing groups, unfortunately.
The RegEx constructor as well as other methods from the RegEx class accept RegexOptions flags.
For example:
Regex.Matches(input, pattern, RegexOptions.ExplicitCapture)
answered Nov 12 '18 at 21:44
elgonzoelgonzo
4,49921323
add a comment |
If you want to avoid the clumsiness of (?: ) and turn ( ) groups into non-capturing groups, use the RegexOptions.ExplicitCapture option. Only named groups ((?<name>subexpression)) will be captured if this option is being used.
However, you cannot turn non-capturing groups (?: ) into capturing groups, unfortunately.
The RegEx constructor as well as other methods from the RegEx class accept RegexOptions flags.
For example:
Regex.Matches(input, pattern, RegexOptions.ExplicitCapture)
answered Nov 12 '18 at 21:44
elgonzoelgonzo
4,49921323
If you want to avoid the clumsiness of (?: ) and turn ( ) groups into non-capturing groups, use the RegexOptions.ExplicitCapture option. Only named groups ((?<name>subexpression)) will be captured if this option is being used.
However, you cannot turn non-capturing groups (?: ) into capturing groups, unfortunately.
The RegEx constructor as well as other methods from the RegEx class accept RegexOptions flags.
For example:
Regex.Matches(input, pattern, RegexOptions.ExplicitCapture)
answered Nov 12 '18 at 21:44
elgonzoelgonzo
4,49921323
answered Nov 12 '18 at 21:44
elgonzoelgonzo
4,49921323
answered Nov 12 '18 at 21:44
elgonzoelgonzo
4,49921323
answered Nov 12 '18 at 21:44
elgonzoelgonzo
4,49921323
4,49921323
add a comment |
add a comment |
In any language that supports named capture groups you can simply use them for what you want captured, and ignore the numbered ones.
my $string = q(Available from v5.10 in Perl.);
$string =~ /([A-Z].+?)(?<v>[0-9.]+)s+(.*?)./;
say "Version: $+{v}";
After the regex the capture is available in %+ hash, inside the regex in k<name> or g{name}.
The downside is that you still capture all that other stuff (what hurts efficiency a little), while the upside is that you still capture all that other stuff (what helps flexibility, if some of it turns needed).
answered Nov 13 '18 at 1:05
zdimzdim
32.3k32041
add a comment |
In any language that supports named capture groups you can simply use them for what you want captured, and ignore the numbered ones.
my $string = q(Available from v5.10 in Perl.);
$string =~ /([A-Z].+?)(?<v>[0-9.]+)s+(.*?)./;
say "Version: $+{v}";
After the regex the capture is available in %+ hash, inside the regex in k<name> or g{name}.
The downside is that you still capture all that other stuff (what hurts efficiency a little), while the upside is that you still capture all that other stuff (what helps flexibility, if some of it turns needed).
answered Nov 13 '18 at 1:05
zdimzdim
32.3k32041
add a comment |
In any language that supports named capture groups you can simply use them for what you want captured, and ignore the numbered ones.
my $string = q(Available from v5.10 in Perl.);
$string =~ /([A-Z].+?)(?<v>[0-9.]+)s+(.*?)./;
say "Version: $+{v}";
After the regex the capture is available in %+ hash, inside the regex in k<name> or g{name}.
The downside is that you still capture all that other stuff (what hurts efficiency a little), while the upside is that you still capture all that other stuff (what helps flexibility, if some of it turns needed).
answered Nov 13 '18 at 1:05
zdimzdim
32.3k32041
In any language that supports named capture groups you can simply use them for what you want captured, and ignore the numbered ones.
my $string = q(Available from v5.10 in Perl.);
$string =~ /([A-Z].+?)(?<v>[0-9.]+)s+(.*?)./;
say "Version: $+{v}";
After the regex the capture is available in %+ hash, inside the regex in k<name> or g{name}.
The downside is that you still capture all that other stuff (what hurts efficiency a little), while the upside is that you still capture all that other stuff (what helps flexibility, if some of it turns needed).
answered Nov 13 '18 at 1:05
zdimzdim
32.3k32041
edited Nov 13 '18 at 2:20
answered Nov 13 '18 at 1:05
zdimzdim
32.3k32041
answered Nov 13 '18 at 1:05
zdimzdim
32.3k32041
answered Nov 13 '18 at 1:05
zdimzdim
32.3k32041
32.3k32041
add a comment |
add a comment |
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You could perhaps instantiate your RegEx object with
RegexOptions.ExplicitCapture(docs.microsoft.com/en-gb/dotnet/api/…)– elgonzo
Nov 12 '18 at 21:40
@elgonzo That looks good, if you put it as an answer I can accept.
– Jez
Nov 12 '18 at 21:41