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Non-greedy capturing parenthesis

By : user2174534
Date : October 17 2020, 08:10 PM
like below fixes the issue All of your quantifiers are greedy; you need to add ? to make them non-greedy. In this specific case, you need to be careful, because if you don't ensure it must match the GET query separately, non-greediness will omit the b in db too. There are two decent options here:
Explicitly non-greedy: /^mysql:\/\/(.+):(.+)@(.+)\/(.+?)(?:\?.*)?$/ (You need to group the ? with the rest of GET query; if it's optional by itself, the non-greedy code will stop early, ignore the optional ?, and just shove everything into the match on the greedy .*) Greedy, but excluding ? from the things it's willing to match: /^mysql:\/\/(.+):(.+)@(.+)\/([^?]+)(?:\?.*)?$/ Since a ? can't occur in a legal URL except when splitting the GET query, we swap from .+ to [^?]+ to keep everything until the ?
code :

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Greedy not capturing the first event

By : coder
Date : March 29 2020, 07:55 AM
Does that help greedy effects parent-child relationships, not sibling relationships.
See This Fiddle For Example
code :

Capturing group is too greedy

By : Rudi Harsono
Date : March 29 2020, 07:55 AM
it fixes the issue Keep in mind that a regex engine will try all that is possible to make the pattern succeed. Since you use several .*? in your pattern, you let a lot of flexibility to the regex engine to pursue this purpose. The pattern must be more binding.
To do that, you can replace all the .*? with [^>]*
code :
(<ac:image[^>]*> \s* <ri:attachment[^>]*> )     # group 1
 \s* <ri:page[^>]*/> \s*                        # what you need to remove
(</ri:attachment> \s* </ac:image>)              # group 2

Non-capturing group is not greedy?

By : Junio de Souza
Date : March 29 2020, 07:55 AM
Hope that helps I want to remove everything after the first –
code :
(.+?) (?:– .+)

Java regexp - adding parenthesis makes it greedy?

By : Nam Phuong
Date : March 29 2020, 07:55 AM
this will help In order for a quantifier (?, *, + or {n,m}) to be reluctant (non-greedy or lazy), it must by followed by a ?. Therefore, the pattern \\d{1,2}? is reluctant.
On the other hand, (\\d{1,2})? is composed of two levels of greedy quantifiers:

Non greedy regex within parenthesis and contains text

By : Mailley
Date : March 29 2020, 07:55 AM
around this issue Instead of allowing all characters, you can allow all characters except the closed parenthesis by using [^\)] where the . is now.
code :
re.findall(r'\(([^\)]*meh[^\)]*?)\)', test)
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