What Does The "?:^" Regular Expression Mean?
Solution 1:
When working with groups, you often have several options that modify the behavior of the group:
(foo) // default behavior, matches "foo" and stores a back-reference
(?:foo) // non-capturing group: matches "foo", but doesn't store a back-ref
(?i:foo) // matches "foo" case-insensitively
(?=foo) // matches "foo", but does not advance the current position// ("positive zero-width look-ahead assertion")
(?!foo) // matches anything but "foo", and does not advance the position // ("negative zero-width look-ahead assertion")
to name a few.
They all begin with "?", which is the way to indicate a group modifier. The question mark has nothing to do with optionality in this case.
It simply says:
(?:^foo) // match "foo" at the start of the line, but do not store a back-ref
Sometimes it's just overkill to store a back-reference to some part of the match that you are not going to use anyway. When the group is there only to make a complex expression atomic (e.g. it should either match or fail as a whole), storing a back-reference is an unnecessary waste of resources that can even slow down the regex a bit. And sometimes, you just want to be group 1 the first group relevant to you, instead of the first group in the regex.
Solution 2:
You're probably seeing it in this context
(?:...)
It means that the group won't be captured or used for back-references.
EDIT: To reflect your modified question:
(?:^|....)
means "match the beginning of the line or match ..." but don't capture the group or use it for back-references.
Solution 3:
(?:some stuff)
means that you don't want to match the expression in the parentheses separately. Normally the pieces of a regexp grouped in parentheses are grouped and can be referenced individually (this is called using backreferences).
Solution 4:
Short Answer
It flags the (parenthetical) group as a non-capturing group.
Details About This Particular Expression
The notation for a non-capturing group is:
(?:<expresson>)
In the instance you presented, the caret (^) is part of the expression not part of the capturing group notation. And this instance it's not a special character either.
It looks like they're using an 'or' operator (the pipe) with the caret. So they're looking to match something that is a caret or whatever was on the right of the pipe, but not capture the expression as a group (accomplished with the ?:
in the beginning of the grouping characters.
In General
Non-capturing groups allow you to group an expression in a way that won't be back-refernceable, and will also increase performance of the expression.
Solution 5:
"(?:x) Matches 'x' but does not remember the match."
https://developer.mozilla.org/en/Core_JavaScript_1.5_Guide/Regular_Expressions
Post a Comment for "What Does The "?:^" Regular Expression Mean?"