Regular expression to stop at first match

Question

My regex pattern looks something like   lt xxxx location  file path level1 level2  xxxx some  xxx  gt    I am only interested in the part in quotes assigned to location  Shouldn t it be as easy as below without the greedy switch       location             Does not seem to work

User · Accepted Answer

You need to make your regular expression non-greedy, because by default, "(.*)" will match all of "file path/level1/level2" xxx some="xxx".

Instead you can make your dot-star non-greedy, which will make it match as few characters as possible:

/location="(.*?)"/

Adding a ? on a quantifier (?, * or +) makes it non-greedy.

User · Answer

Use of Lazy quantifiers   with no global flag is the answer    Eg      If you had global flag  g then  it would have matched all the lowest length matches as below

User · Answer

location        will match from the   after location  until the   after some  xxx unless you make it non-greedy  So you either need      i e  make it non-greedy  or better replace    with

User · Answer

Because you are using quantified subpattern and as descried in Perl Doc       By default  a quantified subpattern is  greedy   that is  it will   match as many times as possible  given a particular starting location    while still allowing the rest of the pattern to match  If you want it   to match the minimum number of times possible  follow the quantifier   with a       Note that the meanings don t change  just the    greediness                 Match 0 or more times  not greedily  minimum matches              Match 1 or more times  not greedily   Thus  to allow your quantified pattern to make minimum match  follow it by       location

User · Answer

import regex text    ask her to call Mary back when she comes back                             p   r   i   s call     back  for match in regex finditer p  str text        print  match group 1    Output  Mary

User · Answer

Use non-greedy matching  if your engine supports it  Add the   inside the capture    location

User · Answer

The other answers here fail to spell out a full solution for regex versions which don t support non-greedy matching  The greedy quantifiers           etc  are a Perl 5 extension which isn t supported in traditional regular expressions  If your stopping condition is a single character  the solution is easy  instead of a     b  you can match a  ab  b  i e specify a character class which excludes the starting and ending delimiiters  In the more general case  you can painstakingly construct an expression like start    e  e    n  n    d    end  to capture a match between start and the first occurrence of end  Notice how the subexpression with nested parentheses spells out a number of alternatives which between them allow e only if it isn t followed  by nd and so forth  and also take care to cover the empty string as one alternative which doesn t match whatever is disallowed at that particular point  Of course  the correct approach in most cases is to use a proper parser for the format you are trying to parse  but sometimes  maybe one isn t available  or maybe the specialized tool you are using is insisting on a regular expression and nothing else

User · Answer

How about    location               This avoids the unlimited search with    and will match exactly to the first quote

User · Answer

Here s another way   Here s the one you want  This is lazy   s S     The first item    s S      location           s S   Replace with   1  Explaination  https   regex101 com r ZcqcUm 2    For completeness  this gets the last one  This is greedy   s S    The last item   s S     location             s S    Replace with   1  Explaination  https   regex101 com r LXSPDp 3    There s only 1 difference between these two regular expressions and that is the

[regex] Regular expression to stop at first match

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