Solution for Regular expression to stop at first match
is Given Below:
My regex pattern looks something like
<xxxx location="file path/level1/level2" xxxx some="xxx">
I am only interested in the part in quotes assigned to location. Shouldn’t it be as easy as below without the greedy switch?
Does not seem to work.
You need to make your regular expression lazy/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:
? on a quantifier (
+) makes it non-greedy.
location="(.*)" will match from the
location= until the
some="xxx unless you make it non-greedy.
So you either need
.*? (i.e. make it non-greedy by adding
?) or better replace
[^"]Matches any character except for a ” <quotation-mark>
- More generic:
[^abc]– Matches any character except for an a, b or c
This avoids the unlimited search with .* and will match exactly to the first quote.
Use non-greedy matching, if your engine supports it. Add the ? inside the capture.
Use of Lazy quantifiers
? with no global flag is the answer.
Here’s another way.
Here’s the one you want. This is lazy
The first item:
[sS]*?(?:location="[^"]*")[sS]* Replace with:
For completeness, this gets the last one. This is greedy
The last item:
There’s only 1 difference between these two regular expressions and that is the
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
*? //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
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
you can match
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
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.
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))