regex - Regular expression to stop at first match

Solution:

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:

/location="(.*?)"/
Copy code

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

Note: this is only available in regex engines which implement the Perl 5 extensions (Java, Ruby, Python, etc) but not in "traditional" regex engines (including JavaScript, Awk,sed,grep without-P, etc.).

Solution:

location="(.*)" will match from the" afterlocation= until the" aftersome="xxx unless you make it non-greedy.

So you either need.*? (i.e. make it non-greedy by adding?) or better replace.* with[^"]*.

• [^"] Matches any character except for a " <quotation-mark>
• More generic:[^abc] - Matches any character except for an a, b or c

Solution:

How about

.*location="([^"]*)".*
Copy code

This avoids the unlimited search with .* and will match exactly to the first quote.

Solution:

Use non-greedy matching, if your engine supports it. Add the ? inside the capture.

/location="(.*?)"/
Copy code

Solution:

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.

Solution:

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?

Solution:

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
Copy code

Thus, to allow your quantified pattern to make minimum match, follow it by? :

/location="(.*?)"/
Copy code

Solution:

The other answers here fail to spell out a full solution for regex versions which don't support non-greedy matching. The greedy quantifiers ({-code-1},{-code-2} 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({-code-1})b
Copy code

you can match

{-code-4}
Copy code

i.e specify a character class which excludes the {-code-6}ing and {-code-7}ing delimiiters.

In the more general case, you can painstakingly construct an expression like

{-code-5}
Copy code

to capture a match between{-code-6} and the first occurrence of{-code-7}. Notice how the subexpression with nested parentheses spells out a number of alternatives which between them allowe only if it isn't followed bynd 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.

Solution:

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))
Copy code

Output: Mary