RegEx: Match nth occurrence

Question

I have the following string:

_name=aVlTcWRjVG1YeDhucWdEbVFrN3pSOHZ5QTRjOEJZZmZUZXNIYW1PV2RGOWYrczBhVWRmdVJTMUxYazVBOE8zQ3JNMmNVKzJLM2JJTzFON3FiLzFHUE0xY0pkdz09LS1jbkkwaWoxUUl3YVhMMkhtZHpaOW13PT0"%"3D--57356371d167f"

I want to match everything between = and the end " (note there are other quotes after this so I can't just select the last " ) .

I tried using _name=(.*?)" but there are other quotes in the string as well. Is there a way to match the 3rd quote? I tried _name=(.*?)"{3} but the {3} matches for the quotes back to back, ie """

You can try it here

Answer 1

You can use this regex:

\b_name=(?:[^"]*"){3}

RegEx Demo

RegEx Details:

• \\b_name : Match full word _name :
• = : Match a =
• (?:[^"]*"){3} : Match 0 or more non- " characters followed by a " . Repeat this group 3 times.

Answer 2

If want to match everything between the first and the third(!) double quote (the third isn't necessarily the last, you told), you can use a pattern like this:

$string = '_name=foo"bar"test" more text"';
// This pattern will not include the last " (note the 2, not 3)
$pattern = '/_name=((.*?"){2}.*?)"/';

preg_match($pattern, $string, $m);
echo $m[1];

Output:

foo"bar"test

---

Original answer:

I'm not sure if I got you correctly, but it sounds like you want to perform a so called greedy match , meaning you want to match the string until the last " regardless whether the string contains multiple " s.

To perform a greedy match, just drop the ? , like this:

_name=(.*)"

You can try it here: https://regex101.com/r/uC5eO9/2