PHP Regex, Failure to skip brackets

Question

I want to find a pattern " and b_start in (ANYTHING)" and replace it with an empty string. The following code of mine does not work; i'm afraid it's due to the brackets in the pattern. Can anyone help to point out what's wrong with it? Many thanks.

$a=" and b_start in (2,3,5,6...) and b_end in (4,5,3,4,5...)";
$b=preg_replace("/ and b_start in \(.*\)/", "", $a);

I expect $b to return " and b_end in (4,5,3,4,5...)";

Make the `$.*$` non greedy `$.*?$` – The fourth bird Jul 28 '19 at 11:23 — The fourth bird, Jul 28 '19 at 11:23

score 0 · Accepted Answer · answered Jul 28 '19 at 11:23

You need to make you regex lazy, by default it is greedy in nature so it will match as much as possible Regex Ref

/ and b_start in \(.*?\)/

$a=" and b_start in (2,3,5,6...) and b_end in (4,5,3,4,5...)";
$b=preg_replace("/ and b_start in (.*?)/", "", $a);

echo $b;

PHP Regex, Failure to skip brackets

1 Answers1