Remove the text outside the brackets

问题:

i.e.

$text = 'remove this text (keep this text and 123)';  echo preg_replace('', '', $text); 

It should output:

(keep this text and 123) 

回答1:

Take anything found within the brackets, put it in a capture group and keep that only, like this:

echo preg_replace('/^.*(\(.*\)).*$/', '$1', $text); 

回答2:

This will do it: (and works with nested () as well)

$re = '/[^()]*+(\((?:[^()]++|(?1))*\))[^()]*+/'; $text = preg_replace($re, '$1', $text); 

Here are a couple test cases:

Input: Non-nested case: 'remove1 (keep1) remove2 (keep2) remove3' Nested case:     'remove1 ((keep1) keep2 (keep3)) remove2'  Output: Non-nested case: '(keep1)(keep2)' Nested case:     '(keep1) keep2 (keep3)' 

回答3:

Here the 'non preg_replace' way:

<?  $text = 'remove this text (keep this text)' ;  $start = strpos($text,"(") ;  $end = strpos($text,")") ;   echo substr($text,$start+1,$end-$start-1) ; // without brackets echo substr($text,$start,$end-$start+1) ; // brackets included  ?> 

Note:
- This extracts only the first pair of brackets.
- Replace strpos() with of strrpos() to get the last pair of brackets.
- Nested brackets cause trouble.

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