PHP preg_match & preg_replace outputing wrong

Question

<?php
$string = "[img image:left]1.jpg[/img]Example Text 1[img image:left]2.jpg[/img] Example Text 2";
preg_match("/\[img\s*[^>]+\s*\](.*?)\[\/\s*img\]/i", $string, $match);
$result = preg_replace("/\[img\s*[^>]+\s*\](.*?)\[\/\s*img\]/i", $match['1'], $string);
echo $result;
?>

When using this code it should output 1.jpg, Example Text 1, 2.jpg, Example Text 2.

But however it shows only 2.jpg, Example Text 2.

I dont know what i'm doing wrong.

Thanks, i'll rephrase it the best i can, i'm new to coding didnt knew how to correctly write my question — Carlos Santos
– Carlos Santos, Commented Jun 2, 2014 at 22:34
@CarlosSantos: Your question was perfectly understandable, I was only speaking about the format. — Casimir et Hippolyte
– Casimir et Hippolyte, Commented Jun 2, 2014 at 23:27

Sam · Accepted Answer · 2014-06-02 22:27:09Z

2

There are two fundamental issues:

you don't need to use a preg_match() and a preg_replace(), you can just use preg_replace() and reference your capture groups in the substitution
it looks like you copy pasted some code from HTML regex, and have [^>]+ inside of your [img], which says 1+ non-> characters..it should really be [^\]]+, 1+ non-] characters

Final solution:

$string = "[img image:left]1.jpg[/img]Example Text 1[img image:left]2.jpg[/img] Example Text 2";
$string = preg_replace("/\[img\s*[^\]]+\s*\](.*?)\[\/\s*img\]/i", ' \1 ', $string);

Demo: RegEx and PHP

answered Jun 2, 2014 at 22:27

Sam

20.5k2 gold badges48 silver badges71 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

PHP preg_match & preg_replace outputing wrong

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related