删除两个方括号之间的所有内容

Question

I've been trying to remove some text... between two sets of brackts... for two hours i have tried everything... i've been to existing questions here and the answers don't work for me... so here goes what i have

 [attachment=0:randomID]<!-- ia0 -->randomIMGnam.png<!-- ia0 -->[/attachment:randomID]

i really want to remove all of this from the beginning of a string i was able to remove everything inside the brackets but failed everytime to get rid of the image name

Yes this is from phpbb i've pulled it from my DB no problem but don't want it to be displayed when i echo it.

thanks in advance i really hope I really hope someone can help

edit: what i've tried 1. $extension_pos = strrpos($entry, '<!-- ia0 -->'); // find position of the last dot, so where the extension starts $output = substr($entry, 0, $extension_pos) . '' . substr($entry, $extension_pos); $extension_pos = strrpos($entry, '<!-- ia0 -->'); // find position of the last dot, so where the extension starts $output = substr($entry, 0, $extension_pos) . '' . substr($entry, $extension_pos);

2. $output= preg_replace('#\\].*?\\[#', '', $entry);

1. $output = preg_replace('/\\[[^]]*\\]/', '', $entry);

2. $output explode(']', $entry);

3. $imagename = preg_replace('#([attachment.*?]).*?([/attachment.*?])#', '$1$2', $entry);

Answer 1

You can use this regex to replace:

$string = ' [attachment=0:randomID]<!-- ia0 -->randomIMGnam.png<!-- ia0 -->[/attachment:randomID]';
$string = preg_replace('/\[(.*?)\]/', '', $string);

Answer 2

You could use regular expression as in example:

<?php
    $string = 'test [attachment=0:randomID]randomIMGnam.png[/attachment:randomID] test2 [something]
    test3

    [/something] test4';


    echo preg_replace('#(\[(.*)\](.*)\[/.*\])#Us','',$string);
    // output test test2 test4 


?>

Answer 3

Using regex might be heavy for this kind of task.
You could instead use a simple reasoning, whenever you meet a open bracket increment a counter by one, whenever you meet a close bracket decrement the counter by one.

And as long as your counter is > 0 just ignore the characters.