Remove part of string in PHP

Question

How can i replace unused section in content. For example:

$content = "
    Test content

    [section]
        Section Content
    [section]

    End.
";

$content = preg_replace('/\[section\](.*)\[section\]/', '', $content);

Thanks

add the s modifier to match newlines with .. Also don't forget to make your expression ungreedy /\[section\](.*?)\[section\]/s — HamZa
– HamZa, Commented Nov 23, 2013 at 22:10

hwnd · Accepted Answer · 2013-11-24 00:03:55Z

3

The s modifier allows a dot metacharacter in the pattern to match all characters, including newlines.

The i modifier is used for case-insensitive matching allowing both upper and lower case letters. By adding the quantifier ? it makes for a non greedy match and matches the least amount possible.

$content = preg_replace('/\[section\].*?\[section\]/si', '', $content);

See Demo

edited Nov 24, 2013 at 0:03

answered Nov 23, 2013 at 22:19

hwnd

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1 Comment

hwnd Over a year ago

@adoweb If this worked, its nice to accept an answer and close the question.

cronoklee · Accepted Answer · 2013-11-24 00:08:24Z

2

What's the regex for? This should work and will run a lot quicker:

$content = explode('[section]',$content);
$content = $content[0].$content[2];

answered Nov 24, 2013 at 0:08

cronoklee

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