如何在PHP中使用正则表达式删除特定字符之间的字符串？

Question

I have string like below,

$string = "test coontevt [gallery include=\"12,24\"] first [gallery include=\"12,24\"] second";

i need to remove the string starts with [gallery to first ocuurance of it's ] .

i already use this one,

$string12 = preg_replace('/[gallery.+?)+(/])/i', '', $string);

but i get empty string only.

Finally i want result for the above string is,

$string ="test coontevt first second".

How can i do this using regular expression?.

plz help me?

Answer 1

The character [ is a regex meta-character. TO match a literal [ you need to escape it.

$string12 = preg_replace('/\[gallery.+?\]/i', '', $string);

or

$string12 = preg_replace('/\[gallery[^\]]+\]/i', '', $string);

Answer 2

You need to escape the square brackets

$string12 = preg_replace('/\[gallery.+?\]/i', '', $string);

The round brackets are unnecessary so I removed them, also the quantifier between those brackets and the forward slash before the last square bracket.

To avoid multiple space in the result, I would match also the surrounding spaces and replace with 1 space.

\\s+\\[gallery.+?\\]\\s+ and replace with one space

$string12 = preg_replace('/\s+\[gallery.+?\]\s+/i', ' ', $string);

See this expression here online on Regexr

Answer 3

Try it like this:

$string12 = preg_replace('/\[gallery[^\]]+\]/i', '', $string);

[^\\]]+ means that there can be one or more character that is not ] . And there is no need for any ( and ) if you don't want to use the backreferences.