Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

preg_replace how to replace only matching xxx($1)yyy pattern inside the selector

I'm trying to use a regular expression to erase only the matching part of an string. I'm using the preg_replace function and have tried to delete the matching text by putting parentheses around the matching portion. Example:

preg_replace('/text1(text2)text3/is','',$html);

This replaces the entire string with '' though. I only want to erase text2, but leave text1 and text3 intact. How can I match and replace just the part of the string that matches?

like image 204
PartySoft Avatar asked Feb 04 '11 12:02

PartySoft


3 Answers

Use backreferences (i.e. brackets) to keep only the parts of the expression that you want to remember. You can recall the contents in the replacement string by using $1, $2, etc.:

preg_replace('/(text1)text2(text3)/is','$1$2',$html);
like image 116
Mansoor Siddiqui Avatar answered Oct 01 '22 22:10

Mansoor Siddiqui


There is an alternative to using text1 and text3 in the match pattern and then putting them back in via the replacement string. You can use assertions like this:

preg_replace('/(?<=text1)(text2)(?=text3)/', "", $txt);

This way the regular expression looks just for the presence, but does not take the two strings into account when applying the replacement.

http://www.regular-expressions.info/lookaround.html for more information.

like image 39
mario Avatar answered Oct 02 '22 00:10

mario


Try this:

$text = preg_replace("'(text1)text2(text3)'is", "$1$2", $text);

Hope it works!

Edit: changed \\1\\2 to $1$2 which is the recommended way.

like image 25
aorcsik Avatar answered Oct 02 '22 00:10

aorcsik