PHP Preg_replace with a * in the pattern

Escape the * with a \ and always do your patterns in single quotes, not double quotes otherwise you wind up escaping the escape characters as well and regexes look bad enough to start with.

$pattern = '3,\*';

Here you go, a better example

<?php

$pattern = '#(3),(\*)#';

$data = "Here is a test 3,* string";

echo  preg_replace( $pattern, '$1,$1', $data );

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produces

Here is a test 3,3 string

Thank you for your help so far.
The code works in isolation but you look at this test
It really does strange things to the string....

Any more help would get a huge :D from me... thanks in advance

<?php
	$test = 'test test ix="2,*" test test ix="3,*" test test ix="4,*" test test';
		
	$i=0;
	$find[$i]='#ix="(.*),(\*)"#';
	$replace[$i]='ix="$1,$1"';
	
	echo "<br>Before:".$test;
	
	$t = preg_replace($find, $replace, $test);
	echo "<br>After:".$t;	

?>

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Oh - it was a HUGE :D not a HUG..... I misread it.

My apologies.....

:-O

Hurray it works

Faith in mankind = RESTORED!

Hugz + points

Glad you're sorted. Thanks for the hugz as well as the points

:-D

One additional note - regex is greedy and will match as much as possible to a pattern like .* but you can tell it not to be greedy by using .*? so in your pattern I could have used

$find[$i]='#ix="(.*?),(\*)"#';

but if that test is always a number then it is safer to be specific which is why I replaced .* with [0-9]*