asked on

PHP Preg_replace with a * in the pattern

Hello experts.

I have a small bit of code...

ix="3,*"
and i am trying to do a simple preg_replace so that it becomes
ix="3,3"

Basically whatever the first item is i want it to replace the star.
I think the star is causing me issues as i cant get anything to work.

Any help would be gratefully accepted ...

thanks

Beverley Portlock

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,\*';

Beverley Portlock

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

handypam

ASKER

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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ASKER CERTIFIED SOLUTION

Beverley Portlock

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Beverley Portlock

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

My apologies.....

:-O

handypam

ASKER

Hurray it works

Faith in mankind = RESTORED!

Hugz + points

Beverley Portlock

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]*