replace a number random form a string in PHP

Question

I have this code

<?php

$source[]="clock=1,time=1,stamp=3,color=33";
$source[]="clock=2,time=1,stamp=1,color=61";

$label="clock";

$what=$label."=(\d)";
$this="clock=0";

for($i=0; $i<3; $i++)
{
$new_source=preg_replace( $what,$this,$source[$i],$count);
echo $new_source;
};

?>

I need to replace $label=1; or $label=x with $label=0 but the x is variable.

From the look of it, it's actually replacing clock=1 with clock=clock=0 — xzyfer
– xzyfer, Commented Mar 24, 2011 at 13:01
No, it is not. It is matching clock=n and replacing it with clock=0 — Richard Parnaby-King
– Richard Parnaby-King, Commented Mar 24, 2011 at 13:03

Richard Parnaby-King · Accepted Answer · 2011-03-24 13:02:13Z

1

First off the bat, don't use $this. Rename it to, for example, $replacement. $this is a reserved variable in PHP5.

Next, you don't need $count in pre_replace.

Finally, wrap $what in a delimiter (such as /).

$new_source=preg_replace( '/'.$what.'/',$replacement,$source[$i]);

answered Mar 24, 2011 at 13:02

Richard Parnaby-King

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1 Comment

Bogdan C. Over a year ago

Thanx I'll try this right away. PS: $this was just for example :) i don't use it

Imi Borbas · Accepted Answer · 2011-03-24 13:03:34Z

0

<?php

$source[]="clock=1,time=1,stamp=3,color=33";
$source[]="clock=2,time=1,stamp=1,color=61";

$label="clock";

$what="/$label=(\d)/";
$to="clock=0";

for($i=0; $i<3; $i++)
{
$new_source=preg_replace( $what,$to,$source[$i]);
echo $new_source;
};

This is now correct. The problems were:

You used a variable named $this to store the replace target string, which is incorrect because PHP refers to the current object with the variable $this in an object-oriented context.
You passed a parameter named $count to preg_replace which had no value.

answered Mar 24, 2011 at 13:03

Imi Borbas

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