Trouble with preg_replace for smilies

[DragonHighLord]

Ok, I'm trying the usual smilies replace function (and I am trying it from scratch so I learn more rather then just getting a code snippet somewhere) and I'm having trouble with it.

My database has the the following fields: id, code, image. The code is the the part they type to get a smilie and the image is just that, the .gif name of the image

Here is the function I have:

function display_smilies($text)
{
global $config;

//get smilies code from db
$query = sql_query( "SELECT * FROM table_smilies" );
while ( $smilies = sql_fetch_array($query) )
{
	$search = $smilies["code"];
	$replace = "<img src=\"/images/smilies/{$smilies["image"]}\">";
	$text = preg_replace($search, $replace, $text);
             echo "<br>code is: $search, pic: $replace";
}

return $text;

}

Now , I put in that echo statement at the end there to see what was going on, the code gets displayed properky and the actual smilie image gets displayed behind it, so all seems to be working ok. However I gets tons of errors pointing to the function file on the line number that contains the preg_replace command

Warning: No ending delimiter ';' found in

Warning: No ending delimiter ':' found in

Warning: Compilation failed: nothing to repeat at offset 0 in

And for those that have code of 8), I get
Warning: Delimiter must not be alphanumeric or backslash in

[DragonHighLord]

anybody have ideas?

[laserlight]

What if you try:

$text = preg_replace('/' . $search . '/', $replace, $text);

my guess is that the pattern doesnt have the pattern delimiters.

[PyroX]

In this case, preg_replace is acting exactly as it should, since it is for regular expressions. It wants to know where the end to the regex starting with ; is.

IMO, you should be using str_replace() instead.

That will fix your problem, and be faster to boot.

[DragonHighLord]

I did the strr_replace instead and all works perfect, thanks