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[code]
$string = preg_replace('#\[\[(.+?)\|(.+?)\]\]#is', '<a href="$1" title="$2">$2</a>', $string);
[/code]

[[http://google.com/|google.com]]

will output:

<a href="http://google.com/" title="google.com">google.com</a>

In order to use other bbcode or images inplace for the link I need this:

[code]
$string = preg_replace('#\[\[(.+?)\|(.+?)\|(.+?)\]\]#is', '<a href="$1" title="$3">$2</a>', $string);
[/code]

But how can I write it so it won't need both $2 and $3 to spit out the url. Like if $2 and $3 isn't there, it would just use $1 for as a replacement? Or do I have to write 2 regex patterns?
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