Problem with function call inside preg_replace

Question

Does not work, the $1-value is lost when calling the function:

echo preg_replace('"\b(http://\S+)"', '<a href="$1">'.findTopDomain('$1').'</a>', $text);

Works fine, outputs: stackoverflow.com

echo preg_replace('"\b(http://\S+)"', '<a href="$1">'.findTopDomain('http://stackoverflow.com/questions/ask').'</a>' , $text);

I need to send the $1 value to a function from within preg_replace. What am I doing wrong?

Gumbo · Accepted Answer · 2009-12-20 16:41:51Z

2

You need to set the e modifier to have the substitution expression to be executed:

preg_replace('"\b(http://\S+)"e', '"<a href=\\"$1\\">".findTopDomain("$1")."</a>"', $text)

Note that your substitution now has to be a valid PHP expression. In this case the expression would be evaluated to:

"<a href=\"$1\">".findTopDomain("$1")."</a>"

And don’t forget to escape the output with at least htmlspecialchars:

preg_replace('"\b(http://\S+)"e', '"<a href=\\"".htmlspecialchars("$1")."\\">".htmlspecialchars(findTopDomain("$1"))."</a>"', $text)

answered Dec 20, 2009 at 16:31

Gumbo

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Comments

Pekka · Accepted Answer · 2009-12-20 16:31:53Z

2

Perform a regular expression search and replace using a callback

answered Dec 20, 2009 at 16:31

Pekka

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