PHP preg_replace without eval

Posted by Alec Smart on Stack Overflow See other posts from Stack Overflow or by Alec Smart
Published on 2010-06-03T11:21:54Z Indexed on 2010/06/03 11:24 UTC
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Am trying to use something like:

$newdata = preg_replace($pattern, $replacement, $data);

Now my replacement is something like

$pattern = "/START(.*?)END/is";
$replacement = "START $config END";

Now, $config contains contents like

array('Test\\\'s Page')

The problem is that after I write the content, $newdata becomes

START array('Test\\'s Page') END

As you see above a single \ goes missing because it gets evaluated. How do I avoid that?

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