preg_replace regex not working as expected

Question

So i'm trying to do the following and it's not working. not sure why:

$block = preg_replace('#\{([A-Z0-9\-_]+)\}#', "<?php " . $this->compile_var(\\1) . " ?>", $block);

What i'm using that for is to turn this input:

<link href="{VAR_TEMPLATE_PATH}css/common.css" rel="stylesheet" type="text/css">

into:

 <link href="<?php $this->page_vars["TEMPLATE_PATH"] ?>css/common.css" rel="stylesheet" type="text/css">

The actual conversion is fine, thats all contained in compile_var, but preg_replace is no longer turning \\1 into "VAR_TEMPLATE_PATH", which is used to do when inside a string.

Instead it is passing "\1" as the argument for compile_var?

Why is this happening all of a sudden? and how can i fix it?

Cheers!

Try \1 instead of \\1?

brandonscript
– brandonscript

2013-12-03 06:11:27 +00:00
Commented Dec 3, 2013 at 6:11 — brandonscript
– brandonscript, Commented Dec 3, 2013 at 6:11

anubhava · Accepted Answer · 2013-12-03 07:19:35Z

1

Use it like this:

$block = '<link href="{VAR_TEMPLATE_PATH}css/common.css" rel="stylesheet" type="text/css">';
$block=preg_replace('#\{(?:VAR_)?([\w-]+)\}#', '<?php $this->compile_var("\1") ?>', $block);
echo $block;

OUTPUT:

<link href="<?php $this->compile_var("TEMPLATE_PATH") ?>css/common.css" rel="stylesheet" type="text/css">

If you want VAR_ also in final output then use:

$block = preg_replace('#\{([\w-]+)\}#', '<?php $this->compile_var("\1") ?>', $block);

answered Dec 3, 2013 at 7:19

anubhava

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Comments

brandonscript · Accepted Answer · 2013-12-03 06:13:13Z

0

Try using \1 in compile_var instead.

<?php " . $this->compile_var(\1) . " ?>

Here's a working example: http://regex101.com/r/rM0jD1

answered Dec 3, 2013 at 6:13

brandonscript

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1 Comment

Dylan Lundy Over a year ago

That just throws a syntax error: Parse error: syntax error, unexpected '\' (T_NS_SEPARATOR), expecting identifier (T_STRING) in

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