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Replace with Dynamic variable in preg_replace

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Feb 20.2012

Hi
I’m trying the following code:

[code]$t = ’12<– AB_C –>’;
$AB_C = ‘abc’;
echo preg_replace(‘/<– ([A-Z_]+) –>/’, “$$1”, $t);[/code]

I want to get “12abc” , but it outputs: 12$AB_C , so, it not recognize the replacement as dynamic variable.
Is it any way to use the matched word in preg_replace() as a variable, or dynamic variable?

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PHP

4 Comments(s) ↴

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@hyperionXSFeb 20.2012 — Use replace() instead
[CODE]
 $t = '12<-- AB_C -->';
$AB_C = 'abc';
alert( $t.replace(/<-- AB_C -->/i, $AB_C) )
[/CODE]
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@MarPloauthorFeb 20.2012 — Use replace() instead
[CODE]
 $t = '12<-- AB_C -->';
$AB_C = 'abc';
alert( $t.replace(/<-- AB_C -->/i, $AB_C) )
[/CODE][/QUOTE]
Hi

alert() is not in PHP, your example is wrong.

AB_C is not known, it is got with regexp.
Copy linkTweet thisAlerts:
@hyperionXSFeb 20.2012 — Oh, sorry, I thought it was JavaScript.

For PHP use str_replace
[CODE]
 $t = '12<-- AB_C -->';
$AB_C = 'abc';
echo str_replace('<-- AB_C -->', $AB_C, $t)
[/CODE]
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@MarPloauthorFeb 20.2012 — For those who look for a solution to this problem, the '/e' flag, which evalates the replacement, solved the problem, and returns the results i want, using:
preg_replace('/&lt;-- ([A-Z_]+) --&gt;/e', "$$1", $t);
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