PHP create_function returns false

Question

Situation:

This works:

$functionCode = 'return ($myvar1 == "something") && ($myvar2 != "");';
$newfunc = create_function($functionParamsConcat, $functionCode);

But the problem is that the funcitonCode is dynamic, and the boolean expression inside it can vary (it is taken from the database). The confussing part is that that when i retrieve the SAME functionCode from the database and assign it to the variable, the $newfunc returned by create_function is always false.

Any ideas? Thank you

Are you sure a quote isn't being escaped? More importantly, why would you build functions from the database? If someone got the chance to SQL inject you, they could do some serious damage. — Jeremy Harris, Oct 28 '13 at 18:37
I tryied using PHP built in function addslashes to escape the whole functionCode, failing, $newfunc is always false. (forget about sqlinjection in this case though I appreciate your concern) — mpinvidio, Oct 28 '13 at 18:47

score 0 · Answer 1 · answered Nov 18 '13 at 16:13

0

$functionCode = 'return ($myvar1 == "something") && ($myvar2 != "");'; $newfunc = create_function($functionParamsConcat, $functionCode);

in checking don't use = as it will set it, but use == or ===
for ! use !==

answered Nov 18 '13 at 16:13

Mohammed Al Ashaal

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What do you mean? I don't see the difference from your code to mine. Thanks. – mpinvidio Nov 24 '13 at 14:09
i said don't use != but use !== – Mohammed Al Ashaal Dec 06 '13 at 08:38

PHP create_function returns false

1 Answers1