Php Variable as conditional assignment with eval() [duplicate]

Question

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php variable as conditional assignment

I am trying to concatenate a conditional assignment through php variables like this

$cndtnal='&& $x==4';

if($y==5 eval($cndtnal)){
   print 'Hello World';
}

But I am getting a Parse error: syntax error, unexpected $end in : eval()'d code on line 1.

I have tried too :

$cndtnal='&& $x==4';

if (eval('$y==5'.$cndtnal)){
       print 'Hello World';
    }

But I get the same error.

Thanks.

Your last question about this has an answer -- does that answer not work? — user895378
– user895378, Commented Aug 29, 2012 at 4:46

uınbɐɥs · Accepted Answer · 2012-08-29 04:51:20Z

0

You're missing return and a semicolon.

But really, you should avoid eval.

<?php
$foo = 1;
$bar = 2;
$test = ' && $bar == 2';
if(eval('return $foo == 1' . $test . ';')) {
    echo "That horribly eval()'d code worked!";
}
?>

answered Aug 29, 2012 at 4:51

uınbɐɥs

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