How to assign conditional answer into variable in eval() in PHP?

Question

we execute dynamic string IF condition in PHP eval() function but the answer is not assign in variable.

Please look below code :

$variable_1 = "if(3>4)";
$variable_1 = "1";
$variable_1 = "0";

$if_val = "$variable_1{ $return_if = $variable_2 } else { $return_if = $variable_3 }";
$final_if_ans = eval($if_val);

Aside from your 3 variables having the same name, you can print out the resulting code by printing $if_val to see what you're actually running. $return_if is not defined either. — Qirel
– Qirel, Commented Aug 8, 2018 at 5:08
You should be really careful with using eval(), it can be a security risk and should be avoided at all costs! — Qirel
– Qirel, Commented Aug 8, 2018 at 5:10
In $if_val, $return_if is not defined and also you've missed ; in each condition which will also throw error. — Abhay Padda
– Abhay Padda, Commented Aug 8, 2018 at 5:16

Abhay Padda · Accepted Answer · 2018-08-08 05:12:28Z

3

Use the following code to get the desired result:

$variable_1 = "if(3>4)";
$variable_2 = "1";
$variable_3 = "0";

$if_val = "$variable_1{ return $variable_2; } else { return $variable_3; }";
$final_if_ans = eval($if_val); 
echo $final_if_ans;

Important: The eval() language construct is very dangerous because it allows execution of arbitrary PHP code. Its use thus is discouraged. If you have carefully verified that there is no other option than to use this construct, pay special attention not to pass any user provided data into it without properly validating it beforehand.

answered Aug 8, 2018 at 5:12

Abhay Padda

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Comments

Qirel · Accepted Answer · 2018-08-08 05:02:08Z

0

You declared all variables as $variable_1. Change to $variable_2 and $variable_3

Change this

$variable_1 = "if(3>4)";
$variable_1 = "1";
$variable_1 = "0";

to

$variable_1 = "if(3>4)";
$variable_2 = "1";
$variable_3 = "0";

edited Aug 8, 2018 at 5:02

Qirel

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answered Aug 8, 2018 at 4:56

Geomi George

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