Pass variable as condition for if statement in php

Question

I have dynamic condition for if statement that stored as a variable. How to make if statement worked using that variable as as its condition?

I have tried using eval function but it's still didn't work

Here is my code:

$a = 2;
$b = 2;
$c = '$a==$b';
$d = eval('return $c;');

if($d === true){
    echo 'yes';
}
else{
    echo 'no';
}

It should be return 'yes', but my current code return 'no'.

score 1 · Accepted Answer · answered May 26 '19 at 05:21

1

You just need to use double quotes instead of single quote

<?php
$a = 2;
$b = 2;
$c = "$a==$b";
$d = eval("return $c;");

if($d === true){
    echo 'yes';
}
else{
    echo 'no';
}
?>

answered May 26 '19 at 05:21

vivek modi

Thank you, I never thought that using single or double quotes will affect much in eval function. – Kevin A.S. May 26 '19 at 05:24
Always remember single quote can not understand variable and double quote can understand the variable – vivek modi May 26 '19 at 05:31
And please check the correct answer tick – vivek modi May 26 '19 at 05:32

score 0 · Answer 2 · answered May 26 '19 at 05:20

0

After some experiments, I have found this solution. So what I need to do is put the eval function inside if condition:

<?php
$a = 2;
$b = 2;
$c = '$a==$b';

if(eval("return $c;")){ //must using double quotes
    echo 'yes';
}
else{
    echo 'no';
}

answered May 26 '19 at 05:20

Kevin A.S.

2 Answers2