If statement inside variables use without eval() method [duplicate] - php

Question

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How to dynamically pass condition of if statement
(2 answers)
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I try to check a condition using php variable like..
if ($amount. $row['condition']. $row['amount']) {
return true;
} else {
return false;
}
but the condition always return true value ,i also echo the single variable it will show the value and also do dd($amount. $rom['condition']. $rom['amount']); it will show "500>1000" ,i remove the double qutes and manually past value in if condition it will show false with this condition, how to solve and write this condition ?

Answer 1

I would use a match or switch like so:
<?php
# match php 8+
$result = match ($row['condition']) {
'==' => $amount == $row['amount'],
'<' => $amount < $row['amount'],
'>' => $amount > $row['amount'],
};
# switch
switch ($row['condition']) {
case '==':
$result = $amount == $row['amount'];
break;
case '<':
$result = $amount < $row['amount'];
break;
case '>':
$result = $amount > $row['amount'];
break;
};
var_dump($result);
Try online

Answer 2

I've never faced such scenario, but I would imagine something like:
if ($row['condition'] == "==") {
if ($amount == $row['amount'])
return true
else
return false
}
if ($row['condition'] == ">") {
if ($amount > $row['amount'])
return true
else
return false
}
if ($row['condition'] == "<") {
if ($amount < $row['amount'])
return true
else
return false
}

Answer 3

Use eval()
$condition = $amount. $row['condition']. $row['amount'];
if (eval("return $condition;")) {
return true;
} else {
return false;
}
Important Note: Before using eval() make sure you have sanitized user input while inserting into database.
Docs for input sanitization : https://www.cloudways.com/blog/prevent-laravel-xss-exploits/