PHP Variable being overwritten in if statement

Question

Below is my code. The posted statement is working & bringing variables through.

However, the if statement $table="1" is overwriting the $table variable with the value 1... so the check is always true.

Can anyone help me fix this?

Thank you!

<?php
    print_r($_POST);


    $table = $_POST['origintable'];
    $database = $_POST['origindatabase'];

    if ($table = "1") { 
        echo "<br> &nbsp;&nbsp;&nbsp;&nbsp; Database name: <b>$database</b> Table name:<b>$table</b>.";
    }
    else {
        echo "Have a good night!";
    }
?>

That is an assignment operation not a comparison. It succeeds in the if because assignment expressions evaluate to their value. So your code check if 1 was a truthy value after assigning it to $table — Mark Fox, Apr 03 '18 at 05:53
That is sort of the expected behavior with the code you have. — ArtisticPhoenix, Apr 03 '18 at 05:54

Rahul · Accepted Answer · 2018-04-03T05:53:10.967

3

you need to use == operator because = is an assignment operator and == is comparison operator or you can use === if you want to check same type with comparison example given below

if ($table == "1") {

example

if (1 == "1") {
    // it will return true
}
if (1 === "1") {
     // it will return false
}

edited Apr 03 '18 at 05:53

answered Apr 03 '18 at 05:48

Rahul

1,617
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score 1 · Answer 2 · answered Apr 03 '18 at 05:47

1

try this, compare == operator

if ($table == "1") { 
    echo "<br> &nbsp;&nbsp;&nbsp;&nbsp; Database name: <b>$database</b> Table name:<b>$table</b>.";
}
else {
    echo "Have a good night!";
}

answered Apr 03 '18 at 05:47

Rp9

1,955
2
23
31

PHP Variable being overwritten in if statement

2 Answers2