How to include a php file in php with condition

Question

While running my code on localhost I had a problem with the include command.

Here's my code:

<?php
 $res = 2; // this also can be change to any number. it is based on user input, but for simply the problem i make it to be set manually
 If ($res =1,){

    $open = include ("weekend.php");
}
else{
    $open = include ("weekday.php");
}
echo $open;
?>

I expected the output to be weekday.php, but the output is weekend.php. It works fine if I use $res = 1.

correct your if condition......... If ($res =1) changes into if ($res == 1) — Inzamam Idrees
– Inzamam Idrees, Commented Jan 1, 2019 at 6:34

Mureinik · Accepted Answer · 2019-01-01 06:34:24Z

2

Besides the obvious typos (the capital I in If, the comma at the end of the condition), you're using the wrong operator. = is the assignment operator. In order to check equality, you should use the == operator:

if ($res == 1) {
    // ---^
    $open = include ("weekend.php");
}
else {
    $open = include ("weekday.php");
}

answered Jan 1, 2019 at 6:34

Mureinik

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