Compare int variable after assign in php

Question

I have a simple php code :

<?php
$page = 2;
$max = $page * 10 + 1;
$min = $page - 11;
$counter = 1;

if($counter > $min && $counter < $max) // so, it will be if(1 > 9 && 1 < 20)
{
    echo "true!";
}
else
{
    echo "false!";
}
?>

If I run this code, It always echo true. However, If I change the assign like this :

$max = 21;
$min = 9;

The code work fine and echo false. Where did I wrong?

Edit : The code work fine if value of counter >= 9 :(

P/S : sorry for bad English.

You have to change your $counter to get a false response. Your $min is -9 and your $max is 21. So any value of $counter from -8 to 20 will return true, otherwise false. But with that kind of static values your question makes no sense. — Tom Freudenberg
– Tom Freudenberg, Commented Oct 24, 2015 at 2:11

someOne · Accepted Answer · 2015-10-24 04:01:05Z

1

$page = 2;
$min = $page - 11;

That means $min is -9, not 9.

edited Oct 24, 2015 at 4:01

someOne

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answered Oct 24, 2015 at 2:05

user3277192

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1 Comment

Hai Hoang Over a year ago

Ah, I see it. My bad :D

Jonathan Lam · Accepted Answer · 2015-10-24 02:05:28Z

1

In your code, the line:

$min = $page - 11;

sets $min to -9, not 9 (because $page is 2).

Therefore, $counter > $min && $counter < $max would be true, because 1 > -9 AND 1 < 21.

answered Oct 24, 2015 at 2:05

Jonathan Lam

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1 Comment

Hai Hoang Over a year ago

Ah, I see it. My bad :D

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