php integer

[Revan]

How can I check if an integer is an even number?
I want to have a submit button and for each of the users that clicks it, it adds their usernames to a field in a table.
However, I cannot have the # of users to be an uneven number (IE 1, 3, 5, 7 etc).
So there is my problem.

Anyone?

[Tekton]

I'm sure there are prolly cleaner ways to write this, but....

if($value % 2 == 0) 
  { 
    // it's even
  }else{ 
    $value += 1;  // or whatever 
  }

[Colin F]

you could do something like this... if you don't find a better way

if (($number/2) == (intval(($number+1)/2))
{
//is even number
}
else
{
//is uneven number
}

[Revan]

Cheers people
I tried both of them first with a single figured number, and they both worked fine.
I then tried with 20000000000000000000 just for the hell of it. Colins code said this was uneven, while Tektons said even.
Then I tried with 20000000000000000001 which reversedly Colins code called uneven, while Tektons called it even.


Now I dont need such a high number, just 100 or so. And they both say thats even XD
Thanks again guys

[filburt1]

Originally Posted by Revan

Cheers people
I tried both of them first with a single figured number, and they both worked fine.
I then tried with 20000000000000000000 just for the hell of it. Colins code said this was uneven, while Tektons said even.
Then I tried with 20000000000000000001 which reversedly Colins code called uneven, while Tektons called it even.


Now I dont need such a high number, just 100 or so. And they both say thats even XD
Thanks again guys

Tekton's method is by far more preferable (modulus division).

PHP Code:

if ($number % 2 == 0) 


...evaulates to true if the number is evenly divisible by 2 (i.e., even).

[Dean C]

Also it's always nice to write a function for these things

PHP Code:

function is_even($num)
{
        if($number % 2 == 0)
        {
                return true;
        }
        else
        {
                return false;
        }
} 


Then to check if it's even use this:

PHP Code:

$number = 2;
if(is_even($number))
{
       echo 'it\'s even!';
}
else
{
        echo 'it\'s not even!';
} 


[Revan]

Added the fucntion for future usage
But I feel pretty sure that the arguement in the function ($num in above) and the function coding ($number in above) should be equal hehe

[Dean C]

Yep you're write, I've been programming all day so that's my excuse