PHP check if is integer

Try-

if ((string)(int) $calc === (string)$calc) {
  //it is an integer
}else{
  //it is a float
}

Demo

As CodeBird pointed out in a comment to the question, floating points can exhibit unexpected behaviour due to precision "errors".

e.g.

<?php
$x = 1.4-0.5;
$z = 0.9;

echo $x, ' ', $z, ' ', $x==$z ? 'yes':'no';

prints on my machine (win8, x64 but 32bit build of php)

0.9 0.9 no

took a while to find a (hopefully correct) example that is a) relevant to this question and b) obvious (I think x / y * y is obvious enough).

again this was tested on a 32bit build on a 64bit windows 8

<?php
$y = 0.01; // some mambojambo here... 
for($i=1; $i<31; $i++) { // ... because ...
    $y += 0.01; // ... just writing ...
} // ... $y = 0.31000 didn't work

$x = 5.0 / $y;
$x *= $y;

echo 'x=', $x, "\r\n";
var_dump((int)$x==$x);

and the output is

x=5
bool(false)

Depending on what you're trying to achieve it might be necessary to check if the value is within a certain range of an integer (or it might be just a marginalia on the other side of the spectrum ;-) ), e.g.

function is_intval($x, $epsilon = 0.00001) {
    $x = abs($x - round($x));
    return $x < $epsilon;
};

and you might also take a look at some arbitrary precision library, e.g. the bcmath extension where you can set "the scale of precision".

You can do it using ((int) $var == $var)

$var = 9;
echo ((int) $var == $var) ? 'true' : 'false';
//Will print true;
$var = 9.6;
echo ((int) $var == $var) ? 'true' : 'false';
//Will print false;

Basically you check if the int value of $var equal to $var

round() will return a float. This is because you can set the number of decimals.

You could use a regex:

if(preg_match('~^[0-9]+$~', $calc))

PHP will convert $calc automatically into a string when passing it to preg_match().

You can use number_format() to convert number into correct format and then work like this

$count = (float)(44.28);

$multiple = (float)(0.36);

$calc = $count / $multiple;
//$calc = 44.28 / 0.36 = 123

$calc = number_format($calc, 2, '.', '');

if(($calc) == round($calc))
die("is integer");
else
die("is not integer");

Demo

Ok I guess I'am pretty late to the party but this is a alternative using fmod() which is a modulo operation. I simply store the fraction after the calculation of 2 variables and check if they are > 0 which would imply it is a float.

<?php

  class booHoo{
     public function __construct($numberUno, $numberDos) {
        $this->numberUno= $numberUno;
        $this->numberDos= $numberDos;
     }

     public function compare() {
       $fraction = fmod($this->numberUno, $this->numberDos);
       if($fraction > 0) {
         echo 'is floating point';
       } else {
         echo 'is Integer';
       }
     }
   }

$check= new booHoo(5, 0.26);
$check->compare();

Eval here

Edit: Reminder Fmod will use a division to compare numbers the whole documentation can be found here

if (empty($calc - (int)$calc))
{
    return true; // is int
}else{
    return false; // is no int
}

Try this:

//$calc = 123;
$calc = 123.110;
if(ceil($calc) == $calc)
{
    die("is integer");
}
else
{
    die("is float");
}

you may use the is_int() function at the place of round() function.

if(is_int($calc)) { 
die('is integer');
} else {
die('is float);
}

I think it would help you

A more unorthodox way of checking if a float is also an integer:

// ctype returns bool from a string and that is why use strval
$result = ctype_digit(strval($float));