Question

PHP has an intval() function that will convert a string to an integer. However I want to check that the string is an integer beforehand, so that I can give a helpful error message to the user if it’s wrong. PHP has is_int(), but that returns false for string like "2".

PHP has the is_numeric() function, but that will return true if the number is a double. I want something that will return false for a double, but true for an int.

e.g.:

my_is_int("2") == TRUE
my_is_int("2.1") == FALSE

Answer 1

How about using ctype_digit?

From the manual:

<?php
$strings = array('1820.20', '10002', 'wsl!12');
foreach ($strings as $testcase) {
    if (ctype_digit($testcase)) {
        echo "The string $testcase consists of all digits.\n";
    } else {
        echo "The string $testcase does not consist of all digits.\n";
    }
}
?>

The above example will output:

The string 1820.20 does not consist of all digits.
The string 10002 consists of all digits.
The string wsl!12 does not consist of all digits.

This will only work if your input is always a string:

$numeric_string = '42';
$integer        = 42;

ctype_digit($numeric_string);  // true
ctype_digit($integer);         // false

If your input might be of type int, then combine ctype_digit with is_int.

If you care about negative numbers, then you’ll need to check the input for a preceding -, and if so, call ctype_digit on a substr of the input string. Something like this would do it:

function my_is_int($input) {
  if ($input[0] == '-') {
    return ctype_digit(substr($input, 1));
  }
  return ctype_digit($input);
}