PHP validating integers

Question

I was wondering, what would be the best way to validate an integer. I'd like this to work with strings as well, so I could to something like

(string)+00003 -> (int)3 (valid)

(string)-027 -> (int)-27 (valid)

(int)33 -> (int)33 (valid)

(string)'33a' -> (FALSE) (invalid)

That is what i've go so far:

function parseInt($int){
    //If $int already is integer, return it
    if(is_int($int)){return $int;}
//If not, convert it to string
$int=(string)$int;
//If we have '+' or '-' at the beginning of the string, remove them
$validate = ($int[0] === '-' || $int[0] === '+')?substr($int, 1):$int;
//If $validate matches pattern 0-9 convert $int to integer and return it
//otherwise return false
return preg_match('/^[0-9]+$/', $validate)?(int)$int:FALSE;
}

As far as I tested, this function works, but it looks like a clumsy workaround.

Is there any better way to write this kind of function. I've also tried

filter_var($foo, FILTER_VALIDATE_INT);

but it won't accept values like '0003', '-0' etc.

Bart S. · Accepted Answer · 2010-04-04 16:00:30Z

6

You could try ctype_digit or is_numeric

answered Apr 4, 2010 at 16:00

Bart S.

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1 Comment

Mikk Over a year ago

Thanks I'll try to combine ctype_digit or is_numeric with something else. Not exactly what I was looking for, but close enough.

TheHippo · Accepted Answer · 2010-11-04 18:58:09Z

2

there is a native function called intval(). Is that's what you're looking for?

edited Nov 4, 2010 at 18:58

TheHippo

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answered Apr 4, 2010 at 15:55

iNPUTmice

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1 Comment

davethegr8 Over a year ago

That just returns the value of the input as an integer, it doesn't do any validation. Objects that are not integers will return 0.

confiq · Accepted Answer · 2010-04-04 16:02:17Z

1

it won't accept "0003" cause it's not 'clear' integer. Integer can't start with zero but it can be negative (notice that you remove '-' with your function). As said before me, use ctype_digit or is_numeric

answered Apr 4, 2010 at 16:02

confiq

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1 Comment

Mikk Over a year ago

Actually '-' stays (because $validate get's value substr($int,1) or just $int, and (int)$int is returned), my bad - probably comments in my code weren't clear enough.

hobodave · Accepted Answer · 2010-04-04 16:04:36Z

0

You could use intval as Jacob suggests.

However, if you want a Locale aware int validation class, then I suggest using the Zend Framework's Zend_Validate_Int validator.

Usage

<?php
$v = new Zend_Validate_Int();
if ($v->isValid($myVal)) {
    // yay
} else {
    // fail
}

Note that because of the pick-and-choose structure of the Zend Framework, you don't need to use this for your entire application. The Validators have minimal dependencies on Zend_Locale, for locale aware validation, and Zend_Registry.

answered Apr 4, 2010 at 16:04

hobodave

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Comments

davethegr8 · Accepted Answer · 2010-04-04 16:07:02Z

0

Quick and dirty way, won't handle '0003' as far as I know.

function parseInt($in) {
    return ($in == intval($in) ? $in : false);
}

answered Apr 4, 2010 at 16:07

davethegr8

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Comments

kguest · Accepted Answer · 2010-04-04 21:18:41Z

0

You could also use PEAR's Validate package; http://pear.php.net/manual/en/package.validate.validate.number.php

Various option are:

decimal (mixed) - Decimal chars or false when decimal not allowed. For example ",." to allow both "." and ",".
dec_prec (int) - Number of allowed decimals.
min (float) - Minimum value.
max (float) - Maximum value.

answered Apr 4, 2010 at 21:18

kguest

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