Shortest way to check if a variable contains positive integer using PHP?

Question

I want to check if user input a positive integer number.

1    = true
+10  = true
.1   = false
-1   = false
10.5 = false


Just a positive number. 
No characters.
No special character.
No dot.
No minus sign.

I tried is_int() function but it is returning false even on positive integers. Is there a string to int problem?

what should it return for "+3928742938749283479823749283749" ? — user187291
– user187291, Commented Apr 1, 2010 at 11:24
@stereofrog: textbox length validation may block you for this value in my application :) — Naveed
– Naveed, Commented Apr 1, 2010 at 11:41
your likely missing stereofrogs point. A signed Integer has a range from -2147483648 to 2147483647. Everything before or after that is technically a float — Gordon
– Gordon, Commented Apr 1, 2010 at 11:45
@Gordon: Yes I have got stereofrog's point but I was joking that validation will not allow you to pass integer range :) — Naveed
– Naveed, Commented Apr 1, 2010 at 11:53

Yacoby · Accepted Answer · 2010-04-01 11:21:36Z

36

Something like this should work. Cast the value to an integer and compare it with its original form (As we use == rather than === PHP ignores the type when checking equality). Then as we know it is an integer we test that it is > 0. (Depending on your definition of positive you may want >= 0)

$num = "20";

if ( (int)$num == $num && (int)$num > 0 )

edited Apr 1, 2010 at 11:21

answered Apr 1, 2010 at 11:16

Yacoby

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4 Comments

Yacoby Over a year ago

@NAVEED It may be because you are doing int($num). Casting is done by surrounding the type you want to cast to in brackets. In this case (int)$num.

Kirzilla Over a year ago

(int)$num == $num (int)'abcde' == 'abcde' will return true

Awena Over a year ago

What if my $num = "0.2" ; then it returns false... How can I check decimals? My numbers are in the following format "+0.2", "-0.8" ect...

Kenny Over a year ago

I solved it this way: $num = (int)$num >= 0 ? (int)$num : 0;

Community · Accepted Answer · 2023-11-17 19:32:19Z

31

Try the native Filter function*

filter_var($value, FILTER_VALIDATE_INT, array(
    'options' => array('min_range' => 1)
));

^{* if you just want to make sure the input string consists of an arbitrary length digit sequence, use a regex with [0-9] or [\d+]}

Examples with filter_var:

var_dump( filter_var(1, FILTER_VALIDATE_INT) ); // int(1)

var_dump( filter_var('1', FILTER_VALIDATE_INT) ); // int(1)

var_dump( filter_var('+10', FILTER_VALIDATE_INT) ); // int(10)

var_dump( filter_var(.1, FILTER_VALIDATE_INT) ); // bool(false)

var_dump( filter_var('.1', FILTER_VALIDATE_INT) ); // bool(false)

var_dump( filter_var(-1, FILTER_VALIDATE_INT, 
    array('options' => array('min_range' => 1))) ); // bool(false)

var_dump( filter_var('-1', FILTER_VALIDATE_INT, 
    array('options' => array('min_range' => 1))) ); // bool(false)

var_dump( filter_var('2147483648', FILTER_VALIDATE_INT) ); // bool(false)

var_dump( filter_var('0xFF', FILTER_VALIDATE_INT) ); // bool(false)

var_dump( filter_var(0xFF, FILTER_VALIDATE_INT) ); // int(255)

edited Nov 17, 2023 at 19:32

CommunityBot

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answered Apr 1, 2010 at 11:39

Gordon

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7 Comments

Naveed Over a year ago

and what about abcd, 123d, 10.5 and (space) ??

Gordon Over a year ago

@NAVEED try yourself and tell me :) I predict all false

Naveed Over a year ago

You have to wait.. leaving ;)

Fake Code Monkey Rashid Over a year ago

+1: This is how I'd do it but one thing to keep in mind is that 0 won't validate because PHP automatically casts it into false. Use === to avoid automatic type casting in evaluation.

Bryan Miller Over a year ago

I ran 'abcd', '123d', 10.5, and space (' ') and they all came back as false using simply: if (filter_var($int, FILTER_VALIDATE_INT)) { echo("Integer is valid"); } else { echo("Integer is not valid"); }

|

Imabigfatseacow · Accepted Answer · 2010-11-07 15:53:17Z

3

I use a regular expression. Very simple if you think about it. You need more punctuation marks if you want to make a number not a whole positive integer (minus sign and a period). So this just makes sure the only thing you have are numbers 0-9 for the value.

if(!ereg('^[0-9]+$', $value)) {
  $errors .= "This is not a positive whole number";
}

You could add another part on there to make sure it is less than a certain amount of characters as well. Hope this helps.

answered Nov 7, 2010 at 15:53

Imabigfatseacow

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Comments

Decko · Accepted Answer · 2010-11-19 13:03:28Z

3

I would say this is the best way

if (is_int($num) && $num > 0)

as typecasting to an int is very slow.

answered Nov 19, 2010 at 13:03

Decko

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2 Comments

PhoneixS Over a year ago

But get in mind with this answer that if $num is an integer string as $num = '123', is_int($num) return false

Dawid Over a year ago

I use is_numeric() to check if the given string is a valid number

Nopcea Flavius · Accepted Answer · 2020-11-09 13:01:36Z

1

the easiest way is:

if intval($x) > 0 {
 echo "true"
}

edited Nov 9, 2020 at 13:01

answered Apr 1, 2010 at 11:57

Nopcea Flavius

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1 Comment

Gordon Over a year ago

intval('1e10'); // 1 and intval('420000000000000000000'); // 2147483647

hram908 · Accepted Answer · 2013-01-08 22:35:15Z

0

if(!preg_match('/^[0-9]+$/', $input)) {

Deprecated: Function ereg() is deprecated

answered Jan 8, 2013 at 22:35

hram908

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