[SOLVED] is_int VS is_numeric noobish question

[asmith]

hey guys

 

when i check this :

 

if (is_int($_POST[number]))

 

it always return false , seems like $_POST always send a string .

is_int will return true for 2, NOT for "2" . 

 

how can i check an input that it is numeric ?  ( is_numeric allow float too but i  just whole numbers.)

and i don't mean by preg_match .

 

thanks in advance

 

[emehrkay]

is_numeric will take "2" and cast it as a number and then test to see if it is a number

 

Return Values

 

Returns TRUE if var is a number or a numeric string, FALSE otherwise.

 

maybe you need to do

 

if(is_numeric($_POST['number']) && is_int($_POST['number'])){}

 

 

[corbin]

$number = $_POST['number'];
$int = (int) $number;
if($int == $number) {
    //integer
}

//OR....

if(ctype_digit($_POST['number'])) {
    //integer
}

[corbin]

is_numeric will take "2" and cast it as a number and then test to see if it is a number

 

Return Values

 

Returns TRUE if var is a number or a numeric string, FALSE otherwise.

 

maybe you need to do

 

if(is_numeric($_POST['number']) && is_int($_POST['number'])){}

 

 

 

That will always return false....

 

is_numeric does the casting internally and the variable isn't passed by reference.  That means, the second time, it will still be a string....

[asmith]

ctype_digit !

that was a function i had never heard of  , works perfect .

 

thanks guys for taking time

 

why this function is not so popular anyway ?

 

 

[corbin]

I don't know.... I hadn't ever heard of it either until the other day when someone posted about it in a thread similar to this thread.

 

Converting the types and comparing (forcing int) is actually a little bit faster, but it's obviously more code, and it's like .00000000000000001 seconds faster.

[asmith]

thanks for the information 

[emehrkay]

is_numeric will take "2" and cast it as a number and then test to see if it is a number

 

Return Values

 

Returns TRUE if var is a number or a numeric string, FALSE otherwise.

 

maybe you need to do

 

if(is_numeric($_POST['number']) && is_int($_POST['number'])){}

 

 

 

 

That will always return false....

 

is_numeric does the casting internally and the variable isn't passed by reference.  That means, the second time, it will still be a string....

 

i think you are confused. you want to change what is held in the $_POST var by casting it as an integer

$int = (int) $number;

not me. I want to check to see if what is in there is a number, and if it is an integer

 

I doubted my logic since you brought it into question and created a function to test what I was saying. Run it

 

<?php


function numberAndIntegerCheck($x){
    if(is_numeric($x) && is_int($x)){
        echo '<br />$x is a number and an integer!';
    }else{
        echo '<br />fail!';
    }
}

numberAndIntegerCheck(4);
numberAndIntegerCheck("4");
?>

[corbin]

C:\Users\Corbin\Desktop>php post.php

<br />$x is a number and an integer!<br />fail!

C:\Users\Corbin\Desktop>

 

"That will always return false...."

 

Unless you do like you said and cast the $_POST vars, that will indeed always return false, since $_POST variables are strings.

 

If you meant doing that, like you said in your second post, then I guess I missed that part.

[emehrkay]

$number = $_POST['number'];
$int = (int) $number;
if($int == $number) {
     //integer
}

 

additionally, if you want to correct code, this doesnt make any sense.

 

you are casting $number as an intger and using a == to check against dataypes - wrong! when checking datatypes you use the triple equal sign ===

 

and you dont even need to compare the two vars, just trying to cast a new datatype is check enough

 

$x = "asfdasf4";
if((integer) $x){
    echo 'that is a number';
}

[corbin]

Ahhh I didn't know you could do an if on a cast.

 

 

Also,

 

$number = $_POST['number'];
$int = (int) $number;
if($int === $number) {
     //this would always be false
}

 

$number would be a string, and $int would be an integer, thus, they need to be compared with == so that $int will be converted back to a string for comparison.

 

Edit:  (Actually, it might convert $number to a integer, in which case, this code would break since if $number failed at conversion the first time and was set to zero, it would fail the second time, and the two 0s would match.)