[SOLVED] Checking if a variable is an integer?

Warptweet · August 24, 2007

I have a form that asks my users for the width and height of their submission in pixels, but I want to prevent my users from entering a letter or a word to try and trick the system.

Is there a way of checking if my variable $variable is an integer or a string (or even a combination)?

I'm pretty sure this is very easy, but for some reason I can't put my finger on it.

dbo · August 24, 2007

$flag = is_int($var);

or

$matches = preg_match("^\d+$", $var);

if( $matches != 0 ) $flag = intval($var);

trq · August 24, 2007

is_int won't work because all data retrieved via POST or GET is actually a string. Use is_numeric().

teng84 · August 24, 2007

is_int won't work because all data retrieved via POST or GET is actually a string. Use is_numeric().

then after using thorpes suggestion use int

sample

<?php
echo (int) ( (0.1+0.7) * 10 ); // echoes 7!
?>

note numeric accept" + " and etc....see thorpes link

dbo · August 24, 2007

Doh! I always use regex anyways. But no matter what you do I suggest after doing your check you do a cast to the datatype that it should be. Apparently regex has some oddities where $ doesn't necessarily mean the end of the line.

Warptweet · August 24, 2007

$flag = is_int($var);

or

$matches = preg_match("^\d+$", $var);

if( $matches != 0 ) $flag = intval($var);

Okay, apparently is_int won't work.

So for your second code, how do I make it execute code inside { and } if it IS an int?

Or, does that make the variable an integer, so I don't need to do an if statement?

dbo · August 24, 2007

Oh wow, I really gave you bad code, whoops! As others pointed out apparently is_int wont work.

Try this.

function IsInteger($val)
{
  $matches = preg_match("^\d+$", $val);
  if( $matches != 0 )
  {
     return true;
  }
  return false;
}

usage:

$val = "5";
$is_int = IsInteger($val);
if( $is_int )
{
  $val = intval($val);
} 
else
{
  echo "ERROR! Not an int!";
}

dbo · August 24, 2007

Actually instead of \d+ you should probably use [0-9]+

dbo · August 24, 2007

Hrmm. A thought occured to me so I'm going to throw it out here and see what some of the other people think. I think it may not be a "best practice" but here it is anyways.

Try this.

function IsInteger(&$val)
{
  $matches = preg_match("^\d+$", $val);
  if( $matches != 0 )
  {
     $val = intval($val);
     return true;
  }
  return false;
}

usage:

$val = "5";
$is_int = IsInteger($val);
if( $is_int )
{
  //Process normally... we don't need to cast since we passed by reference
} 
else
{
  echo "ERROR! Not an int!";
}

MadTechie · August 24, 2007

why not keep it simple

IE

$X = "7.4"; 
$isint = (bool)($X == ((int)$X));
var_dump($isint);

Warptweet · August 24, 2007

None of your code works.

How about keep this even simpler?

How can I check if a number is divisable by 1? And if it isnt, make $error .= "You entered an incorrect width.";

After all, a letter cannot be divisable by 1, therefore it should be a logical method.

MadTechie · August 24, 2007

try

$X = "test"; 
$isint = (bool)(($X == ((int)$X)) && is_numeric($X));
var_dump($isint);
echo "$X";

dbo · August 24, 2007

What do you mean it doesn't work? The first code I spewed was the suck but those last 2 should work fine. And yeah the madmans code should work fine, but its not perty! Whatever you do, turn it into a function and reuse the code. Don't write that stuff by hand each time.

On another note.... any thoughts from anyone about passing that by reference and doing the cast?

Warptweet · August 24, 2007

Sorry for being so inexperienced, but what do I do with madtechies code?

Which part of it can I put $error .= "Invalid Width.";

Which part of it can I put echo "Your width is valid!";

dbo · August 24, 2007

Create a file called testme.php and put this code in it.

<?php

function IsInteger(&$val)
{
  $matches = preg_match("^\d+$", $val);
  if( $matches != 0 )
  {
     $val = intval($val);
     return true;
  }
  return false;
}

?>

<HTML>
<HEAD>
<TITLE>Testing...</TITLE>
</HEAD>

<BODY>

<?php
$test1 = "5";
$test2 = "Doh";

echo $test1 . " is ";

$is_int = IsInteger($test1);
if( $is_int )
{
  echo " an integer!<br />";   
} 
else
{
  echo " is NOT an integer!<br />";
}

echo "<br /><br />";

echo $test2 . " is ";

$is_int = IsInteger($test2);
if( $is_int )
{
  echo " an integer!<br />";   
} 
else
{
  echo " is NOT an integer!<br />";
}
?>

</BODY>
</HTML>

Put this on your webserver and then point your browser to the file.

MadTechie · August 24, 2007

$X =$_GET['width']; 
$isint = (bool)(($X == ((int)$X)) && is_numeric($X));
echo ($isint)?"":"invalid width";

EDIT: leaves it in dbo capable hand

dbo · August 24, 2007

Make one modification to the code I previously gave you.

Change

$matches = preg_match("^\d+$", $val);

to

$matches = preg_match("^[0-9]+$", $val);

Warptweet · August 24, 2007

http://warptweet.com/developzone/code.php

Doesn't work.

MadTechie · August 24, 2007

use

$matches = preg_match("/^[0-9]+$/", $val);

dbo · August 24, 2007

Correct.

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[SOLVED] Checking if a variable is an integer?

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