Re: [PHP] problem with intval and !=

[Cesar Cordovez]

The other purpose of this procedure is if the user types "12asdasd" into 
$value, $number will be assigned "12" so $value and $number are not the 
same, meaning $value is not a number.  Which is correct, but the 
procedure is not working correctly in PHP when it should.  (It works in 
c, c++, fortran and, believe it or not HyperCard!!!)

Cesar.



Cesar Cordovez wrote:

Curt:

My fault!  You are right, but thats not what I want. $value comes from a 
form filled by a user.  I want $value to be an integer.  Not a float. So 
if the user types "12.3" the system has to send an error msg.

Therefore the procedure.

By-the-way, Im using PHP 4.3.3 (on windows XP profesional.  Don't say a 
thing!!! I rather work on my mac!)  =)

Cesar

Curt Zirzow wrote:

* Thus wrote Cesar Cordovez ([EMAIL PROTECTED]):

Can somebody explain me why is this happening?

$value = "12.3";
$number = intval($value);
echo "Number: $number, Value: $value<br>";
// echoes: Number: 12, Value: 12.3


I'm guessing you wondering why the .3 is removed.  Thats because
12.3 isn't an integer but a float.
$number = floatval($value);


if ($number != $value) {
    echo "Bad";
} else {
    echo "Good";  // echoes Good!!!!!!
}
The previous should be echoing "Bad".  (I think!)


correct. If you use the floatval() it will echo "Good".

Curt


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