php function integer parameter problem

Question

function is_zipcode_valid($zipcode){ ... }

if I call that function with is_zipcode_valid(08004); my parameter 08004 gets in as 8004, basically it removed all precending 0's.

How can I get around that problem?

thanks

Answer 1

Pass it as a string and modify your function to expect a string instead:

is_zipcode_valid('08004');

It's not possible to preserve leading zeros when using integers. If an integer value has a leading zero, it is interpreted in base 8. Here is an example:

echo 010; // Output is 8.

Answer 2

PHP treats integers as base8 numbers if a leading zero is present. Pass your number as a string

Answer 3

As everyone says pass it in as a string.

if php5 you can make sure it is string going in by casting

$zipcode = (string) $zipcode; 

or

$zipcode = "$zipcode"; 




  function is_zipcode_valid( $zipcode){ 

   is_string($zipcode) //returns true
 }

See Php.Net

Note: Instead of casting a variable to a string , it is also possible to enclose the variable in double quotes.

//create zipcode from number function

    function form_zipcode($zipcode,$desiredLength){ 
              if(str_len($zipcode) != $desiredLength)
                  {
                       return str_pad($zipcode,$desiredLength,"0",STR_PAD_LEFT);
                  }
                else
                 {
                  return $zipcode;
                  }
       }

Using form_zipcode(55,5) would return "00055";

Answer 4

Sorry typo last night (GMT here)

    <?php
echo form_zipcode("55",5);

function form_zipcode($zipcode,$desiredLength){ 
              if(strlen($zipcode) != $desiredLength)
                  {
                       return str_pad($zipcode,$desiredLength,"0",STR_PAD_LEFT);
                  }
                else
                 {
                  return $zipcode;
                  }
       }
 ?>

Works as expected returns 00055 That not what you require??