PHP - Length of a number containing 0's to the left - php

Question

I'm trying to validate a number by it's length. This number has to have 4 digits so it passes the validation. The problem is when this number has 0's to it's left, like 0035.
Right now I'm at this:
echo (strlen ((string) 0025 ));
Which gives a total of 2, but I want this to count the 0's to it's left, so it gives me a total of 4.
Clearly the cast of the integer to string is not working, how can i do this?

Answer 1

You can't do that way, a left zero means the number is octal and not decimal, you can use sprintf() to do that.
Example:
echo strlen(sprintf("%04d", 25));
Live Test:
http://codepad.viper-7.com/VQr7Xz
Comment Answer:
I don't want to add the 0s to the number, i want to detect if the
number has 0s. If the number received is 25, it's not a valid number.
If it is 0025 it is valid. What i want is to validate only numbers
with 4 digits. – Cláudio Ribeiro
Cláudio, numbers have infinite left zeros, although a user has explicitly type 2 or 3 left zeros there are more hidden left zeros, it's a math basic, this is why it's impossible to know how many left zeros the user has typed if you receive an integer variable. If the variable has a constant size and you want to know how many left zeros it has you can do this:
<?php
$int = 25;
echo 4 - strlen($int);
Live test: http://codepad.viper-7.com/fT2jSn
But if you the variable has variable length it must be a string type instead of a numeric type.
An example where the variable received is a string:
<?php
$strs = array("0025","000035","01","2");
foreach($strs as $str)
{
preg_match("/^0+/", $str, $matches);
echo strlen(#$matches[0]);
echo "<br>";
}
Live Test: http://codepad.viper-7.com/BTRTgR

Answer 2

That should work:
$str = "0025";
if( is_numeric($str) && strlen($str) == 4)
{
echo "pass";
}

Answer 3

If it's a number, not a string, the number doesn't have digits. It has a value. You can format that value into a string with 4 digits which is left padded with 0s. But to validate whether a number has 4 digits is nonsense, since the number value has no formatting. The value only becomes "4 digits" when you format it as base 10 number. Until then the value is a value which can be expressed in a multitude of bases and has a different number of "digits" in all of them.
You either want to format the number to a 0-padded 4 digit string, or you want to check whether the value is between 0 and 9999 (or 1000 and 9999 if it has to be exactly "4 digits").
if (0 <= $num && $num <= 9999) {
$numStr = sprintf('%04d', $num);
} else {
trigger_error('Number out of range');
}