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