PHP - Get length of digits in a number - php

Question

I would like to ask how I can get the length of digits in an Integer. For example:
$num = 245354;
$numlength = mb_strlen($num);
$numlength should be 6 in this example. Somehow I can't manage it to work?
Thanks
EDIT: The example code above --^ and its respective method mb_strlen(); works just fine.

Answer 1

Maybe:
$num = 245354;
$numlength = strlen((string)$num);

Answer 2

Accepted answer won't work with the big numbers. The better way to calculate the length of any number is to invoke floor(log10($num) + 1) with a check for 0.
$num = 12357;
echo $num !== 0 ? floor(log10($num) + 1) : 1; // prints 5
It has multiple advantages. It's faster, you don't do the casting of types, it works on big numbers, it works with different number systems like bin, hex, oct.
The equation does the logarithm with base 10 then makes the floor of it and adds 1.
This solution can work independently on the base, so if you want to calculate the length of binary or hex just change the base of the logarithm.
Working fiddle

Answer 3

The accepted solution presents a problem when evaluating negative numbers.
It works with a positive number:
$num = 245354;
$numlength = strlen((string)$num);
// Result: 6
But with a negative number, the (-) is added to the count:
$num = -245354;
$numlength = strlen((string)$num);
// Result: 7
Quick workaround:
$num = -245354;
$numlength = strlen((string)abs($num));
// Result: 6

Answer 4

More elegant way :)
ceil(log10($num));

Answer 5

You could also use some basic math!
$digits = (int)(log($num,10)+1)
<?php
$num = 123;
$num2 = 1234;
$num3 = 12345;
function digits($num){
return (int) (log($num, 10) + 1);
}
echo "\n $num: " . digits($num); // 123: 3
echo "\n $num2:" . digits($num2); // 1234: 4
echo "\n $num3:" . digits($num3); // 12345: 5
echo "\n";

Answer 6

Another way to find out the length of a number in digits would be to divide the integer part of the number to 10 until it becomes 0.
Example:
2021/10 = 202.1
202/10 = 20.2
20/10 = 2
2/10 = 0.2
Code:
function numberGetLength($number) {
$count = 0;
while (intval($number) > 0) {
$number = intval($number) / 10;
$count += 1;
}
return $count
}

Answer 7

Just using some version of (int)(log($num,10)+1) fails for 10, 100, 1000, etc. It counts the number 10 as 1 digit, 100 as two digits, etc. It also fails with 0 or any negative number.
If you must use math (and the number is non-negative), use:
$numlength = (int)(log($num+1, 10)+1);
Or for a math solution that counts the digits in positive OR negative numbers:
$numlength = ($num>=0) ? (int)(log($num+1, 10)+1) : (int)(log(1-$num, 10)+1);
But the strlen solution is just about as fast in PHP.

Answer 8

In PHP types are loosely set and guessed, if you want to see something as a string if it is an integer, float, and (i have not tried this) bool then #Gorjunav is the most correct answer.
Reset the variable as a string
$stringNum = (string) $num;
Then you can go anything string related you want with it! And vice-versa for changing a string to an int
$number = (int) $stringNum;
and so on...

Answer 9

count only integer value
`<?php
$n1 =12345;
$n2 =123454.55;
$n3 =12345564.557;
echo "The Number you Type: ".$n1."<br>";
$count = 0;
while ($n1 != 0)
{
$n1 = $n1 / 10;
$n1 = intval($n1);
++$count;
}
echo "The Digit in a Number: ".$count;
}
?>`

Answer 10

echo strlen((string) abs($num)); // using **abs** it'll work with negative integers as well

Answer 11

Tested in PHP 4.4.9 - 8.0.0
$array = array(-1, 0, -0, 1, 4, 9, 10, -10, 20, -20, 100, -100);
foreach( $array as $key => $num ){
echo $key."\t{$num}\t=>\t".($num !== 0 ? floor(log10(abs($num)) + 1) : 1)."\n";
}
/* Output:
0 -1 => 1
1 0 => 1
2 0 => 1
3 1 => 1
4 4 => 1
5 9 => 1
6 10 => 2
7 -10 => 2
8 20 => 2
9 -20 => 2
10 100 => 3
11 -100 => 3
*/

Answer 12

The following function work for either integers or floats (works with PHP7+):
function digitsCount($number): int
{
$number = abs($number);
$numberParts = explode(".", $number);
return
strlen($numberParts[0]) +
(strlen($numberParts[1] ?? 0));
}