Strange behavior with PHP's str_pad() and numbers

Question

$caseid is (usually) a 5-digit number (it could be lower). I want a leading 0 so that the number is 6 digits long:

<?php
$caseid=12345;
echo str_pad(strval($caseid), 6-strlen(strval($caseid)), '0', STR_PAD_LEFT); ?>

This doesn't work as expected (displays the same number). Instead if I add 5 to the second argument of str_pad , it works.

<?php echo str_pad(strval($caseid), 11-strlen(strval($caseid)), '0', STR_PAD_LEFT); ?>

It works. What is the error here?

Answer 1

你不需要计算你想要它的总字符数的差异。

str_pad(strval($caseid), 6, '0', STR_PAD_LEFT);

Answer 2

I was once bothered by str_pad() s behavior too. I got tired of it, and returned to old stable sprintf() .

<?php
header('Content-Type: text/plain');

$test = [
     1,
     12,
     123,
     1234,
     12345,
     123456
];

foreach($test as $n){
    echo sprintf('%06d' . PHP_EOL, $n);
}
?>

Result:

000001
000012
000123
001234
012345
123456

Maybe, it is not an answer, but I hope it might help somebody.

Answer 3

See the function prototype:

string str_pad (
        string $input , 
        int $pad_length 
        [, string $pad_string = " " 
        [, int $pad_type = STR_PAD_RIGHT ]]

)

Pad a string to a certain length with another string , $pad_length means the length you want.

str_pad in php.net