PHP numeric left-side auto-fill 0

Copy codeThe Code is as follows:
<? Php
$ SourceNumber = "1 ";
$ NewNumber = substr (strval ($ sourceNumber + 1000), 1, 3 );
Echo "$ newNumber ";
?>

At this time, the following error occurs: 001.
If you want to increase the number of digits, you can increase the number by 1000, and then increase the number by 3.
For example, if I want to add "4 0" 03rd rows, it will become like this.
Copy codeThe Code is as follows:
<? Php
$ NewNumber = substr (strval ($ sourceNumber + 100000 );
?>

In fact, to display a few digits in total, add the number of zeros after $ sourceNumber + 1, and change the last number to the number of digits.

Better method:
String str_pad (string $ input, int $ pad_length [, string $ pad_string [, int $ pad_type])

Copy codeThe Code is as follows:
<? Php
$ Input = "Alien ";
Echo str_pad ($ input, 10 );
// Produces "Alien"
Echo str_pad ($ input, 10, "-=", STR_PAD_LEFT );
// Produces "-=-Alien"
Echo str_pad ($ input, 10, "_", STR_PAD_BOTH );
// Produces "_ Alien ___"
Echo str_pad ($ input, 6 ,"___");
// Produces "Alien _"
?>


Fill in the length of the string. Pad it with pad_string. It is filled on the right by default. If STR_PAD_LEFT is filled on the left, STR_PAD_BOTH is supplemented on both sides.
The next time we use str_pad, it is built-in and certainly faster than custom.
/*
I think the above method is not very good. Let's introduce a method I wrote. The method function is as follows, so that when you want the result 001, the method is dispRepair ('1', 3, '0 ')
Function: complement function
Str: original string
Type: type. 0 indicates post-completion, and 1 indicates pre-completion.
Len: New String Length
Msg: Fill in characters
*/
Copy codeThe Code is as follows:
Function dispRepair ($ str, $ len, $ msg, $ type = '1 '){
$ Length = $ len-strlen ($ str );
If ($ length <1) return $ str;
If ($ type = 1 ){
$ Str = str_repeat ($ msg, $ length). $ str;
} Else {
$ Str. = str_repeat ($ msg, $ length );
}
Return $ str;
}