PHP automatically adds 0 to the left side of numbers

PHP automatically adds 0 to the left side of numbers_PHP tutorial

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Release： 2016-07-21 15:53:28

Original

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Copy the code The code is as follows:

 $sourceNumber = "1"; 
 $newNumber = substr (strval($sourceNumber+1000),1,3); 
 echo "$newNumber"; 
?> 

This time it will appear: 001
If you want To increase the number of digits, you can increase 1000, and then increase 3 as well.
Example: If I want to add "4 zeros", line 03 will look like this.

Copy code The code is as follows:

$newNumber = substr(strval($sourceNumber+100000 ),1,5); 
?> 

In fact, how many digits do you want to display in total? Just add as many 0s after $sourceNumber+1, and change the last number directly. to display several digits.

A better way:
string str_pad ( string $input, int $pad_length [, string $pad_string [, int $pad_type]] )

Copy the code The code is as follows:

$input = "Alien"; 
echo str_pad($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_" 
?> ; 

Complete the length of the string. Use pad_string to complement. The default is to pad on the right. If STR_PAD_LEFT, pad to the left. STR_PAD_BOTH pads on both sides together.
Use str_pad next time. After all, it is built-in and will definitely be faster than custom one.
/*
I don’t think your method above is very good. Let me introduce a method I wrote. The method function is as follows, so when you want the result 001, the method: dispRepair('1',3,'0')
Function: fill-in function
str: original string
type: type, 0 is back-padded, 1 is front-padded
len: new string length
msg: padding characters
*/

Copy code The 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; 
} 