$sourceNumber = "1";
$newNumber = substr(strval($sourceNumber+1000),1,3);
echo "$newNumber";
?>
这个时候就会出现:001
如果要增加位数的话可以将1000加大,然后把3也加大。
举例:如果我要补上 "4个0" 第03行 就要变成这样。
$newNumber = substr(strval($sourceNumber+100000),1,5);
?>
其实就是总共要显示几位数字,就把$sourceNumber+1后面补上多少个0,最后一个数字就直接改成显示几位数字。
更好的方法:
string str_pad ( string $input, int $pad_length [, string $pad_string [, int $pad_type]] )
$input = "Alien";
echo str_pad($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($input, 6 , "___");
// produces "Alien_"
?>
补齐字符串的长度.以pad_string 补.默认补在右边,如果STR_PAD_LEFT就补到左边,STR_PAD_BOTH两边一起补。
下次用str_pad,毕竟是内置的,肯定比自定义的快。
/*
你上面的方法我觉得不怎么好,介绍一下我写的一个方法。方法函数如下,这样当你要的结果001的话,方法:dispRepair('1',3,'0')
功能:补位函数
str:原字符串
type:类型,0为后补,1为前补
len:新字符串长度
msg:填补字符
*/
function dispRepair($str,$len,$msg,$type='1') {
$length = $len - strlen($str);
if($lengthif ($type == 1) {
$str = str_repeat($msg,$length).$str;
} else {
$str .= str_repeat($msg,$length);
}
return $str;
}

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