2015-02-26 99 views

回答

17

简短的回答:sprintf('%05.2f', 1);将给期望的结果01.00

注意如何%02%05取代。

说明

forum post我指出了正确的方向:第一个数字并没有表示前导零的数量还是总charaters的小数点分隔符左边的数量,但总数的结果字符串中的字符!

sprintf('%02.2f', 1);产生至少十进制分隔符 “.” 加上精度至少2个字符。由于这已经是3个字符,所以开始时%02不起作用。为了得到一个人需要3个字符添加对精度和小数点分隔符所需 “2个前导零”,使之成为sprintf('%05.2f', 1);

一些代码

$num = 42.0815; 

function printFloatWithLeadingZeros($num, $precision = 2, $leadingZeros = 0){ 
    $decimalSeperator = "."; 
    $adjustedLeadingZeros = $leadingZeros + mb_strlen($decimalSeperator) + $precision; 
    $pattern = "%0{$adjustedLeadingZeros}{$decimalSeperator}{$precision}f"; 
    return sprintf($pattern,$num); 
} 

for($i = 0; $i <= 6; $i++){ 
    echo "$i max. leading zeros on $num = ".printFloatWithLeadingZeros($num,2,$i)."\n"; 
} 

输出

0 max. leading zeros on 42.0815 = 42.08 
1 max. leading zeros on 42.0815 = 42.08 
2 max. leading zeros on 42.0815 = 42.08 
3 max. leading zeros on 42.0815 = 042.08 
4 max. leading zeros on 42.0815 = 0042.08 
5 max. leading zeros on 42.0815 = 00042.08 
6 max. leading zeros on 42.0815 = 000042.08