简短的回答:sprintf('%05.2f', 1);
将给期望的结果01.00
注意如何%02
被%05
取代。
说明
这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!
例
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