簡短的回答:sprintf('%05.2f', 1);
將給期望的結果01.00
注意如何%02
被%05
取代。
說明
這forum post我指出了正確的方向:第一個數字並沒有表示前導零的數量還是總charaters的小數點分隔符左邊的數量,但總數的結果字符串中的字符!
例
sprintf('%02.2f', 1);
產生至少十進制分隔符 「.
」 加上精度至少2個字符。由於這已經是3個字符,所以開始時%02
不起作用。爲了得到一個人需要3個字符添加對精度和小數點分隔符所需 「2個前導零」,使之成爲sprintf('%05.2f', 1);
一些代碼
$num = 42.0815;
function printFloatWithLeadingZeros($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";
}
輸出
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