2015-02-26 141 views

回答

17

簡短的回答:sprintf('%05.2f', 1);將給期望的結果01.00

注意如何%02%05取代。

說明

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!

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