就個人而言,我會寫一個更通用的程序。我已經做了兩個方面:
更傳統的方法:
// Pass in 'value' you want to pad, and 'len' as total length of string
// to be returned to you. For example, value=24, len=6 would return 000024.
public static function padIntWithLeadingZeros2(value:int, len:uint):String
{
var paddedValue:String = value.toString();
if (paddedValue.length < len)
{
for (var i:int = 0, numOfZeros:int = (len - paddedValue.length); i < numOfZeros; i++)
{
paddedValue = "0" + paddedValue;
}
}
return paddedValue;
}
我做的自己的風格:
// Pass in 'value' you want to pad, and 'len' as total length of string
// to be returned to you. For example, value=24, len=6 would return 000024.
public static function padIntWithLeadingZeros(value:int, len:uint):String
{
var paddedValue:String = value.toString();
if (paddedValue.length < len)
{
var leadingZeros:String = "0000000000";
paddedValue = leadingZeros.substring(0, (len - paddedValue.length)) + paddedValue;
}
return paddedValue;
}
原來,一個來講是其他高效流逝的時間來執行該功能。所以,這只是一個偏好問題。
R. Grimes
mhmm我想,可能有一個更簡單的方法,但沒關係。謝謝。 – mate64 2011-02-15 08:24:22