我如何才能改变我的DateTime变量“s”的时间?
DateTime s = some datetime;
我如何才能改变我的DateTime变量“s”的时间?
DateTime s = some datetime;
当前回答
好了,我要深入介绍我的建议,一个扩展方法:
public static DateTime ChangeTime(this DateTime dateTime, int hours, int minutes, int seconds, int milliseconds)
{
return new DateTime(
dateTime.Year,
dateTime.Month,
dateTime.Day,
hours,
minutes,
seconds,
milliseconds,
dateTime.Kind);
}
然后调用:
DateTime myDate = DateTime.Now.ChangeTime(10,10,10,0);
重要的是要注意,这个扩展返回一个新的日期对象,所以你不能这样做:
DateTime myDate = DateTime.Now;
myDate.ChangeTime(10,10,10,0);
但是你可以这样做:
DateTime myDate = DateTime.Now;
myDate = myDate.ChangeTime(10,10,10,0);
其他回答
最好的解决方案是:
currdate.AddMilliseconds(currdate.Millisecond * -1).AddSeconds(currdate.Second * -1).AddMinutes(currdate.Minute * -1).AddHours(currdate.Hour * -1);
DateTime出了什么问题。AddSeconds方法,您可以添加或减去秒?
由于DateTime是不可变的,所以当需要更改日期组件时,必须创建一个新实例。不幸的是,没有内置功能来设置DateTime实例的各个组件。
使用以下扩展方法
public static DateTime SetPart(this DateTime dateTime, int? year, int? month, int? day, int? hour, int? minute, int? second)
{
return new DateTime(
year ?? dateTime.Year,
month ?? dateTime.Month,
day ?? dateTime.Day,
hour ?? dateTime.Hour,
minute ?? dateTime.Minute,
second ?? dateTime.Second
);
}
public static DateTime SetYear(this DateTime dateTime, int year)
{
return dateTime.SetPart(year, null, null, null, null, null);
}
public static DateTime SetMonth(this DateTime dateTime, int month)
{
return dateTime.SetPart(null, month, null, null, null, null);
}
public static DateTime SetDay(this DateTime dateTime, int day)
{
return dateTime.SetPart(null, null, day, null, null, null);
}
public static DateTime SetHour(this DateTime dateTime, int hour)
{
return dateTime.SetPart(null, null, null, hour, null, null);
}
public static DateTime SetMinute(this DateTime dateTime, int minute)
{
return dateTime.SetPart(null, null, null, null, minute, null);
}
public static DateTime SetSecond(this DateTime dateTime, int second)
{
return dateTime.SetPart(null, null, null, null, null, second);
}
您可以设置单独的DateTime组件,如
var now = DateTime.Now;
now.SetSecond(0);
Using an extencion to DateTime:
public enum eTimeFragment
{
hours,
minutes,
seconds,
milliseconds
}
public static DateTime ClearTimeFrom(this DateTime dateToClear, eTimeFragment etf)
{
DateTime dtRet = dateToClear;
switch (etf)
{
case eTimeFragment.hours:
dtRet = dateToClear.Date;
break;
case eTimeFragment.minutes:
dtRet = dateToClear.AddMinutes(dateToClear.Minute * -1);
dtRet = dtRet.ClearTimeFrom(eTimeFragment.seconds);
break;
case eTimeFragment.seconds:
dtRet = dateToClear.AddSeconds(dateToClear.Second * -1);
dtRet = dtRet.ClearTimeFrom(eTimeFragment.milliseconds);
break;
case eTimeFragment.milliseconds:
dtRet = dateToClear.AddMilliseconds(dateToClear.Millisecond * -1);
break;
}
return dtRet;
}
像这样使用:
游戏机。WriteLine (DateTime。Now ClearTimeFrom (eTimeFragment时间)。
这个必须返回: 2016-06-06 00:00:00.000
这难道不能解决你的问题吗?
Dateime dt = DateTime.Now;
dt = dt.AddSeconds(10);