碰巧看到这篇文章，因为我正在寻找相同的功能，这可能会做什么人想要的。取原日期，更换时间部分

DateTime dayOpen = DateTime.Parse(processDay.ToShortDateString() + " 05:00 AM");

int year = 2012;
int month = 12;
int day = 24;
int hour = 0;
int min = 0;
int second = 23;
DateTime dt = new DateTime(year, month, day, hour, min, second);

DateTime出了什么问题。AddSeconds方法，您可以添加或减去秒?

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);

这是一种贫民区的方法，但它很有效:)

DateTime dt = DateTime.Now; //get a DateTime variable for the example
string newSecondsValue = "00";
dt = Convert.ToDateTime(dt.ToString("MM/dd/yyyy hh:mm:" + newSecondsValue));

 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