我需要删除日期时间的时间部分,或者可能有以下格式的日期在对象形式,而不是在字符串形式。

06/26/2009 00:00:00:000

我不能使用任何字符串转换方法,因为我需要对象形式的日期。

我尝试先将DateTime转换为字符串,从它中删除特定的时间日期,但它添加了12:00:00 AM,只要我将它转换回DateTime对象。


当前回答

我写了一个DateOnly结构。它在皮肤下使用一个DateTime,但没有时间部分被公开:

using System;

public struct DateOnly : IComparable, IFormattable, IComparable<DateOnly>, IEquatable<DateOnly>
{

    private DateTime _dateValue;

    public int CompareTo(object obj)
    {
        if (obj == null)
        {
            return 1;
        }

        DateOnly otherDateOnly = (DateOnly)obj;
        if (otherDateOnly != null)
        {
            return ToDateTime().CompareTo(otherDateOnly.ToDateTime());
        }
        else
        {
            throw new ArgumentException("Object is not a DateOnly");
        }
    }

    int IComparable<DateOnly>.CompareTo(DateOnly other)
    {
        return this.CompareToOfT(other);
    }
    public int CompareToOfT(DateOnly other)
    {
        // If other is not a valid object reference, this instance is greater.
        if (other == new DateOnly())
        {
            return 1;
        }
        return this.ToDateTime().CompareTo(other.ToDateTime());
    }

    bool IEquatable<DateOnly>.Equals(DateOnly other)
    {
        return this.EqualsOfT(other);
    }
    public bool EqualsOfT(DateOnly other)
    {
        if (other == new DateOnly())
        {
            return false;
        }

        if (this.Year == other.Year && this.Month == other.Month && this.Day == other.Day)
        {
            return true;
        }
        else
        {
            return false;
        }
    }

    public static DateOnly Now()
    {
        return new DateOnly(DateTime.Now.Year, DateTime.Now.Month, DateTime.Now.Day);
    }

    public static bool TryParse(string s, ref DateOnly result)
    {
        DateTime dateValue = default(DateTime);
        if (DateTime.TryParse(s, out dateValue))
        {
            result = new DateOnly(dateValue.Year, dateValue.Month, dateValue.Day);
            return true;
        }
        else
        {
            return false;
        }
    }

    public static DateOnly Parse(string s)
    {
        DateTime dateValue = default(DateTime);
        dateValue = DateTime.Parse(s);
        return new DateOnly(dateValue.Year, dateValue.Month, dateValue.Day);
    }

    public static DateOnly ParseExact(string s, string format)
    {
        CultureInfo provider = CultureInfo.InvariantCulture;
        DateTime dateValue = default(DateTime);
        dateValue = DateTime.ParseExact(s, format, provider);
        return new DateOnly(dateValue.Year, dateValue.Month, dateValue.Day);
    }

    public DateOnly(int yearValue, int monthValue, int dayValue) : this()
    {
        Year = yearValue;
        Month = monthValue;
        Day = dayValue;
    }

    public DateOnly AddDays(double value)
    {
        DateTime d = new DateTime(this.Year, this.Month, this.Day);
        d = d.AddDays(value);
        return new DateOnly(d.Year, d.Month, d.Day);
    }

    public DateOnly AddMonths(int months)
    {
        DateTime d = new DateTime(this.Year, this.Month, this.Day);
        d = d.AddMonths(months);
        return new DateOnly(d.Year, d.Month, d.Day);
    }

    public DateOnly AddYears(int years)
    {
        DateTime d = new DateTime(this.Year, this.Month, this.Day);
        d = d.AddYears(years);
        return new DateOnly(d.Year, d.Month, d.Day);
    }

    public DayOfWeek DayOfWeek
    {
        get
        {
            return _dateValue.DayOfWeek;
        }
    }

    public DateTime ToDateTime()
    {
        return _dateValue;
    }

    public int Year
    {
        get
        {
            return _dateValue.Year;
        }
        set
        {
            _dateValue = new DateTime(value, Month, Day);
        }
    }

    public int Month
    {
        get
        {
            return _dateValue.Month;
        }
        set
        {
            _dateValue = new DateTime(Year, value, Day);
        }
    }

    public int Day
    {
        get
        {
            return _dateValue.Day;
        }
        set
        {
            _dateValue = new DateTime(Year, Month, value);
        }
    }

    public static bool operator == (DateOnly aDateOnly1, DateOnly aDateOnly2)
    {
        return (aDateOnly1.ToDateTime() == aDateOnly2.ToDateTime());
    }

    public static bool operator != (DateOnly aDateOnly1, DateOnly aDateOnly2)
    {
        return (aDateOnly1.ToDateTime() != aDateOnly2.ToDateTime());
    }

    public static bool operator > (DateOnly aDateOnly1, DateOnly aDateOnly2)
    {
        return (aDateOnly1.ToDateTime() > aDateOnly2.ToDateTime());
    }

    public static bool operator < (DateOnly aDateOnly1, DateOnly aDateOnly2)
    {
        return (aDateOnly1.ToDateTime() < aDateOnly2.ToDateTime());
    }

    public static bool operator >= (DateOnly aDateOnly1, DateOnly aDateOnly2)
    {
        return (aDateOnly1.ToDateTime() >= aDateOnly2.ToDateTime());
    }

    public static bool operator <= (DateOnly aDateOnly1, DateOnly aDateOnly2)
    {
        return (aDateOnly1.ToDateTime() <= aDateOnly2.ToDateTime());
    }

    public static TimeSpan operator - (DateOnly aDateOnly1, DateOnly aDateOnly2)
    {
        return (aDateOnly1.ToDateTime() - aDateOnly2.ToDateTime());
    }


    public override string ToString()
    {
        return _dateValue.ToShortDateString();
    }

    public string ToString(string format)
    {
        return _dateValue.ToString(format);
    }

    public string ToString(string fmt, IFormatProvider provider)
    {
        return string.Format("{0:" + fmt + "}", _dateValue);
    }

    public string ToShortDateString()
    {
        return _dateValue.ToShortDateString();
    }

    public string ToDbFormat()
    {
        return string.Format("{0:yyyy-MM-dd}", _dateValue);
    }
}

这是从VB转换过来的。NET,所以抱歉,如果一些转换不是100%

其他回答

试试这个,如果你使用DateTimeOffset,它也会照顾时区

date1 = date1.LocalDateTime.Date;

如果你需要增加工作时间,可以这样做:

date1 = date1.LocalDateTime.Date;
date1 = date1.AddHours(23).AddMinutes(59).AddSeconds(59);

下面是另一个使用String的方法。格式

    DateTime todaysDate = DateTime.UtcNow;

    string dateString = String.Format("{0:dd/MM/yyyy}", todaysDate);

    Console.WriteLine("Date with Time: "+ todaysDate.ToString());

    Console.WriteLine("Date Only : " + dateString);

输出:

Date with Time: 9/4/2016 11:42:16 AM

Date Only : 04/09/2016

如果日期时间存储在数据库中,这也适用。

有关更多日期和时间格式,请检查这些链接:

参考1

参考2

希望有帮助。

我认为你会这样:DateTime onlyDate = DateTime. today . date; 或者,这是相同的DateTime onlyDate = yourDateTime.Date; 所以使用属性Date。

我知道这是一个有很多答案的老帖子,但我还没有见过这种删除时间部分的方法。假设您有一个名为myDate的DateTime变量,其中日期和时间部分。你可以从它创建一个新的DateTime对象,没有时间部分,使用这个构造函数:

public DateTime(int year, int month, int day);

是这样的:

myDate = new DateTime(myDate.Year, myDate.Month, myDate.Day);

通过这种方式,您可以基于旧的DateTime对象创建一个新的DateTime对象,其中00:00:00作为时间部分。

这可以简单地这样做:

var dateOnly = new DateTime(dateTime.Year, dateTime.Month, dateTime.Day)