日期时间 - 获取下周二答案

【问题标题】：Datetime - Get next tuesday日期时间 - 获取下周二
【发布时间】：2011-09-14 19:46:55
【问题描述】：

如何获得下周二的日期？

在 PHP 中，就像strtotime('next tuesday'); 一样简单。

如何在 .NET 中实现类似的功能

【问题讨论】：

ASP.NET 是一组 Web 技术。 C# 是一种语言。您确实需要从纯 .NET 的角度来考虑这一点。现在，对于“下周二”——是“今天之后的第一个周二”吗？如果是星期一，有人说“下星期二见”，我希望这意味着 8 天而不是 1 天。如果今天是星期二呢？您需要一天中的什么时间？
如果今天是星期二，你想找出下一个星期二的日期吗？或者今天是星期一，你想从星期一开始找到第二个星期二？
最近的星期二是任何特定的日子。
@brenjtL：如果已经是星期二？
如果已经是星期二，那么就在同一天

标签： c# .net date

【解决方案1】：

包含或排除当前日期的非常简单的示例，您指定您感兴趣的日期和星期几。

public static class DateTimeExtensions
{
    /// <summary>
    /// Gets the next date.
    /// </summary>
    /// <param name="date">The date to inspected.</param>
    /// <param name="dayOfWeek">The day of week you want to get.</param>
    /// <param name="exclDate">if set to <c>true</c> the current date will be excluded and include next occurrence.</param>
    /// <returns></returns>
    public static DateTime GetNextDate(this DateTime date, DayOfWeek dayOfWeek, bool exclDate = true)
    {
        //note: first we need to check if the date wants to move back by date - Today, + diff might move it forward or backwards to Today
        //eg: date - Today = 0 - 1 = -1, so have to move it forward
        var diff = dayOfWeek - date.DayOfWeek;
        var ddiff = date.Date.Subtract(DateTime.Today).Days + diff;

        //note: ddiff < 0 : date calculates to past, so move forward, even if the date is really old, it will just move 7 days from date passed in
        //note: ddiff >= (exclDate ? 6 : 7) && diff < 0 : date is into the future, so calculated future weekday, based on date
        if (ddiff < 0 || ddiff >= (exclDate ? 6 : 7) && diff < 0)
            diff += 7; 

        //note: now we can get safe values between 0 - 6, especially if past dates is being used
        diff = diff % 7;

        //note: if diff is 0 and we are excluding the date passed, we will add 7 days, eg: 1 week
        diff += diff == 0 & exclDate ? 7 : 0;

        return date.AddDays(diff);
    }
}

一些测试用例

[TestMethod]
    public void TestNextDate()
    {
        var date = new DateTime(2013, 7, 15);
        var start = date;
        //testing same month - forwardOnly
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Tuesday)); //16
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Wednesday)); //17
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Thursday)); //18
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Friday)); //19
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Saturday)); //20
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Sunday)); //21
        Assert.AreEqual(start.AddDays(1), date.GetNextDate(DayOfWeek.Monday)); //22

        //testing same month - include date
        Assert.AreEqual(start = date, date.GetNextDate(DayOfWeek.Monday, false)); //15
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Tuesday, false)); //16
        Assert.AreEqual(start.AddDays(1), date.GetNextDate(DayOfWeek.Wednesday, false)); //17

        //testing month change - forwardOnly
        date = new DateTime(2013, 7, 29);
        start = date;
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Tuesday)); //30
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Wednesday)); //31
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Thursday)); //2013/09/01-month increased
        Assert.AreEqual(start.AddDays(1), date.GetNextDate(DayOfWeek.Friday)); //02

        //testing year change
        date = new DateTime(2013, 12, 30);
        start = date;
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Tuesday)); //31
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Wednesday)); //2014/01/01 - year increased
        Assert.AreEqual(start = start.AddDays(1), date.GetNextDate(DayOfWeek.Thursday)); //02
    }

【讨论】：

经过一些广泛的测试后，我对原始答案进行了额外的更改。现在，这将根据使用的日期、过去、现在和未来安全地计算第二天。前面的所有示例都很棒，但在某些条件下都失败了。我没有使它成为一个单行语句，以便可以对计算所做的事情进行额外的 cmets。 Jon Skeet 的正面案例很棒，虽然我的案例是从一个日期向后移动 1 天，但仍然大于今天，如果它移动到今天或昨天怎么办......这解决了它。

【解决方案2】：

Objective C 版本：

+(NSInteger) daysUntilNextWeekday: (NSDate*)startDate withTargetWeekday: (NSInteger) targetWeekday
{
    NSInteger startWeekday = [[NSCalendar currentCalendar] component:NSCalendarUnitWeekday fromDate:startDate];
    return (targetWeekday - startWeekday + 7) % 7;
}

【讨论】：

很酷的答案，但最初的问题是关于 .NET。

【解决方案3】：

它也可以是一个扩展，这完全取决于

public static class DateTimeExtensions
{
    public static IEnumerable<DateTime> Next(this DateTime date, DayOfWeek day)
    {
        // This loop feels expensive and useless, but the point is IEnumerable
        while(true)
        {
            if (date.DayOfWeek == day)
            {
                yield return date;
            }
            date = date.AddDays(1);
        }
    }
}

用法

    var today = DateTime.Today;
    foreach(var monday in today.Next(DayOfWeek.Monday))
    {
        Console.WriteLine(monday);
        Console.ReadKey();
    }

【讨论】：

没有退出子句的循环收益回报似乎有点冒险。

【解决方案4】：

现在是单线风格 - 以防您需要将其作为参数传递给某些机制。

DateTime.Now.AddDays(((int)yourDate.DayOfWeek - (int)DateTime.Now.DayOfWeek + 7) % 7).Day

在这种特定情况下：

DateTime.Now.AddDays(((int)DayOfWeek.Tuesday - (int)DateTime.Now.DayOfWeek + 7) % 7).Day

【讨论】：

【解决方案5】：

@Jon Skeet 好答案。

前一天：

private DateTime GetPrevWeekday(DateTime start, DayOfWeek day) {
    // The (... - 7) % 7 ensures we end up with a value in the range [0, 6]
    int daysToRemove = ((int) day - (int) start.DayOfWeek - 7) % 7;
    return start.AddDays(daysToRemove);
}

谢谢！！

【讨论】：

请注意，此解决方案涉及传递给模运算符的负数。 Wikipedia article about the modulo operator 说 “当 a 或 n 为负数时，天真的定义就会失效，编程语言在定义这些值的方式上会有所不同。” 虽然这可能在 C# 中有效，但在数学上更 '获得相同结果的可靠解决方案是像这样交换 DayOfWeek 值：int daysToSubtract = -(((int)dateTime.DayOfWeek - (int)day + 7) % 7);

【解决方案6】：

DateTime nexttuesday=DateTime.Today.AddDays(1);

while(nexttuesday.DayOfWeek!=DayOfWeek.Tuesday)
   nexttuesday = nexttuesday.AddDays(1);

【讨论】：

【解决方案7】：

对于这个问题有更少冗长和更聪明/优雅的解决方案，但以下 C# 函数在许多情况下都非常有效。

/// <summary>
/// Find the closest weekday to the given date
/// </summary>
/// <param name="includeStartDate">if the supplied date is on the specified day of the week, return that date or continue to the next date</param>
/// <param name="searchForward">search forward or backward from the supplied date. if a null parameter is given, the closest weekday (ie in either direction) is returned</param>
public static DateTime ClosestWeekDay(this DateTime date, DayOfWeek weekday, bool includeStartDate = true, bool? searchForward=true)
{
    if (!searchForward.HasValue && !includeStartDate) 
    {
        throw new ArgumentException("if searching in both directions, start date must be a valid result");
    }
    var day = date.DayOfWeek;
    int add = ((int)weekday - (int)day);
    if (searchForward.HasValue)
    {
        if (add < 0 && searchForward.Value)
        {
            add += 7;
        }
        else if (add > 0 && !searchForward.Value)
        {
            add -= 7;
        }
        else if (add == 0 && !includeStartDate)
        {
            add = searchForward.Value ? 7 : -7;
        }
    }
    else if (add < -3) 
    {
        add += 7; 
    }
    else if (add > 3)
    {
        add -= 7;
    }
    return date.AddDays(add);
}

【讨论】：

唯一实现为 DateTime 扩展的答案。虽然其他解决方案都可以工作，但将其作为扩展方法会产生最容易使用的代码。

【解决方案8】：

正如我在 cmets 中提到的，“下周二”可以表示多种含义，但这段代码给出了“下周二发生，或者今天如果已经是周二”：

DateTime today = DateTime.Today;
// The (... + 7) % 7 ensures we end up with a value in the range [0, 6]
int daysUntilTuesday = ((int) DayOfWeek.Tuesday - (int) today.DayOfWeek + 7) % 7;
DateTime nextTuesday = today.AddDays(daysUntilTuesday);

如果您想在已经是星期二的情况下给出“一周的时间”，您可以使用：

// This finds the next Monday (or today if it's Monday) and then adds a day... so the
// result is in the range [1-7]
int daysUntilTuesday = (((int) DayOfWeek.Monday - (int) today.DayOfWeek + 7) % 7) + 1;

...或者您可以使用原始公式，但从明天开始：

DateTime tomorrow = DateTime.Today.AddDays(1);
// The (... + 7) % 7 ensures we end up with a value in the range [0, 6]
int daysUntilTuesday = ((int) DayOfWeek.Tuesday - (int) tomorrow.DayOfWeek + 7) % 7;
DateTime nextTuesday = tomorrow.AddDays(daysUntilTuesday);

编辑：只是为了让它变得漂亮和多才多艺：

public static DateTime GetNextWeekday(DateTime start, DayOfWeek day)
{
    // The (... + 7) % 7 ensures we end up with a value in the range [0, 6]
    int daysToAdd = ((int) day - (int) start.DayOfWeek + 7) % 7;
    return start.AddDays(daysToAdd);
}

所以要获得“今天或未来 6 天”的值：

DateTime nextTuesday = GetNextWeekday(DateTime.Today, DayOfWeek.Tuesday);

获取“下周二不包括今天”的值：

DateTime nextTuesday = GetNextWeekday(DateTime.Today.AddDays(1), DayOfWeek.Tuesday);

【讨论】：

哇，我只是想知道我如何才能在下周二之前获得第 n 天，然后你用一个很好的例子更新了你的答案。谢谢
很难选择正确的答案。但你的似乎是最多才多艺的，你让它很容易理解。谢谢你的帮助。
@brenjt：实际上我会说 Sven 的用途更广泛，因为您可以指定星期几，但这是您的决定 :)（我现在编辑了我的版本以提供更通用的版本.)
+7)%7 解决方案非常好。虽然我没有使用它的原因是因为它有点微优化并且太容易出错（以及牺牲一些可读性），但恕我直言。
单元测试：[TestMethod] public void ShouldGetNextSaturday() { var now = DateTime.Now; var test = GetNextWeekday(DateTime.Today, DayOfWeek.Saturday); Assert.IsTrue(now.Day

【解决方案9】：

DateTime nextTuesday = DateTime.Today.AddDays(((int)DateTime.Today.DayOfWeek - (int)DayOfWeek.Tuesday) + 7);

【讨论】：

如果今天是星期一，您提供的答案将是从星期二开始的一周，而不是明天。

【解决方案10】：

这应该可以解决问题：

static DateTime GetNextWeekday(DayOfWeek day)
{
    DateTime result = DateTime.Now.AddDays(1);
    while( result.DayOfWeek != day )
        result = result.AddDays(1);
    return result;
}

【讨论】：

很好的回应，如果今天是星期二（哈哈），今天或下星期二会返回吗？
这将在下周二返回。如果您希望它今天返回，只需从第一行中删除 .AddDays(1)，这样它还会检查 DateTime.Now 本身。