我怎样才能在周三和下周周三的C＃中找到日期：
public Form1()
{
InitializeComponent();
CurrentDate.Text = "Today's Date: " + DateTime.Now.ToString("dd/MM/yyyy");
CurrentRent.Text = "Current Rent Date: "; // last wednesday
NextRent.Text = "Next Rent Date: "; // next wednesday
}
答案 0 :(得分：32)
要查找下周三，请继续添加天数，直至找到。要查找上一个星期三，只需减去天数，直到找到一个。
DateTime nextWednesday = DateTime.Now.AddDays(1);
while (nextWednesday.DayOfWeek != DayOfWeek.Wednesday)
nextWednesday = nextWednesday.AddDays(1);
DateTime lastWednesday = DateTime.Now.AddDays(-1);
while (lastWednesday.DayOfWeek != DayOfWeek.Wednesday)
lastWednesday = lastWednesday.AddDays(-1);
答案 1 :(得分：5)
DateTime.Now.AddDays(7)和DateTime.Now.AddDays(-7)是你如何做算术的。如果不是，那么您需要做的是使用DayOfWeek属性来确定确定哪一天是“星期三”所需的天数（正数和负数）。然后，您可以将该值传递到AddDays。
例如，如果今天是星期二，你将在上周三AddDays(-6)和下周三AddDays(8)。
我会告诉你计算这些的任务。
答案 2 :(得分：3)
使用AddDays例程：
// increment by the number of offset days to get the correct date
DayOfWeek desiredDay = DayOfWeek.Wednesday;
int offsetAmount = (int) desiredDay - (int) DateTime.Now.DayOfWeek;
DateTime lastWeekWednesday = DateTime.Now.AddDays(-7 + offsetAmount);
DateTime nextWeekWednesday = DateTime.Now.AddDays(7 + offsetAmount);
应该这样做！
注意：如果是星期一，“上周三”会给你发生的最后一个星期三，但是“下周三”会给你9天后的星期三！如果您希望在两天内获得星期三，则需要使用“％”运算符。这意味着第二个“nextweek”语句将显示为“（7 + offsetAmount）％7”。
答案 3 :(得分：2)
您可以使用它来计算它：
DateTime day = DateTime.Today;
while (day.DayOfWeek != DayOfWeek.Wednesday)
day = day.AddDays(-1);
var currentRent = day;
var nextRent = day.AddDays(7);
请注意，如果今天是星期三，则今天会显示currentRent，而今天不会显示nextRent。如果你想要反转，你可以颠倒逻辑。
DateTime day = DateTime.Today;
while (day.DayOfWeek != DayOfWeek.Wednesday)
day = day.AddDays(1);
var currentRent = day.AddDays(-7);
var nextRent = day;
答案 4 :(得分：1)
您需要使用DayOfWeek枚举以及switch语句的if-else结构来确定添加/减去日期的天数。这是繁琐的编码，但很简单。
DateTime nextRent;
DateTime lastRent;
DateTime today = DateTime.Now;
if (today.DayOfWeek == DayOfWeek.Wednesday)
{
nextRent = today.AddDays(7);
lastRent = today.AddDays(-7);
}
else if (today.DayOfWeek == DayOfWeek.Thursday)
{
nextRent = today.AddDays(6);
lastRent = today.AddDays(-8);
}
//ect for all days
答案 5 :(得分：1)
这会奏效。您需要计算您提供的日期与最近的星期三之间的天数差异，并根据差异是否大于零来计算上一个/下个星期三。
int difference = date.DayOfWeek - DayOfWeek.Wednesday;
DateTime lastWednesday = difference > 0 ? date.AddDays(-1 * difference) : date.AddDays(-1 * (7 + difference));
DateTime nextWednesday = lastWednesday.AddDays(7);
答案 6 :(得分：1)
根据Servy的回答，这是一个扩展方法，该方法将返回所需的日期/日期：
public static DateTime GetPrevious(this DateTime date, DayOfWeek dayOfWeek)
{
var lastDay = date.AddDays(-1);
while (lastDay.DayOfWeek != dayOfWeek)
{
lastDay = lastDay.AddDays(-1);
}
return lastDay;
}
public static DateTime GetNext(this DateTime date, DayOfWeek dayOfWeek)
{
var nextDay = date.AddDays(+1);
while (nextDay.DayOfWeek != dayOfWeek)
{
nextDay = nextDay.AddDays(+1);
}
return nextDay;
}
答案 7 :(得分：1)
某些答案的问题之一是DayoyOfWeek是一个值为0-6的枚举。如果一周的后一天等于星期日（枚举值= 0），而您给出了星期二（枚举值= 2），而您想要下一个星期二，则仅执行计算就会出错。
我在自己正在处理的项目中使用了两种方法供自己使用的类。
public static class DateHelper
{
public static DateTime GetDateForLastDayOfWeek(DayOfWeek DOW, DateTime DATE)
{
int adjustment = ((int)DATE.DayOfWeek < (int)DOW ? 7 : 0);
return DATE.AddDays(0- (((int)(DATE.DayOfWeek) + adjustment) - (int)DOW));
}
public static DateTime GetDateForNextDayOfWeek(DayOfWeek DOW, DateTime DATE)
{
int adjustment = ((int)DATE.DayOfWeek < (int)DOW ? 0 : 7);
return DATE.AddDays(((int)DOW) - ((int)(DATE.DayOfWeek)) + adjustment);
}
}
XUnit测试证明了上面的代码有效。
public class DateHelperUnitTests
{
[Theory]
[InlineData(2020, 1, 7, 2020, 1, 7)]
[InlineData(2020, 1, 8, 2020, 1, 7)]
[InlineData(2020, 1, 9, 2020, 1, 7)]
[InlineData(2020, 1, 10, 2020, 1, 7)]
[InlineData(2020, 1, 11, 2020, 1, 7)]
[InlineData(2020, 1, 12, 2020, 1, 7)]
[InlineData(2020, 1, 13, 2020, 1, 7)]
[InlineData(2020, 1, 14, 2020, 1, 14)]
[InlineData(2020, 1, 15, 2020, 1, 14)]
public void GetDateForLastDayOfWeek_MultipleValues_Pass(
int InputYear, int InputMonth, int InputDay,
int ExpectedYear, int ExpectedMonth, int ExpectedDay)
{
DateTime DateToTest = new DateTime(InputYear, InputMonth, InputDay);
DateTime NewDate = DateHelper.GetDateForLastDayOfWeek(DayOfWeek.Tuesday, DateToTest);
DateTime DateExpected = new DateTime(ExpectedYear,ExpectedMonth,ExpectedDay);
Assert.True(0 == DateTime.Compare(DateExpected.Date, NewDate.Date));
}
[Theory]
[InlineData(2020, 1, 7, 2020, 1, 14)]
[InlineData(2020, 1, 8, 2020, 1, 14)]
[InlineData(2020, 1, 9, 2020, 1, 14)]
[InlineData(2020, 1, 10, 2020, 1, 14)]
[InlineData(2020, 1, 11, 2020, 1, 14)]
[InlineData(2020, 1, 12, 2020, 1, 14)]
[InlineData(2020, 1, 13, 2020, 1, 14)]
[InlineData(2020, 1, 14, 2020, 1, 21)]
[InlineData(2020, 1, 15, 2020, 1, 21)]
public void GetDateForNextDayOfWeek_MultipleValues_Pass(
int InputYear, int InputMonth, int InputDay,
int ExpectedYear, int ExpectedMonth, int ExpectedDay)
{
DateTime DateToTest = new DateTime(InputYear, InputMonth, InputDay);
DateTime NewDate = DateHelper.GetDateForNextDayOfWeek(DayOfWeek.Tuesday, DateToTest);
DateTime DateExpected = new DateTime(ExpectedYear, ExpectedMonth, ExpectedDay);
Assert.True(0 == DateTime.Compare(DateExpected.Date, NewDate.Date));
}
}
答案 8 :(得分：0)
来自Servy的改进答案。您需要在while循环中将nextWednesday或lastWednesday日期实际设置为新日期，否则它将进入无限循环
DateTime nextWednesday = DateTime.Now.AddDays(1);
while (nextWednesday.DayOfWeek != DayOfWeek.Wednesday)
nextWednesday = nextWednesday.AddDays(1);
DateTime lastWednesday = DateTime.Now.AddDays(-1);
while (lastWednesday.DayOfWeek != DayOfWeek.Wednesday)
lastWelastWednesday.AddDays(-1);
答案 9 :(得分：0)
您可以创建2个DateTime扩展方法，可以与DayOfWeek参数一起使用：
public static class DateTimeExtension
{
public static DateTime GetPreviousWeekDay(this DateTime currentDate, DayOfWeek dow)
{
int currentDay = (int)currentDate.DayOfWeek, gotoDay = (int)dow;
return currentDate.AddDays(-7).AddDays(gotoDay-currentDay);
}
public static DateTime GetNextWeekDay(this DateTime currentDate, DayOfWeek dow)
{
int currentDay = (int)currentDate.DayOfWeek, gotoDay = (int)dow;
return currentDate.AddDays(7).AddDays(gotoDay - currentDay);
}
}
然后可以按如下方式使用：
DateTime testDate = new DateTime(2017, 01, 21);
Console.WriteLine(testDate.GetPreviousWeekDay(DayOfWeek.Wednesday));
Console.WriteLine(testDate.GetNextWeekDay(DayOfWeek.Wednesday));
答案 10 :(得分：0)
private static DateTime FindPreviousDayOfWeek(DateTime fromDate, DayOfWeek findDay,
bool skipSame = false)
{
if (fromDate.DayOfWeek < findDay)
fromDate = fromDate.AddDays(-((int)fromDate.DayOfWeek - 1 + (int)findDay));
else if (fromDate.DayOfWeek > findDay)
fromDate = fromDate.AddDays(-((int)fromDate.DayOfWeek - (int)findDay));
else if (fromDate.DayOfWeek == findDay && skipSame == true)
fromDate = fromDate.AddDays(-7);
return fromDate;
}
如果将saem作为所需的日期，则skipSame变量不包括当前日期。
答案 11 :(得分：0)
这是一线实现的目标。理解它实际上是如何工作的，但确实如此：
下周三获取：
dt.AddDays(-(int)(dt.AddDays(-4).DayOfWeek) + 6);
上周三获取
dt.AddDays(-(int)(dt.AddDays(-3).DayOfWeek));
在两种情况下，它都会返回星期三本身的日期。 这适用于任何工作日，只需调整AddDays（）调用中的数字即可。 例如星期五：
下周五获得
dt.AddDays(-(int)(dt.AddDays(-6).DayOfWeek) + 6);
上周五获得
dt.AddDays(-(int)(dt.AddDays(-5).DayOfWeek));
答案 12 :(得分：0)
此扩展方法应该在每周的任何一天都可以解决问题
public static class DateTimeExtensions
{
public static DateTime LastDayOfWeek(this DateTime _date, DayOfWeek dayofweek)
{
return _date.AddDays(-1 * ((_date.DayOfWeek - dayofweek) % 7)).Date;
}
public static DateTime NextDayOfWeek(this DateTime _date, DayOfWeek dayofweek)
{
return _date.LastDayOfWeek(dayofweek).AddDays(7).Date;
}
}
用法
var lastWendsday = DateTime.Now.LastDayOfWeek(DayOfWeek.Wednesday);