给定日期范围，我需要知道该范围内有多少个星期一（或星期二、星期三等）。
我目前正在使用 C#。
ThePeeje
问问题
33567 次
14 回答
50
试试这个：
static int CountDays(DayOfWeek day, DateTime start, DateTime end)
{
TimeSpan ts = end - start; // Total duration
int count = (int)Math.Floor(ts.TotalDays / 7); // Number of whole weeks
int remainder = (int)(ts.TotalDays % 7); // Number of remaining days
int sinceLastDay = (int)(end.DayOfWeek - day); // Number of days since last [day]
if (sinceLastDay < 0) sinceLastDay += 7; // Adjust for negative days since last [day]
// If the days in excess of an even week are greater than or equal to the number days since the last [day], then count this one, too.
if (remainder >= sinceLastDay) count++;
return count;
}
于 2008-10-29T21:40:02.273 回答
22
由于您使用的是 C#，如果您使用的是 C#3.0，则可以使用 LINQ。
假设您有一个 Array/List/IQueryable 等，其中包含您的日期作为 DateTime 类型：
DateTime[] dates = { new DateTime(2008,10,6), new DateTime(2008,10,7)}; //etc....
var mondays = dates.Where(d => d.DayOfWeek == DayOfWeek.Monday); // = {10/6/2008}
添加：
不确定您是否意味着对它们进行分组并计算它们，但这里也是在 LINQ 中执行此操作的方法：
var datesgrouped = from d in dates
group d by d.DayOfWeek into grouped
select new { WeekDay = grouped.Key, Days = grouped };
foreach (var g in datesgrouped)
{
Console.Write (String.Format("{0} : {1}", g.WeekDay,g.Days.Count());
}
于 2008-10-29T20:47:32.053 回答
20
看看计算星期几的不同算法很有趣，@Gabe Hollombe 在这个主题上指向 WP 是一个好主意（我记得大约 20 年前在 COBOL 中实现了Zeller 的同余），但它更像是当他们问现在几点时，递给某人一张时钟的蓝图。
在 C# 中：
private int CountMondays(DateTime startDate, DateTime endDate)
{
int mondayCount = 0;
for (DateTime dt = startDate; dt < endDate; dt = dt.AddDays(1.0))
{
if (dt.DayOfWeek == DayOfWeek.Monday)
{
mondayCount++;
}
}
return mondayCount;
}
这当然不会评估“星期一”的结束日期，因此如果需要，请让 for 循环评估
dt < endDate.AddDays(1.0)
于 2008-10-29T20:50:10.800 回答
5
这是一些伪代码：
DifferenceInDays(Start, End) / 7 // Integer division discarding remainder
+ 1 if DayOfWeek(Start) <= DayImLookingFor
+ 1 if DayOfWeek(End) >= DayImLookingFor
- 1
where以天为单位返回，并DifferenceInDays以整数形式返回星期几。映射使用什么并不重要，只要它不断增加并与.End - StartDayOfWeekDayOfWeekDayImLookingFor
请注意，此算法假定日期范围包含在内。如果End不应该是范围的一部分，则必须稍微调整算法。
翻译成 C# 留给读者作为练习。
于 2008-10-29T20:48:29.883 回答
3
任何特定的语言和日期格式？
如果日期表示为天数，则两个值之间的差加一（天）除以 7 就是大部分答案。如果两个结束日期都是相关日期，则添加一个。
已编辑：将“模 7”更正为“除以 7”-谢谢。那就是整数除法。
于 2008-10-29T20:24:24.253 回答
2
你可以试试这个，如果你想在两个日期之间获得特定的工作日
public List<DateTime> GetSelectedDaysInPeriod(DateTime startDate, DateTime endDate, List<DayOfWeek> daysToCheck)
{
var selectedDates = new List<DateTime>();
if (startDate >= endDate)
return selectedDates; //No days to return
if (daysToCheck == null || daysToCheck.Count == 0)
return selectedDates; //No days to select
try
{
//Get the total number of days between the two dates
var totalDays = (int)endDate.Subtract(startDate).TotalDays;
//So.. we're creating a list of all dates between the two dates:
var allDatesQry = from d in Enumerable.Range(1, totalDays)
select new DateTime(
startDate.AddDays(d).Year,
startDate.AddDays(d).Month,
startDate.AddDays(d).Day);
//And extracting those weekdays we explicitly wanted to return
var selectedDatesQry = from d in allDatesQry
where daysToCheck.Contains(d.DayOfWeek)
select d;
//Copying the IEnumerable to a List
selectedDates = selectedDatesQry.ToList();
}
catch (Exception ex)
{
//Log error
//...
//And re-throw
throw;
}
return selectedDates;
}
于 2011-12-06T09:02:07.740 回答
1
添加可能的最小数字以使第一天成为星期一。减去可能的最小数字，使最后一天成为星期一。计算天数的差异并除以 7。
于 2008-10-29T20:33:06.917 回答
1
将日期转换为儒略日数，然后做一些数学运算。由于星期一是零 mod 7，你可以这样计算：
JD1=JulianDayOf(the_first_date)
JD2=JulianDayOf(the_second_date)
Round JD1 up to nearest multiple of 7
Round JD2 up to nearest multiple of 7
d = JD2-JD1
nMondays = (JD2-JD1+7)/7 # integer divide
于 2008-10-29T20:50:37.040 回答
1
我今天也有同样的需要。我从cjm函数开始，因为我不了解JonB函数，而且Cyberherbalist函数不是线性的。
我不得不纠正
DifferenceInDays(Start, End) / 7 // Integer division discarding remainder
+ 1 if DayOfWeek(Start) <= DayImLookingFor
+ 1 if DayOfWeek(End) >= DayImLookingFor
- 1
到
DifferenceInDays(Start, End) / 7 // Integer division discarding remainder
+ 1 if DayImLookingFor is between Start.Day and End.Day
如果从开始日开始，我们在 endDay 之前首先遇到 dayImLookingFor，则使用返回 true 的 between 函数。
我通过计算从 startDay 到另外两天的天数来完成 between 函数：
private int CountDays(DateTime start, DateTime end, DayOfWeek selectedDay)
{
if (start.Date > end.Date)
{
return 0;
}
int totalDays = (int)end.Date.Subtract(start.Date).TotalDays;
DayOfWeek startDay = start.DayOfWeek;
DayOfWeek endDay = end.DayOfWeek;
///look if endDay appears before or after the selectedDay when we start from startDay.
int startToEnd = (int)endDay - (int)startDay;
if (startToEnd < 0)
{
startToEnd += 7;
}
int startToSelected = (int)selectedDay - (int)startDay;
if (startToSelected < 0)
{
startToSelected += 7;
}
bool isSelectedBetweenStartAndEnd = startToEnd >= startToSelected;
if (isSelectedBetweenStartAndEnd)
{
return totalDays / 7 + 1;
}
else
{
return totalDays / 7;
}
}
于 2009-08-24T14:45:39.650 回答
1
这将返回一个整数集合，显示在一个日期范围内一周中的每一天发生了多少次
int[] CountDays(DateTime firstDate, DateTime lastDate)
{
var totalDays = lastDate.Date.Subtract(firstDate.Date).TotalDays + 1;
var weeks = (int)Math.Floor(totalDays / 7);
var result = Enumerable.Repeat<int>(weeks, 7).ToArray();
if (totalDays % 7 != 0)
{
int firstDayOfWeek = (int)firstDate.DayOfWeek;
int lastDayOfWeek = (int)lastDate.DayOfWeek;
if (lastDayOfWeek < firstDayOfWeek)
lastDayOfWeek += 7;
for (int dayOfWeek = firstDayOfWeek; dayOfWeek <= lastDayOfWeek; dayOfWeek++)
result[dayOfWeek % 7]++;
}
return result;
}
或稍有变化，可让您执行 FirstDate.TotalDaysOfWeeks(SecondDate) 并返回 Dictionary
public static Dictionary<DayOfWeek, int> TotalDaysOfWeeks(this DateTime firstDate, DateTime lastDate)
{
var totalDays = lastDate.Date.Subtract(firstDate.Date).TotalDays + 1;
var weeks = (int)Math.Floor(totalDays / 7);
var resultArray = Enumerable.Repeat<int>(weeks, 7).ToArray();
if (totalDays % 7 != 0)
{
int firstDayOfWeek = (int)firstDate.DayOfWeek;
int lastDayOfWeek = (int)lastDate.DayOfWeek;
if (lastDayOfWeek < firstDayOfWeek)
lastDayOfWeek += 7;
for (int dayOfWeek = firstDayOfWeek; dayOfWeek <= lastDayOfWeek; dayOfWeek++)
resultArray[dayOfWeek % 7]++;
}
var result = new Dictionary<DayOfWeek, int>();
for (int dayOfWeek = 0; dayOfWeek < 7; dayOfWeek++)
result[(DayOfWeek)dayOfWeek] = resultArray[dayOfWeek];
return result;
}
于 2012-02-23T18:01:36.940 回答
1
修改后的代码在这里有效并由我测试
private int CountDays(DayOfWeek day, DateTime startDate, DateTime endDate)
{
int dayCount = 0;
for (DateTime dt = startDate; dt < endDate; dt = dt.AddDays(1.0))
{
if (dt.DayOfWeek == day)
{
dayCount++;
}
}
return dayCount;
}
例子：
int Days = CountDays(DayOfWeek.Friday, Convert.ToDateTime("2019-07-04"),
Convert.ToDateTime("2019-07-27")).ToString();
于 2019-07-07T05:57:16.717 回答
0
我有一个类似的报告问题。我需要两个日期之间的工作日数。我本可以循环浏览日期并数数，但我的离散数学训练不允许我这样做。这是我在 VBA 中编写的一个函数，用于获取两个日期之间的工作日数。我确定 .net 具有类似的 WeekDay 功能。
1
2 ' WorkDays
3 ' returns the number of working days between two dates
4 Public Function WorkDays(ByVal dtBegin As Date, ByVal dtEnd As Date) As Long
5
6 Dim dtFirstSunday As Date
7 Dim dtLastSaturday As Date
8 Dim lngWorkDays As Long
9
10 ' get first sunday in range
11 dtFirstSunday = dtBegin + ((8 - Weekday(dtBegin)) Mod 7)
12
13 ' get last saturday in range
14 dtLastSaturday = dtEnd - (Weekday(dtEnd) Mod 7)
15
16 ' get work days between first sunday and last saturday
17 lngWorkDays = (((dtLastSaturday - dtFirstSunday) + 1) / 7) * 5
18
19 ' if first sunday is not begin date
20 If dtFirstSunday <> dtBegin Then
21
22 ' assume first sunday is after begin date
23 ' add workdays from begin date to first sunday
24 lngWorkDays = lngWorkDays + (7 - Weekday(dtBegin))
25
26 End If
27
28 ' if last saturday is not end date
29 If dtLastSaturday <> dtEnd Then
30
31 ' assume last saturday is before end date
32 ' add workdays from last saturday to end date
33 lngWorkDays = lngWorkDays + (Weekday(dtEnd) - 1)
34
35 End If
36
37 ' return working days
38 WorkDays = lngWorkDays
39
40 End Function
于 2008-10-29T20:52:36.083 回答
0
private System.Int32 CountDaysOfWeek(System.DayOfWeek dayOfWeek, System.DateTime date1, System.DateTime date2)
{
System.DateTime EndDate;
System.DateTime StartDate;
if (date1 > date2)
{
StartDate = date2;
EndDate = date1;
}
else
{
StartDate = date1;
EndDate = date2;
}
while (StartDate.DayOfWeek != dayOfWeek)
StartDate = StartDate.AddDays(1);
return EndDate.Subtract(StartDate).Days / 7 + 1;
}
于 2008-10-29T21:57:14.517 回答
0
四年后，我想我会做一个测试：
[TestMethod]
public void ShouldFindFridaysInTimeSpan()
{
//reference: http://stackoverflow.com/questions/248273/count-number-of-mondays-in-a-given-date-range
var spanOfSixtyDays = new TimeSpan(60, 0, 0, 0);
var setOfDates = new List<DateTime>(spanOfSixtyDays.Days);
var now = DateTime.Now;
for(int i = 0; i < spanOfSixtyDays.Days; i++)
{
setOfDates.Add(now.AddDays(i));
}
Assert.IsTrue(setOfDates.Count == 60,
"The expected number of days is not here.");
var fridays = setOfDates.Where(i => i.DayOfWeek == DayOfWeek.Friday);
Assert.IsTrue(fridays.Count() > 0,
"The expected Friday days are not here.");
Assert.IsTrue(fridays.First() == setOfDates.First(i => i.DayOfWeek == DayOfWeek.Friday),
"The expected first Friday day is not here.");
Assert.IsTrue(fridays.Last() == setOfDates.Last(i => i.DayOfWeek == DayOfWeek.Friday),
"The expected last Friday day is not here.");
}
我的使用TimeSpan有点矫枉过正——实际上我想TimeSpan直接查询。
于 2012-01-26T22:04:38.940 回答