How do I find the start of the week (both Sunday and Monday) knowing just the current time in C#?
Something like:
DateTime.Now.StartWeek(Monday);
Use an extension method:
public static class DateTimeExtensions
{
public static DateTime StartOfWeek(this DateTime dt, DayOfWeek startOfWeek)
{
int diff = (7 + (dt.DayOfWeek - startOfWeek)) % 7;
return dt.AddDays(-1 * diff).Date;
}
}
Which can be used as follows:
DateTime dt = DateTime.Now.StartOfWeek(DayOfWeek.Monday);
DateTime dt = DateTime.Now.StartOfWeek(DayOfWeek.Sunday);
The quickest way I can come up with is:
var sunday = DateTime.Today.AddDays(-(int)DateTime.Today.DayOfWeek);
If you would like any other day of the week to be your start date, all you need to do is add the DayOfWeek value to the end
var monday = DateTime.Today.AddDays(-(int)DateTime.Today.DayOfWeek + (int)DayOfWeek.Monday);
var tuesday = DateTime.Today.AddDays(-(int)DateTime.Today.DayOfWeek + (int)DayOfWeek.Tuesday);
A little more verbose and culture-aware:
System.Globalization.CultureInfo ci =
System.Threading.Thread.CurrentThread.CurrentCulture;
DayOfWeek fdow = ci.DateTimeFormat.FirstDayOfWeek;
DayOfWeek today = DateTime.Now.DayOfWeek;
DateTime sow = DateTime.Now.AddDays(-(today - fdow)).Date;
Using Fluent DateTime:
var monday = DateTime.Now.Previous(DayOfWeek.Monday);
var sunday = DateTime.Now.Previous(DayOfWeek.Sunday);
Ugly but it at least gives the right dates back
With start of week set by system:
public static DateTime FirstDateInWeek(this DateTime dt)
{
while (dt.DayOfWeek != System.Threading.Thread.CurrentThread.CurrentCulture.DateTimeFormat.FirstDayOfWeek)
dt = dt.AddDays(-1);
return dt;
}
Without:
public static DateTime FirstDateInWeek(this DateTime dt, DayOfWeek weekStartDay)
{
while (dt.DayOfWeek != weekStartDay)
dt = dt.AddDays(-1);
return dt;
}
Let's combine the culture-safe answer and the extension method answer:
public static class DateTimeExtensions
{
public static DateTime StartOfWeek(this DateTime dt, DayOfWeek startOfWeek)
{
System.Globalization.CultureInfo ci = System.Threading.Thread.CurrentThread.CurrentCulture;
DayOfWeek fdow = ci.DateTimeFormat.FirstDayOfWeek;
return DateTime.Today.AddDays(-(DateTime.Today.DayOfWeek- fdow));
}
}
This would give you the preceding Sunday (I think):
DateTime t = DateTime.Now;
t -= new TimeSpan ((int) t.DayOfWeek, 0, 0, 0);
For Monday
DateTime startAtMonday = DateTime.Now.AddDays(DayOfWeek.Monday - DateTime.Now.DayOfWeek);
For Sunday
DateTime startAtSunday = DateTime.Now.AddDays(DayOfWeek.Sunday- DateTime.Now.DayOfWeek);
This may be a bit of a hack, but you can cast the .DayOfWeek property to an int (it's an enum and since its not had its underlying data type changed it defaults to int) and use that to determine the previous start of the week.
It appears the week specified in the DayOfWeek enum starts on Sunday, so if we subtract 1 from this value that'll be equal to how many days the Monday is before the current date. We also need to map the Sunday (0) to equal 7 so given 1 - 7 = -6 the Sunday will map to the previous Monday:-
DateTime now = DateTime.Now;
int dayOfWeek = (int)now.DayOfWeek;
dayOfWeek = dayOfWeek == 0 ? 7 : dayOfWeek;
DateTime startOfWeek = now.AddDays(1 - (int)now.DayOfWeek);
The code for the previous Sunday is simpler as we don't have to make this adjustment:-
DateTime now = DateTime.Now;
int dayOfWeek = (int)now.DayOfWeek;
DateTime startOfWeek = now.AddDays(-(int)now.DayOfWeek);
using System;
using System.Globalization;
namespace MySpace
{
public static class DateTimeExtention
{
// ToDo: Need to provide culturaly neutral versions.
public static DateTime GetStartOfWeek(this DateTime dt)
{
DateTime ndt = dt.Subtract(TimeSpan.FromDays((int)dt.DayOfWeek));
return new DateTime(ndt.Year, ndt.Month, ndt.Day, 0, 0, 0, 0);
}
public static DateTime GetEndOfWeek(this DateTime dt)
{
DateTime ndt = dt.GetStartOfWeek().AddDays(6);
return new DateTime(ndt.Year, ndt.Month, ndt.Day, 23, 59, 59, 999);
}
public static DateTime GetStartOfWeek(this DateTime dt, int year, int week)
{
DateTime dayInWeek = new DateTime(year, 1, 1).AddDays((week - 1) * 7);
return dayInWeek.GetStartOfWeek();
}
public static DateTime GetEndOfWeek(this DateTime dt, int year, int week)
{
DateTime dayInWeek = new DateTime(year, 1, 1).AddDays((week - 1) * 7);
return dayInWeek.GetEndOfWeek();
}
}
}
Putting it all together, with Globalization and allowing for specifying the first day of the week as part of the call we have
public static DateTime StartOfWeek ( this DateTime dt, DayOfWeek? firstDayOfWeek )
{
DayOfWeek fdow;
if ( firstDayOfWeek.HasValue )
{
fdow = firstDayOfWeek.Value;
}
else
{
System.Globalization.CultureInfo ci = System.Threading.Thread.CurrentThread.CurrentCulture;
fdow = ci.DateTimeFormat.FirstDayOfWeek;
}
int diff = dt.DayOfWeek - fdow;
if ( diff < 0 )
{
diff += 7;
}
return dt.AddDays( -1 * diff ).Date;
}
Step 1:
Create a static class
public static class TIMEE
{
public static DateTime StartOfWeek(this DateTime dt, DayOfWeek startOfWeek)
{
int diff = (7 + (dt.DayOfWeek - startOfWeek)) % 7;
return dt.AddDays(-1 * diff).Date;
}
public static DateTime EndOfWeek(this DateTime dt, DayOfWeek startOfWeek)
{
int diff = (7 - (dt.DayOfWeek - startOfWeek)) % 7;
return dt.AddDays(1 * diff).Date;
}
}
Step 2: Use this class to get both start and end day of the week
DateTime dt = TIMEE.StartOfWeek(DateTime.Now ,DayOfWeek.Monday);
DateTime dt1 = TIMEE.EndOfWeek(DateTime.Now, DayOfWeek.Sunday);
var now = System.DateTime.Now;
var result = now.AddDays(-((now.DayOfWeek - System.Threading.Thread.CurrentThread.CurrentCulture.DateTimeFormat.FirstDayOfWeek + 7) % 7)).Date;
This would give you midnight on the first Sunday of the week:
DateTime t = DateTime.Now;
t -= new TimeSpan ((int) t.DayOfWeek, t.Hour, t.Minute, t.Second);
This gives you the first Monday at midnight:
DateTime t = DateTime.Now;
t -= new TimeSpan ((int) t.DayOfWeek - 1, t.Hour, t.Minute, t.Second);
Try with this in C#. With this code you can get both the first date and last date of a given week. Here Sunday is the first day and Saturday is the last day, but you can set both days according to your culture.
DateTime firstDate = GetFirstDateOfWeek(DateTime.Parse("05/09/2012").Date, DayOfWeek.Sunday);
DateTime lastDate = GetLastDateOfWeek(DateTime.Parse("05/09/2012").Date, DayOfWeek.Saturday);
public static DateTime GetFirstDateOfWeek(DateTime dayInWeek, DayOfWeek firstDay)
{
DateTime firstDayInWeek = dayInWeek.Date;
while (firstDayInWeek.DayOfWeek != firstDay)
firstDayInWeek = firstDayInWeek.AddDays(-1);
return firstDayInWeek;
}
public static DateTime GetLastDateOfWeek(DateTime dayInWeek, DayOfWeek firstDay)
{
DateTime lastDayInWeek = dayInWeek.Date;
while (lastDayInWeek.DayOfWeek != firstDay)
lastDayInWeek = lastDayInWeek.AddDays(1);
return lastDayInWeek;
}
I tried several, but I did not solve the issue with a week starting on a Monday, resulting in giving me the coming Monday on a Sunday. So I modified it a bit and got it working with this code:
int delta = DayOfWeek.Monday - DateTime.Now.DayOfWeek;
DateTime monday = DateTime.Now.AddDays(delta == 1 ? -6 : delta);
return monday;
The same for end of the week (in style of Compile This's answer):
public static DateTime EndOfWeek(this DateTime dt)
{
int diff = 7 - (int)dt.DayOfWeek;
diff = diff == 7 ? 0 : diff;
DateTime eow = dt.AddDays(diff).Date;
return new DateTime(eow.Year, eow.Month, eow.Day, 23, 59, 59, 999) { };
}
Thanks for the examples. I needed to always use the "CurrentCulture" first day of the week and for an array I needed to know the exact Daynumber.. so here are my first extensions:
public static class DateTimeExtensions
{
//http://stackoverflow.com/questions/38039/how-can-i-get-the-datetime-for-the-start-of-the-week
//http://stackoverflow.com/questions/1788508/calculate-date-with-monday-as-dayofweek1
public static DateTime StartOfWeek(this DateTime dt)
{
//difference in days
int diff = (int)dt.DayOfWeek - (int)CultureInfo.CurrentCulture.DateTimeFormat.FirstDayOfWeek; //sunday=always0, monday=always1, etc.
//As a result we need to have day 0,1,2,3,4,5,6
if (diff < 0)
{
diff += 7;
}
return dt.AddDays(-1 * diff).Date;
}
public static int DayNoOfWeek(this DateTime dt)
{
//difference in days
int diff = (int)dt.DayOfWeek - (int)CultureInfo.CurrentCulture.DateTimeFormat.FirstDayOfWeek; //sunday=always0, monday=always1, etc.
//As a result we need to have day 0,1,2,3,4,5,6
if (diff < 0)
{
diff += 7;
}
return diff + 1; //Make it 1..7
}
}
Here is a correct solution. The following code works regardless if the first day of the week is a Monday or a Sunday or something else.
public static class DateTimeExtension
{
public static DateTime GetFirstDayOfThisWeek(this DateTime d)
{
CultureInfo ci = System.Threading.Thread.CurrentThread.CurrentCulture;
var first = (int)ci.DateTimeFormat.FirstDayOfWeek;
var current = (int)d.DayOfWeek;
var result = first <= current ?
d.AddDays(-1 * (current - first)) :
d.AddDays(first - current - 7);
return result;
}
}
class Program
{
static void Main()
{
System.Threading.Thread.CurrentThread.CurrentCulture = CultureInfo.GetCultureInfo("en-US");
Console.WriteLine("Current culture set to en-US");
RunTests();
Console.WriteLine();
System.Threading.Thread.CurrentThread.CurrentCulture = CultureInfo.GetCultureInfo("da-DK");
Console.WriteLine("Current culture set to da-DK");
RunTests();
Console.ReadLine();
}
static void RunTests()
{
Console.WriteLine("Today {1}: {0}", DateTime.Today.Date.GetFirstDayOfThisWeek(), DateTime.Today.Date.ToString("yyyy-MM-dd"));
Console.WriteLine("Saturday 2013-03-02: {0}", new DateTime(2013, 3, 2).GetFirstDayOfThisWeek());
Console.WriteLine("Sunday 2013-03-03: {0}", new DateTime(2013, 3, 3).GetFirstDayOfThisWeek());
Console.WriteLine("Monday 2013-03-04: {0}", new DateTime(2013, 3, 4).GetFirstDayOfThisWeek());
}
}
Modulo in C# works bad for -1 mod 7 (it should be 6, but C# returns -1)
so... a "one-liner" solution to this will look like this :)
private static DateTime GetFirstDayOfWeek(DateTime date)
{
return date.AddDays(
-(((int)date.DayOfWeek - 1) -
(int)Math.Floor((double)((int)date.DayOfWeek - 1) / 7) * 7));
}
I did it for Monday, but with similar logic for Sunday.
public static DateTime GetStartOfWeekDate()
{
// Get today's date
DateTime today = DateTime.Today;
// Get the value for today. DayOfWeek is an enum with 0 being Sunday, 1 Monday, etc
var todayDayOfWeek = (int)today.DayOfWeek;
var dateStartOfWeek = today;
// If today is not Monday, then get the date for Monday
if (todayDayOfWeek != 1)
{
// How many days to get back to Monday from today
var daysToStartOfWeek = (todayDayOfWeek - 1);
// Subtract from today's date the number of days to get to Monday
dateStartOfWeek = today.AddDays(-daysToStartOfWeek);
}
return dateStartOfWeek;
}
The following method should return the DateTime that you want. Pass in true for Sunday being the first day of the week, false for Monday:
private DateTime getStartOfWeek(bool useSunday)
{
DateTime now = DateTime.Now;
int dayOfWeek = (int)now.DayOfWeek;
if(!useSunday)
dayOfWeek--;
if(dayOfWeek < 0)
{// day of week is Sunday and we want to use Monday as the start of the week
// Sunday is now the seventh day of the week
dayOfWeek = 6;
}
return now.AddDays(-1 * (double)dayOfWeek);
}
You could use the excellent Umbrella library:
using nVentive.Umbrella.Extensions.Calendar;
DateTime beginning = DateTime.Now.BeginningOfWeek();
However, they do seem to have stored Monday as the first day of the week (see the property nVentive.Umbrella.Extensions.Calendar.DefaultDateTimeCalendarExtensions.WeekBeginsOn), so that previous localized solution is a bit better. Unfortunate.
Edit: looking closer at the question, it looks like Umbrella might actually work for that too:
// Or DateTime.Now.PreviousDay(DayOfWeek.Monday)
DateTime monday = DateTime.Now.PreviousMonday();
DateTime sunday = DateTime.Now.PreviousSunday();
Although it's worth noting that if you ask for the previous Monday on a Monday, it'll give you seven days back. But this is also true if you use BeginningOfWeek, which seems like a bug :(.
Following on from Compile This' answer, use the following method to obtain the date for any day of the week:
public static DateTime GetDayOfWeek(DateTime dateTime, DayOfWeek dayOfWeek)
{
var monday = dateTime.Date.AddDays((7 + (dateTime.DayOfWeek - DayOfWeek.Monday) % 7) * -1);
var diff = dayOfWeek - DayOfWeek.Monday;
if (diff == -1)
{
diff = 6;
}
return monday.AddDays(diff);
}
This will return both the beginning of the week and the end of the week dates:
private string[] GetWeekRange(DateTime dateToCheck)
{
string[] result = new string[2];
TimeSpan duration = new TimeSpan(0, 0, 0, 0); //One day
DateTime dateRangeBegin = dateToCheck;
DateTime dateRangeEnd = DateTime.Today.Add(duration);
dateRangeBegin = dateToCheck.AddDays(-(int)dateToCheck.DayOfWeek);
dateRangeEnd = dateToCheck.AddDays(6 - (int)dateToCheck.DayOfWeek);
result[0] = dateRangeBegin.Date.ToString();
result[1] = dateRangeEnd.Date.ToString();
return result;
}
I have posted the complete code for calculating the begin/end of week, month, quarter and year on my blog
ZamirsBlog
Calculating this way lets you choose which day of the week indicates the start of a new week (in the example I chose Monday).
Note that doing this calculation for a day that is a Monday will give the current Monday and not the previous one.
//Replace with whatever input date you want
DateTime inputDate = DateTime.Now;
//For this example, weeks start on Monday
int startOfWeek = (int)DayOfWeek.Monday;
//Calculate the number of days it has been since the start of the week
int daysSinceStartOfWeek = ((int)inputDate.DayOfWeek + 7 - startOfWeek) % 7;
DateTime previousStartOfWeek = inputDate.AddDays(-daysSinceStartOfWeek);
I work with a lot of schools, so correctly using Monday as the first day of the week is important here.
A lot of the most terse answers here don't work on Sunday -- we often end up returning the date of tomorrow on Sunday, which is not good for running a report on last week's activities.
Here's my solution, which returns last Monday on Sunday, and today on Monday.
// Adding 7 so remainder is always positive; Otherwise % returns -1 on Sunday.
var daysToSubtract = (7 + (int)today.DayOfWeek - (int)DayOfWeek.Monday) % 7;
var monday = today
.AddDays(-daysToSubtract)
.Date;
Remember to use a method parameter for "today" so it's unit testable!!
Here is a combination of a few of the answers. It uses an extension method that allows the culture to be passed in. If one is not passed in, the current culture is used. This will give it maximum flexibility and reuse.
/// <summary>
/// Gets the date of the first day of the week for the date.
/// </summary>
/// <param name="date">The date to be used</param>
/// <param name="cultureInfo">If none is provided, the current culture is used</param>
/// <returns>The date of the beggining of the week based on the culture specifed</returns>
public static DateTime StartOfWeek(this DateTime date, CultureInfo cultureInfo=null) =>
date.AddDays(-1 * (7 + (date.DayOfWeek - (cultureInfo ?? CultureInfo.CurrentCulture).DateTimeFormat.FirstDayOfWeek)) % 7).Date;
Example Usage:
public static void TestFirstDayOfWeekExtension() {
DateTime date = DateTime.Now;
foreach(System.Globalization.CultureInfo culture in CultureInfo.GetCultures(CultureTypes.UserCustomCulture | CultureTypes.SpecificCultures)) {
Console.WriteLine($"{culture.EnglishName}: {date.ToShortDateString()} First Day of week: {date.StartOfWeek(culture).ToShortDateString()}");
}
}
If you want Saturday or Sunday or any day of week, but not exceeding the current week (Sat-Sun), I got you covered with this piece of code.
public static DateTime GetDateInCurrentWeek(this DateTime date, DayOfWeek day)
{
var temp = date;
var limit = (int)date.DayOfWeek;
var returnDate = DateTime.MinValue;
if (date.DayOfWeek == day)
return date;
for (int i = limit; i < 6; i++)
{
temp = temp.AddDays(1);
if (day == temp.DayOfWeek)
{
returnDate = temp;
break;
}
}
if (returnDate == DateTime.MinValue)
{
for (int i = limit; i > -1; i++)
{
date = date.AddDays(-1);
if (day == date.DayOfWeek)
{
returnDate = date;
break;
}
}
}
return returnDate;
}
We like one-liners: Get the difference between the current culture's first day of week and the current day, and then subtract the number of days from the current day:
var weekStartDate = DateTime.Now.AddDays(-((int)now.DayOfWeek - (int)DateTimeFormatInfo.CurrentInfo.FirstDayOfWeek));
Ok, so I have a date stored in UK format (dd/mm/yy) which I need to display in the locale of wherever the user might be.
The issue is that this date can be 000000 (00/00/2000); so I can't convert it to DateTime directly, as DateTime doesn't support 0 values for day or month.
I have this so far:
int dateInt = ddmmyy;
var year = (dateInt % 100) + 2000;
var month = (dateInt / 100) % 100;
var day = (dateInt / 100000);
var result = new DateTime(year, month, day); //2014/00/00 at this point, so breaks.
var resultStr = result.ToString(CultureInfo.InvariantCulture);
return resultStr;
What's the correct way to add support for 0 values initially? I've tried changing the 0 to 1 before converting to DateTime, running the conversion and then replacing with a 0 again; but due to culture variants I see no way that this method can support other cultures, which is the purpose of this conversion to begin with.
Any ideas? I'm guessing this is a common issue.
Is this what you need ?
using System;
namespace ConsoleApplication2
{
class Program
{
static void Main(string[] args)
{
int[] savedDates = new int[] { 000000, 010000, 000013 };
foreach (var item in savedDates)
{
DateTime date = ConvertToDate(item);
Console.WriteLine(item.ToString("D6") + " => " + date.ToShortDateString());
}
Console.ReadLine();
}
private static DateTime ConvertToDate(int item)
{
string temp = item.ToString("D6");
int day = int.Parse(temp.Substring(0, 2));
int month = int.Parse(temp.Substring(2, 2));
int year = int.Parse(temp.Substring(4, 2));
if (day == 0)
day = 1;
if (month == 0)
month = 1;
year += 2000;
return new DateTime(year, month, day);
}
}
}
I would not store dates like this as the methodology for doing so is already provided by the .NET framework.
The best way to store dates would be to use Culture.InvariantCulture for string conversion cases and then convert to local culture for display purposes as necessary. DateTime itself is culture-independent so converting between cultures is very easy.
I have a function which takes two DateTime parameters and I have to add separate offsets to these date. I know that DateTime has a AddDays function to add days to a date and it throws an exception if DateTime is less than MinValue or greater than MaxValue.
Now I want to do a safe check whether adding/subtracting the following number of days to a DateTime can cause over/under flow or not.
safeStartDate = (startDate == DateTime.MinValue || startDate == DateTime.MaxValue) ? startDate : startDate.AddDays(startDateOffset);
safeEndDate = (endDate == DateTime.MaxValue || endDate == DateTime.MinValue) ? endDate : endDate.AddDays(enDateOffset);
By doing this, I am making it one level exception free but date can be DateTime.Max - 1 and while trying to add offset it throws an exception. I am looking a better way that whether the final values over/under flows without doing the actual calculation, in order to prevent exception.
If catch is not called very often you can do:
try
{
safeDate = dt.AddDays(days);
}
catch (ArgumentOutOfRangeException)
{
safeDate = date;
}
Alternatively,
var maxDays = (DateTime.MaxValue - dt).TotalDays;
safeDate = (days <= maxDays) ? dt.AddDays(days) : dt;
Or if there are negative days:
var maxDays = (DateTime.MaxValue - dt).TotalDays;
var minDays = (DateTime.MinValue - dt).TotalDays;
return (minDays <= days && days <= maxDays) ? dt.AddDays(days) : dt;
Or just use the method from Rawling's answer: CanAddDays(dt, days) ? dt.AddDays(days) : dt
The try/catch version is about 25% faster if you don't catch and about 1000x slower if you do. So, if you expected to catch more than about 1 time in every 5000 uses, then use the second version.
You can use the following to check whether you can add a given number of days to a given DateTime without causing an overflow:
bool CanAddDays(DateTime dt, int days)
{
double maxDaysToAdd = (DateTime.MaxValue - dt).TotalDays;
double minDaysToAdd = (DateTime.MinValue - dt).TotalDays;
return days <= maxDaysToAdd && days >= minDaysToAdd;
}
You might consider the following method:
private static DateTime AddDays(DateTime dateTime, int days)
{
var daysTimeSpanTicks = (new TimeSpan(days, 0, 0, 0)).Ticks;
return (days >= 0) ?
(DateTime.MaxValue.Ticks < dateTime.Ticks + daysTimeSpanTicks) ? dateTime : dateTime.AddDays(days) :
(dateTime.Ticks + daysTimeSpanTicks < 0) ? dateTime : dateTime.AddDays(days);
}
A sample usage is:
DateTime date = DateTime.MinValue;
DateTime safe = AddDays(date, -100);
I guess you are looking for something like this
DateTime Now = DateTime.Now;
DateTime Max = DateTime.MaxValue;
Max.Subtract(Now);
int DaysToAdd = 1000;//or something else
if (Max.Day > DaysToAdd) Now.AddDays(DaysToAdd);//add
Here is what I am trying to do:
Given a date, a day of the week, and an integer n, determine whether the date is the nth day of the month.
For example:
input of 1/1/2009,Monday,2
would be false because 1/1/2009 is not the second Monday
input of
11/13/2008,Thursday,2
would return true because it is the second Thursday
How can I improve this implementation?
private bool NthDayOfMonth(DateTime date, DayOfWeek dow, int n)
{
int d = date.Day;
return date.DayOfWeek == dow && (d/ 7 == n || (d/ 7 == (n - 1) && d % 7 > 0));
}
You could change the check of the week so the function would read:
private bool NthDayOfMonth(DateTime date, DayOfWeek dow, int n){
int d = date.Day;
return date.DayOfWeek == dow && (d-1)/7 == (n-1);
}
Other than that, it looks pretty good and efficient.
The answer is from this website. Copy/pasted here in case that site is ever lost.
public static DateTime FindTheNthSpecificWeekday(int year, int month,int nth, System.DayOfWeek day_of_the_week)
{
// validate month value
if(month < 1 || month > 12)
{
throw new ArgumentOutOfRangeException("Invalid month value.");
}
// validate the nth value
if(nth < 0 || nth > 5)
{
throw new ArgumentOutOfRangeException("Invalid nth value.");
}
// start from the first day of the month
DateTime dt = new DateTime(year, month, 1);
// loop until we find our first match day of the week
while(dt.DayOfWeek != day_of_the_week)
{
dt = dt.AddDays(1);
}
if(dt.Month != month)
{
// we skip to the next month, we throw an exception
throw new ArgumentOutOfRangeException(string.Format("The given month has less than {0} {1}s", nth, day_of_the_week));
}
// Complete the gap to the nth week
dt = dt.AddDays((nth - 1) * 7);
return dt;
}
Here is what the MSDN have to say. Its VB, but it translates easily.
It looks like the language supplies date/day methods for a given date. If anybody was interested you can read about Zeller's congruence.
I don't think that's what they wanted you to do but you could find the day of week of the first day of a month from that. Now that I thought about it you could find the day of week for the given day as N and get that modulo 7.
Oh wait, is that the Nth occurance of a day of the week (like Sunday) or like the Nth weekday of the month! Okay I see the examples.
Maybe it would make a difference if you could construct a date such as the 1st of a month..
Given that it is Nth occurance of a day of the week, and that you can't fiddle with whatever datetime datatype, and that you have access to both a get day of week and get day of month functions. Would Sunday be a zero?
1) First, the day of the week would have to match the day of the week given.
2) N would have to be at least 1 and at most 4.
3) The day of the month would range between n*7*dayOfWeek + 1 and n*7*dayOfWeek + 6 for the same n.
- Let me think about that. If Sunday was the first.. 0*7*0+1 = 1 and Saturday the 6th would be 0*7*0+6.
Think 1 and 3 above are sufficient since a get day of month function shouldn't violate 2.
(* first try, this code sucks *)
function isNthGivenDayInMonth(date : dateTime;
dow : dayOfWeek;
N : integer) : boolean;
var B, A : integer (* on or before and after day of month *)
var Day : integer (* day of month *)
begin
B := (N-1)*7 + 1; A := (N-1)*7 + 6;
D := getDayOfMonth(date);
if (dow <> getDayOfWeek(date)
then return(false)
else return( (B <= Day) and (A >= Day) );
end; (* function *)
Hope there isn't a bug in that lol!
[edit: Saturday would have been the 7th, and the upper bound above (N-1)*7 + 7.]
Your solution looks like it would match 2 different weeks? Looks like it would always return zero for Sundays? Should have done pseudocode in C#.. short circuit && is like my if..
hey shouldn't Sunday the first match for N = 1 in months that start on Sunday?
d/ 7 == n
That would result in (either 0 or 1)/7 == 1, that can't be right! Your || catches the (n-1) also, Robert has that. Go with Robert Wagner's answer! It's only 2 lines, short is good! Having (Day-1) mod 7
[edit: (Day-1) div 7]
eliminates my unnecessary variables and 2 lines of setup.
For the record this should be checked for boundary cases and so forth like what if August 31st was a Sunday or Saturday.
[edit: Should have checked the end of week case too. Sorry!]
You can find a function which returns a date for the nth occurrence of particular week day in any month.
See http://chiragrdarji.wordpress.com/2010/08/23/find-second-saturday-and-fourth-saturday-of-month/
In this answer, the following code needs to be flipped:
// Complete the gap to the nth week
dt = dt.AddDays((nth - 1) * 7);
if(dt.Month != month)
{
// we skip to the next month, we throw an exception
throw new ArgumentOutOfRangeException(”The given month has less than ” nth.ToString() ” ”
day_of_the_week.ToString() “s”);
}
Most of the answers above are partially accurate or unnecessarily complex.
You could try this simpler function, which also checks if the given date is the last but Nth day of the month.
public static bool IsNthDayofMonth(this DateTime date, DayOfWeek weekday, int N)
{
if (N > 0)
{
var first = new DateTime(date.Year, date.Month, 1);
return (date.Day - first.Day)/ 7 == N - 1 && date.DayOfWeek == weekday;
}
else
{
var last = new DateTime(date.Year, date.Month, 1).AddMonths(1).AddDays(-1);
return (last.Day - date.Day) / 7 == (Math.Abs(N) - 1) && date.DayOfWeek == weekday;
}
In case you want a list of dates for a span of time (not just one) for the Nth DayOfWeek of a Month, you can use this:
internal static List<DateTime> GetDatesForNthDOWOfMonth(int weekNum, DayOfWeek DOW, DateTime beginDate, DateTime endDate)
{
List<DateTime> datesForNthDOWOfMonth = new List<DateTime>();
int earliestDayOfMonth = 1;
int latestDayOfMonth = 7;
DateTime currentDate = beginDate;
switch (weekNum)
{
case 1:
earliestDayOfMonth = 1;
latestDayOfMonth = 7;
break;
case 2:
earliestDayOfMonth = 8;
latestDayOfMonth = 14;
break;
case 3:
earliestDayOfMonth = 15;
latestDayOfMonth = 21;
break;
case 4:
earliestDayOfMonth = 22;
latestDayOfMonth = 28;
break;
}
while (currentDate < endDate)
{
DateTime dateToInc = currentDate;
DateTime endOfMonth = new DateTime(dateToInc.Year, dateToInc.Month, DateTime.DaysInMonth(dateToInc.Year, dateToInc.Month));
bool dateFound = false;
while (!dateFound)
{
dateFound = dateToInc.DayOfWeek.Equals(DOW);
if (dateFound)
{
if ((dateToInc.Day >= earliestDayOfMonth) &&
(dateToInc.Day <= latestDayOfMonth))
{
datesForNthDOWOfMonth.Add(dateToInc);
}
}
if (dateToInc.Date.Equals(endOfMonth.Date)) continue;
dateToInc = dateToInc.AddDays(1);
}
currentDate = new DateTime(currentDate.Year, currentDate.Month, 1);
currentDate = currentDate.AddMonths(1);
}
return datesForNthDOWOfMonth;
}
...and call it this way:
// This is to get the 1st Monday in each month from today through one year from today
DateTime beg = DateTime.Now;
DateTime end = DateTime.Now.AddYears(1);
List<DateTime> dates = GetDatesForNthDOWOfMonth(1, DayOfWeek.Monday, beg, end);
// To see the list of dateTimes, for verification
foreach (DateTime d in dates)
{
MessageBox.Show(string.Format("Found {0}", d.ToString()));
}
You could get the 2nd Friday of each month like so:
List<DateTime> dates = GetDatesForNthDOWOfMonth(2, DayOfWeek.Friday, beg, end);
...etc.
Here is a simple DateTime extension method to get the nth occurrence in a month.
/// <summary>
/// Gets the Nth occurrence of the specified weekday in the month.
/// </summary>
public static DateTime GetNthOfWeekDayInMonth(this DateTime dt, DayOfWeek dayOfWeek, int n)
{
//Get the first day of the month
DateTime fd = dt.AddDays(-dt.Day).AddDays(1);
// Get the FIRST occurrence of the specified weekday in the month.
fd = dayOfWeek >= fd.DayOfWeek ? fd.AddDays(dayOfWeek - fd.DayOfWeek) : fd.AddDays((7 + ((int)dayOfWeek)) - ((int)fd.DayOfWeek));
// Get the nth occurrence by adding the weeks
fd = fd.AddDays((n-1) * 7);
//Throw exception if you do not want to go past the specified month?
if (fd.Month != dt.Month) throw new Exception($"There is no {n} week in this month");
return fd;
}
Sample:
new DateTime(2022, 05, 12).GetNthOfWeekDayInMonth(DayOfWeek.Tuesday, 2);
// Output: [2022/05/10 00:00:00]
What is the C# optimised version of the following, without using .Net's Timespan or DateTime. How would I NUnit test it?
TimeSpan ts = Date1 - Date2;
int numberOfDays = ts.Days;
You could convert the dates to ticks, substract, then convert the ticks back to days. Though exactly why can't you use TimeSpan? It's likely doing just that.
Something like this:
DateTime a = DateTime.Now;
DateTime b = a.AddDays(2);
// ticks are in hns
long ticks = b.Ticks - a.Ticks;
long seconds = ticks / 10000000;
long minutes = seconds / 60;
long hours = minutes / 60;
long days = hours / 24;
I get back "2" for days.
Date1 and Date2 are what type? In your exameple it looks to be DateTime. You want your Date in what variable if it's not DateTime? You can always have your Date1 and Date2 in String and play with SubString() to get Year, Month and days but that would be a real pain to work with.
The optimized way to do your problem of:
TimeSpan ts = Date1 - Date2;
int numberOfDays = ts.Days;
is
DateTime Date1 = new DateTime(2008,12,01);
DateTime Date2 = new DateTime(2008,12,25);
int numberOfDayBetweenDateThatShouldBe = 25;
Assert.IsTrue((Date2-Date1).TotalDays == numberOfDayBetweenDateThatShouldBe);
thanks, this is homework (asp.net)
The task is to write an asp.net application in c# that takes 2 dates as input from the user, works out the number of days between them and displays that number back to the user. This should not use any of the in-built .net framework DateTime or TimeSpan classes.
and alas my most elegant implementation at production strength
private int CalculateDays(DateTime start, DateTime end )
{
DateTime origin = new DateTime();
return (end - origin).Days - (start - origin).Days;
}
protected void Button1_Click(object sender, EventArgs e)
{
// Parse dates for correctness and range errors, warn as necessary
DateTime start;
DateTime end;
// rough error implement error handling implementation
string errors = string.Empty;
if(!DateTime.TryParse(txt_start_date.Text.Trim(), out start)) errors+="Start date was incorrect";
else if(!DateTime.TryParse(txt_end_date.Text.Trim(), out end)) errors+= (errors.Length>0? errors+= "\n":"") + "End date was incorrect" ;
else if ((end.Day - start.Day) <= 0) errors+= (errors.Length>0? errors+= "\n":"" ) + "End date must be greater than the Start date" ; //CultureInfo.InvariantCulture
else
{
Response.Write(CalculateDays(start, end));
}
Debug.Assert(errors.Length <= 0, errors);
}
Reason being am unable to see how to do this without touching DateTime.Days at least.
edit
okay thought over again and a bit of monkey work, however, no use of DateTime or Timespan anywhere
public static class PrimitiveCal
{
// incremental total days a normal year
private static int[] DaysInYr = new int[] {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365 };
// incremental total days in a leap year
private static int[] DaysInLeapYr = new int[] { 0,31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366 };
public static long GetDays(int[] dt_start, int[] dt_end)
{
int day_diff = dt_end[0] - dt_start[0];
int s_mth = dt_start[1];
int e_mth = dt_end[1];
int s_year = dt_start[2];
int e_year = dt_end[2];
int yr_diff = e_year - s_year;
return day_diff + (yr_diff > 0 ?
// use months as boundaries, cater for leap years
(YrType(e_year)[e_mth - 1] +
(YrType(s_year)[YrType(s_year).Length - 1] - YrType(s_year)[s_mth - 1])) +
(yr_diff == 1 ? 0 : AddMiddleYears(s_year, e_year))
// get month sums in same year
: YrType(e_year)[e_mth - 1] - YrType(e_year)[s_mth - 1]);
}
private static int AddMiddleYears(int s_year, int e_year)
{
int total_days = 0;
for (int i = s_year + 1; i <= e_year - 1; i++) total_days += YrType(i)[YrType(i).Length - 1];
return total_days;
}
private static int[] YrType(int year)
{
return (year % 4 == 0 && year%100!=0) || year % 400 ==0 ? DaysInLeapYr : DaysInYr;
}
}