Calculate date from week number

Question

Anyone know an easy way to get the date of the first day in the week (monday here in Europe). I know the year and the week number? I'm going to do this in C#.

Answer 1

this is my solution when we want to calculate a date given year, week number and day of the week.

int Year = 2014;
int Week = 48;
int DayOfWeek = 4;

DateTime FecIni = new DateTime(Year, 1, 1);
FecIni = FecIni.AddDays(7 * (Week - 1));
if ((int)FecIni.DayOfWeek > DayOfWeek)
{
    while ((int)FecIni.DayOfWeek != DayOfWeek) FecIni = FecIni.AddDays(-1);
}
else
{
    while ((int)FecIni.DayOfWeek != DayOfWeek) FecIni = FecIni.AddDays(1);
}

Answer 2

I used one of the solutions but it gave me wrong results, simply because it counts Sunday as a first day of the week.

I changed:

var firstDay = new DateTime(DateTime.Now.Year, 1, 1).AddDays((weekNumber - 1) * 7);
var lastDay = firstDay.AddDays(6);

to:

var lastDay = new DateTime(DateTime.Now.Year, 1, 1).AddDays((weekNumber) * 7);
var firstDay = lastDay.AddDays(-6);

and now it is working as a charm.

Answer 3

The free Time Period Library for .NET includes the ISO 8601 conform class Week:

// ----------------------------------------------------------------------
public static DateTime GetFirstDayOfWeek( int year, int weekOfYear )
{
  return new Week( year, weekOfYear ).FirstDayStart;
} // GetFirstDayOfWeek

Answer 4

Good news! A pull request adding System.Globalization.ISOWeek to .NET Core was just merged and is currently slated for the 3.0 release. Hopefully it will propagate to the other .NET platforms in a not-too-distant future.

You should be able to use the ISOWeek.ToDateTime(int year, int week, DayOfWeek dayOfWeek) method to calculate this.

You can find the source code here.

Answer 5

Assuming the week number starts at 1

DateTime dt =  new DateTime(YearNumber, 1, 1).AddDays((WeekNumber - 1) * 7 - (WeekNumber == 1 ? 0 : 1));
return dt.AddDays(-(int)dt.DayOfWeek);

This should give you the first day in any given week. I haven't done a lot of testing on it, but looks like it works. It's smaller solution than most other's I found on the web, so wanted to share.