+3  A: 

You could change the check of the week:

int d = date.Day;
return date.DayOfWeek == dow && (d-1)/7 == (n-1);

Other than that, it looks pretty good and efficient.

Robert Wagner
i think that should be (d-1)/7 == (n-1)
Kevin
Thanks for that. Thinking in 0 based counting again.
Robert Wagner
A: 

Here is what the MSDN have to say. Its VB, but it translates easily.

Andrew Bullock
that kinda does the opposite of what I want
Kevin
yeah, but its easily reversed ;)
Andrew Bullock
+1  A: 

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(”The given month has less than ” nth.ToString() ” ”
day_of_the_week.ToString() “s”);
}

// Complete the gap to the nth week
dt = dt.AddDays((nth - 1) * 7);

return dt;
}
George Stocker
this also does the opposite of what I want
Kevin
+1  A: 

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!]

waynecolvin
A: 

In the above example 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”); }

A: 

You can find a function which returns date for nth occurrence of particular week day in any month. http://chiragrdarji.wordpress.com/2010/08/23/find-second-saturday-and-fourth-saturday-of-month/

Chirag Darji