SQL Server: proper use of Select with calendar table

Question

I have created a calendar table that contains all the calendar dates of a year, incl. the corresponding quarter / week / month / day etc. information.

The following Select gives me all Fridays in December. What would be the best way here to get only the second out of these ?

SELECT     *
FROM       Calendar
WHERE      (yearCal = '2014') AND (monthCal = '12') AND (weekDayCal = '6')

Many thanks in advance, Mike.

Answer 1

Assuming that you have a day_of_month type of field, the second Friday must be day 8 to 14.

SELECT     *
FROM       Calendar
WHERE      (yearCal    = 2014)
       AND (monthCal   = 12)
       AND (weekDayCal = 6)
       AND (dayInMonth BETWEEN 8 AND 14)

You could use a Week of month field, but this can get complex depending on what you define as a week. Normally it would by Monday-Sunday or similar, meaning that week1 of month12 may only be 2 days long, and not include a Friday.

Answer 2

If this is a recurring requirement, then you should add a column weekInMonthCal and include it in the constraint clause.

It is possible to count in SQL but, as this involves aggregation in the result set, it is expensive.

Answer 3

You can use ROW_NUMBER() function to order your results (replace the column DateColumn with your column which stores dates. You can then get the second record.

WITH CTE AS
(SELECT     Col1, Col2, Col3, ROW_NUMBER() OVER (ORDER BY DateColumn) RowNumber
FROM       Calendar
WHERE      (yearCal = '2014') AND (monthCal = '12') AND (weekDayCal = '6'))

SELECT Col1, Col2, Col3 FROM CTE
WHERE RowNumber = 2

Answer 4

This would give you the second friday of a given month

SELECT TOP 1 *
FROM
(SELECT     top 2 *
FROM       Calendar
WHERE      (yearCal = '2014') AND (monthCal = '12') AND (weekDayCal = '6')
order by dateCal asc) a
order by dateCal desc