I\'m trying to figure out how to calculate the number of \"Tuesdays\" between two dates in TSQL?
\"Tuesday\"could be any value.
Thank you t-clausen.dk, Saved me few days. To get no of instances of each day:
declare @from datetime= '3/1/2013'
declare @to datetime = '3/31/2013'
select
datediff(day, -7, @to)/7-datediff(day, -6, @from)/7 AS MON,
datediff(day, -6, @to)/7-datediff(day, -5, @from)/7 AS TUE,
datediff(day, -5, @to)/7-datediff(day, -4, @from)/7 AS WED,
datediff(day, -4, @to)/7-datediff(day, -3, @from)/7 AS THU,
datediff(day, -3, @to)/7-datediff(day, -2, @from)/7 AS FRI,
datediff(day, -2, @to)/7-datediff(day, -1, @from)/7 AS SAT,
datediff(day, -1, @to)/7-datediff(day, 0, @from)/7 AS SUN
@t-clausen.dk & Andriy M as response to t-clausen.dks response and comments
The query uses the fact that 1900-01-01 was a monday. And 1900-01-01 is the date 0.
select dateadd(day,0,0)
The second parameter into the datediff
-function is the startdate.
So you are comparing '1899-12-26' with your @to-date and '1899-12-26' is a tuesday
select datename(dw,dateadd(day, 0, -6)), datename(dw, '1899-12-26')
Same thing about the second date that uses the same fact.
As a matter of fact you can compare with any known tuesday and corresponding wednesday (that isnt in the date interval you are investigating).
declare @from datetime= '2011-09-19'
declare @to datetime = '2011-10-15'
select datediff(day, '2011-09-13', @to)/7-datediff(day, '2011-09-14', @from)/7 as [works]
,datediff(day, '2011-10-18', @to)/7-datediff(day, '2011-10-19', @from)/7 as [works too]
,datediff(day, '2011-09-27', @to)/7-datediff(day, '2011-09-28', @from)/7 as [dont work]
Basically the algorithm is "All Tuesdays minus all Wednesdays".
declare @from datetime= '9/20/2011'
declare @to datetime = '9/28/2011'
select datediff(day, -6, @to)/7-datediff(day, -5, @from)/7
Check out this question: Count work days between two dates
There are a few ways you can leverage the answer to that question for yours as well.