Find nth day of month in sql server 2005 - sql

Question

I want to find nth day of given month(eg:- first Sunday,second Tuesday,last friday etc) in sql server 2005.I had used
DATEADD(WEEK, DATEDIFF(WEEK, 0,
DATEADD(DAY, 6 - DATEPART(DAY, GETDATE()), GETDATE())), 6)
it will return first Sunday of a month but don't know how to find the nth day of month I need to return nth day as per the user's request
Can any please help to find this
Any help or suggestions would be grateful
Thanks And Regards

Answer 1

Many data warehouses have a date dimension. Things like dates and day of week are contained in this table. It can be indexed and has relatively good speed since 100 years x 365 days a year is not much data.
Below I use a tally table to create a common table expression for this year. I create another one for November 2014. From there, I use a windowing function to rank order the day of weeks by occurrence.
If you did have all the data in the date dimension, you could even place the occurrence number in the table for that month.
The example below pulls the third Friday of the month of November 2014.
--
-- Why not use a date dimension if you need this functionality often?
--
;
with
cte_Tally
as
(
SELECT ROW_NUMBER() over (order by (select 1)) as rn
FROM master.sys.all_columns c
),
cte_Year
as
(
select
rn as my_doy_num,
dateadd(d, rn, '20140101') as my_day_dte,
month(dateadd(d, rn, '20140101')) as my_month_num,
case month(dateadd(d, rn, '20140101'))
when 1 then 'jan'
when 2 then 'feb'
when 3 then 'mar'
when 4 then 'apr'
when 5 then 'may'
when 6 then 'jun'
when 7 then 'jul'
when 8 then 'aug'
when 9 then 'sep'
when 10 then 'oct'
when 11 then 'nov'
when 12 then 'dec'
else ''
end as my_month_desc,
datepart(dw, dateadd(d, rn, '20140101')) as my_dow_num,
case datepart(dw, dateadd(d, rn, '20140101'))
when 1 then 'sun'
when 2 then 'mon'
when 3 then 'tue'
when 4 then 'wed'
when 5 then 'thu'
when 6 then 'fri'
when 7 then 'sat'
else ''
end as my_dow_desc
from cte_Tally
where rn < 365
),
cte_Nov_2014
as
(
select
my_day_dte,
my_month_desc,
my_dow_desc,
ROW_NUMBER() over (partition by my_dow_desc order by my_day_dte, my_dow_desc) as my_occurence
from cte_Year
where my_month_num = 11
--order by my_dow_desc
)
select *
from cte_Nov_2014
where my_dow_desc = 'fri' and my_occurence = 3

Answer 2

-- 1 find the first nth weekday (Mon/Tue/Wed/Thu/Fri/Sat/Sun) of the month where #dt is in
-- for example, find the 5th Friday of the month of Aug, 2013
declare #nth int=5, #dt datetime='2013-08-12';
declare #dw tinyint =5 -- 1=Mon,2= Tue,3=Wed, 4=Thur, 5=Fri, 6=Sat, 7=Sun
select [1st_nth_wkday]=case when datepart(dw, dateadd(month,
datediff(month,0,#dt),0)+##datefirst-1) >= #dw
then dateadd(day, (#nth-1)*7 + (7-(datepart(dw, dateadd(month,
datediff(month,0,#dt),0)+##datefirst-1)-#dw)), dateadd(month,
datediff(month,0,#dt),0))
else dateadd(day, (#nth-1)*7 + (0-(datepart(dw, dateadd(month,
datediff(month,0,#dt),0)+##datefirst-1)-#dw)), dateadd(month,
datediff(month,0,#dt),0))
end
go
-- 2 find the last nth weekday (Mon/Tue/Wed/Thu/Fri/Sat/Sun) of the month
where #dt is in
-- find the 2nd last Sunday of current month
declare #nth int=2, #dt datetime=current_timestamp;
declare #dw tinyint =7 -- 1=Mon,2= Tue,3=Wed, 4=Thur, 5=Fri, 6=Sat, 7=Sun
select [last_nth_wkday]=case when datepart(dw, dateadd(month,
datediff(month,0,#dt)+1,0)-1+##datefirst-1) >= #dw
then dateadd(day, -(#nth-1)*7 - (datepart(dw, dateadd(month,
datediff(month,0,#dt)+1,0)-1+##datefirst-1)-#dw), dateadd(month,
datediff(month,0,#dt)+1,0)-1)
else dateadd(day, -(#nth)*7 - (datepart(dw, dateadd(month,
datediff(month,0,#dt)+1,0)-1+##datefirst-1)-#dw), dateadd(monI th,
datediff(month,0,#dt)+1,0)-1)
end
go
I actually blogged this here: http://www.sqlservercentral.com/blogs/jeffrey_yao/2014/03/02/fun-in-datetime-calculations/