I want to get month value using week no.
I have week numbers stored in a table with year value.
How to query database to get month value using that week value.
I am using SQL
You can try this:
SELECT DATEPART(m,DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + CAST(t.year as VARCHAR(4))) + (t.week-1), 6))
It depends on how you're classing your week numbers, For example, if we assume that week numbers start on a Monday then we'd have to say that week 1 in 2016 actually started on Monday 28th of December 2015 and finished on Sunday 3rd January 2016. If this is how your week numbers are set up then you can use the method below
Sample Data;
CREATE TABLE #DateTable (WeekNum int, YearNum int)
INSERT INTO #DateTable (WeekNum, YearNum)
VALUES
(1,2016)
,(2,2016)
,(3,2016)
,(4,2016)
,(5,2016)
,(6,2016)
,(7,2016)
We will then cast the week and year into a date, then convert this to a month;
SELECT
WeekNum
,YearNum
,DATEADD(wk, DATEDIFF(wk, 7, '1/1/' + CONVERT(varchar(4),YearNum)) + (WeekNum-1), 7) AS WeekStart
,DATEPART(mm,DATEADD(wk, DATEDIFF(wk, 7, '1/1/' + CONVERT(varchar(4),YearNum)) + (WeekNum-1), 7)) MonthNum
(Edit: updated as source is int)
Gives these results;
WeekNum YearNum WeekStart MonthNum
1 2016 2015-12-28 00:00:00.000 12
2 2016 2016-01-04 00:00:00.000 1
3 2016 2016-01-11 00:00:00.000 1
4 2016 2016-01-18 00:00:00.000 1
5 2016 2016-01-25 00:00:00.000 1
6 2016 2016-02-01 00:00:00.000 2
7 2016 2016-02-08 00:00:00.000 2
You can't go from week number to month because weeks can occur in two different months. For example the 31st Jan 2016 and 1st Feb 2016 are both in week 6.
SELECT DATEPART(WEEK, '2016-01-31')
SELECT DATEPART(WEEK, '2016-02-01')
You can try the query below:
SELECT
[Week],
[Year],
'Output-Month' = MONTH(DATEADD(WEEK, [Week], DATEADD(WEEK, DATEDIFF(WEEK, '19050101', '01/01/' + CAST([Year] AS VARCHAR(4))), '19050101')))
FROM YourTable
1st is to get the 1st day of the year using this:
DATEADD(WEEK, DATEDIFF(WEEK, '19050101', '01/01/' + CAST([Year] AS VARCHAR(4))), '19050101')
2nd is to add your number of week using this:
DATEADD(WEEK, [Week], 'From 1st result')
Last is getting the number of Month using the MONTH function.
Related
I need to get the first day and the last day of the week based on the number of the week, year and month
My week starts on saturday and finish in friday
Example:
Year: 2020
Week: 45
Normal period of week: First day: 2020-10-31 ~ Last day: 2020-11-06
I need return something like
October: First day: 2020-10-31 ~ last day: 2020-10-31
November: First day 2020-11-01 ~ last day: 2020-11-06
my query to return last day of week:
select DATEADD (WEEK, #PcpSemana, DATEADD (YEAR, ('20' + LEFT(#PcpPeriodo,2))-1900, 0)) - 5 as lastDayOfWeek
my query to return first day of week
SELECT WeekStart = DATEADD(DAY,
(CEILING(DATEPART(DAY, DATEADD (WEEK, #PcpSemana, DATEADD (YEAR, ('20' + LEFT(#PcpPeriodo,2))-1900, 0)) - 5) / 7.0) - 1) * 7,
DATEADD(MONTH, DATEDIFF(MONTH, 0,DATEADD (WEEK, #PcpSemana, DATEADD (YEAR, ('20' + LEFT(#PcpPeriodo,2))-1900, 0)) - 5), 0));
I'm using SET DATEFIRST 6
I can't evolve much
PcpPeriodo contains YYMM ( 2011) = 2020 / 11 )
PcpSemana contains weeknumber (45) (01 ~ 53)
I am not sure what your data looks like, but if you have the first day of the week, you can split it among months as:
select weeks.*
from (values (convert(date, '2020-10-31'))) w(weekstart) cross apply
(values (dateadd(day, 6, w.weekstart), eomonth(w.weekstart))
) v(weekend, eom) cross apply
(values (w.weekstart,
case when v1.weekend <= v1.eom then v1.weekend else v1.eom end
),
(case when v1.weekend > v1.eom then dateadd(day, 1, v1.eom) end,
case when v1.weekend > 1.eom then v1.weekend
)
) weeks(weekstart, weekend)
where weeks.weekstart is not null;
This is using apply as a way of storing intermediate results, such as the last day of the month and when the week ends.
DECLARE #d datetime;
SET DATEFIRST 6;
SET #d = '2020-11-01';
WITH weektest as (
select #d as d
union all
select DATEADD(DAY,1,d) from weektest where d<='2020-11-6'
)
SELECT
d,
DATEPART(week, d),
DATEPART(weekday,d),
DATENAME(weekday,d)
from weektest;
output:
d
----------------------- ----------- ----------- ------------------------------
2020-11-01 00:00:00.000 45 2 Sunday
2020-11-02 00:00:00.000 45 3 Monday
2020-11-03 00:00:00.000 45 4 Tuesday
2020-11-04 00:00:00.000 45 5 Wednesday
2020-11-05 00:00:00.000 45 6 Thursday
2020-11-06 00:00:00.000 45 7 Friday
2020-11-07 00:00:00.000 46 1 Saturday
I need to find next final pay period under few conditions. The maturity is 3 months, but when the last payday is till the 3st business day of the current month. Оtherwise is the maturity 4 months. And Saturday and Sunday are not working days. For example: now in August 2019 are 3.08 and 4.08 not working day and when the customer pay the tax till 5.08(Monday - this is the 3st business day for a August) the next payday period is till end of Oktober 2019. Otherwise when the day is 6.08 is the period till end of November
IF #Schema = '1000'
BEGIN
SET #PayPeriod = 3
IF #Payday < DATEADD(DAY, CASE (DATEPART(WEEKDAY, DATEADD(MONTH, DATEDIFF(MONTH, 0, #Payday), 0)) + ##DATEFIRST) % 7
WHEN 6 THEN 2
WHEN 7 THEN 1 ELSE 0 END,
DATEADD(MONTH, DATEDIFF(MONTH, 0, Payday), 3))
SET #PayPeriod = EOMONTH(#Payday, #PayPeriod-1)
ELSE SET #PayPeriod = EOMONTH(#Payday, #PayPeriod)
END
And here is the result. After 06.08.2019 must be the payperiod end of November
2019-08-01 2019-10-31
2019-08-02 2019-10-31
2019-08-03 2019-10-31
2019-08-04 2019-11-30
2019-08-05 2019-11-30 here is the problem!
2019-08-06 2019-11-30
This should work for you:
declare #payday as datetime = '20190602'
;with cte as (
select datefromparts(year(#payday), month(#payday), 1) as monthday
union all
select dateadd(day, 1, monthday)
from cte
where day(monthday) < 10
)
select
case
when count(*) <= 3 then eomonth(#payday, 3)
else eomonth(#payday, 4)
end as MaturityEnd
from cte
where datepart(weekday, monthday) not in (7, 1)
and monthday <= #payday
Here we generate first couple of dates for the same month as in payment date and count business days smaller than payment date. Maturity end date is calculated based on the count of days.
It is not totally clear what you want from your question. I believe what you want is:
Give a date ##DATEFIRST find a date of maturity
3 months later unless
that date falls in the first 3 business days of a month than then
4 months later
Is that correct?
Also what does EOMONTH do? The use of it in your example does not seem to make sense compared to the stated requirements.
I would like to display Monday as the first day of the week and Sunday the last day of the week.
I am using for this report the Report server.
my query:
Select
Datum,
Sum(Prod) as Prod
FROM (
Select
intervaldate as Datum,
Sum(case when TabName = 'Produzierte Dosen' then DisplayUnits else 0 end) as Prod
from vwOeeIntervalCount
where
IntervalDateWeek >= dateadd(wk, datediff(wk, 0, getdate()) - 1, 0)
and IntervalDateWeek < dateadd(wk, datediff(wk, 0, getdate()), 0)
and IntervalDate >= dateadd(day,datediff(day,0,GETDATE())-6,0)
AND IntervalDate < dateadd(day,datediff(day,0,GETDATE()),0)
and CalculationName = 'Packaging'
group by intervaldate
)c
group by Datum
Here the result:
Date |Prod
2018-02-25 00:00:00.000 |1836528
2018-02-26 00:00:00.000 |8131127,99999999
EDIT:
Sorry here is my question.
I would like to display the Monday as the first day of the week. My query displays the Sunday as the first day of the week.
How can I do that?
By default MS SQL Server has configured sunday as first day of a week.
You can set the server config for the first day of a week to any day with the following command:
SET DATEFIRST { number | #number_var }
Value First day of the week is
1 Monday
2 Tuesday
3 Wednesday
4 Thursday
5 Friday
6 Saturday
7 (default, U.S. English) Sunday
Source: https://learn.microsoft.com/en-us/sql/t-sql/statements/set-datefirst-transact-sql
you can use this to the current week:
/*-- Week Start
--1->Sunday --2->Monday....*/
SELECT CAST(DATEADD(dd, -(DATEPART(dw, GETDATE()))+2, GETDATE()) AS DATE) [WeekStart],
CAST(DATEADD(dd, 8-(DATEPART(dw, GETDATE())), GETDATE()) AS DATE) [WeekEnd]
/*-- Week End
--7->Saturday --8-> Sunday...*/
Running this today will produce this result:
WeekStart WeekEnd
---------- ----------
2018-02-26 2018-03-04
if you have dates in your table, you can use them instead of the GETDATE()
I am looking for some T-SQL code that should pick the date which is "One Year back from current date (at the same time last Sunday in the month of January)".
For example:
Current day expected result
2017-02-05 2016-01-31
2017-01-05 2015-01-25
2018-02-19 2017-01-29
2018-01-19 2016-01-31
2019-02-28 2018-01-28
Please note: The year starts from last Sunday in January
I have some T-SQL code which is being used in SQL Server 2014:
select
convert(varchar(10), DATEADD(day, DATEDIFF(day, '19000107', DATEADD(month, DATEDIFF(MONTH, 0, CONVERT(date, CONVERT(VARCHAR(4), (CASE WHEN MONTH(GetDate()) = 1 THEN CONVERT(VARCHAR(4), GetDate(), 112) - 1 ELSE CONVERT(VARCHAR(4), GetDate(), 112) END), 112) + '0101')), 30)) / 7 * 7, '19000107'), 120)
The above code picks the date for current year's (last Sunday in January month). But I want T-SQL code to pick last year's (last Sunday's date in January month) date.
In detail - I want T-SQL code to produce expected result from below table
Current day T-SQL code answer expected result
2017-02-05 2017-01-29 2016-01-31
2017-01-05 2016-01-31 2015-01-25
2018-02-19 2018-01-28 2017-01-29
2018-01-19 2017-01-29 2016-01-31
2019-02-28 2019-01-27 2018-01-28
Any help please.
The best thing for this question is a numbers and date table. This answer shows you how to create one. Such a table is very handsome in many situations...
If I understand this correctly, you want the last Sunday in January of the previous year in all cases? Try this:
DECLARE #dummy TABLE(ID INT IDENTITY,YourDate DATE);
INSERT INTO #dummy VALUES
('2017-02-05'),('2017-01-05'),('2018-02-19'),('2018-01-19'),('2019-02-28');
WITH Years AS
(
SELECT * FROM (VALUES(2010),(2011),(2012),(2013),(2014),(2015),(2016),(2017),(2018),(2019),(2020)) AS t(Yr)
)
,LastSundays AS
(
SELECT Yr AS TheYear
,DATEADD(DAY,(DATEPART(WEEKDAY,LastOfJanuary) % 7)*(-1),LastOfJanuary) AS LastSundayOfJanuary
FROM Years
CROSS APPLY(SELECT CAST(CAST(Yr AS VARCHAR(4)) + '0131' AS DATE)) AS t(LastOfJanuary)
)
SELECT *
FROM #dummy AS d
INNER JOIN LastSundays AS ls ON YEAR(DATEADD(YEAR,-1,d.YourDate))=ls.TheYear;
The result (I do not understand row 2 and 4 completely...)
ID YourDate TheYear LastSundayOfJanuary
1 2017-02-05 2016 2016-01-31
2 2017-01-05 2016 2016-01-31 <--Your sample data is different...
3 2018-02-19 2017 2017-01-29
4 2018-01-19 2017 2017-01-29 <--Your sample data is different...
5 2019-02-28 2018 2018-01-28
Hint You might need to introduce ##DATEFIRST into your calculations...
Here is a way to do it without a date table (which is still a good idea BTW). Tested on all your inputs and it delivers the correct output each time. Obviously you would refactor this a bit as it's longwinded, just to show each step.
/* The input date. */
DECLARE
#input DATE = '2019-02-28';
/* The input date less one year. */
DECLARE
#date_minus_one_year DATE = DATEADD(yy,-1,#input);
/* The year part of the input date less one year. */
DECLARE
#year_date_part INT = DATEPART(yy,#date_minus_one_year);
/* 31 Jan of the previous year. */
DECLARE
#prev_year_jan_eom DATE = CAST(CAST(#year_date_part AS VARCHAR(4))+'-01-31' AS DATE);
/* What day of the week is 31 Jan of the previous year? */
DECLARE
#previous_eom_dw_part INT = DATEPART(dw,#prev_year_jan_eom);
/* Offest 31 Jan to the previous Sunday, won't change if the 31st is itself a Sunday. */
DECLARE
#output DATE = DATEADD(dd,1 - #previous_eom_dw_part,#prev_year_jan_eom);
/* Input and output */
SELECT
#input input
,#output [output];
I didn't think of a way to do it without the conditional in a case. It also uses the trick of casting a numeric year value to a January 1st date.
select case
when
datepart(dayofyear, dt) >
31 - datepart(weekday, dateadd(day, 30, cast(year(dt) as varchar(4))))
then
dateadd(day,
31 - datepart(weekday, dateadd(day, 30, cast(year(dt) as varchar(4)))),
cast(year(dt) as varchar(4))
)
else
dateadd(day,
31 - datepart(weekday, dateadd(day, 30, cast(year(dt) - 1 as varchar(4)))),
cast(year(dt) - 1 as varchar(4))
)
end
from (values
('20100201'), ('20110301'), ('20120401'),
('20130501'), ('20140601'), ('20150701'),
('20160801'), ('20170901'), ('20181001')
) t(dt)
Just for fun (untested)
select
dateadd(week,
-52 * ceil(sign(datediff(day, dt, hs)) + 0.5),
js
)
from
(select <date> dt) as t
cross apply
(
select 31 - datepart(weekday,
datefromparts(year(dt), 1, 31) as js
) t2;
SELECT
convert(varchar(10), DATEADD(day, DATEDIFF(day, '19000107', DATEADD(month, DATEDIFF(MONTH, 0, CONVERT(date, CONVERT(VARCHAR(4), (CASE WHEN MONTH(DATEADD(year,-1,GetDate())) = 1 THEN CONVERT(VARCHAR(4), DATEADD(year,-1,GetDate()), 112) - 1 ELSE CONVERT(VARCHAR(4), DATEADD(year,-1,GetDate()), 112) END), 112) + '0101')), 30)) / 7 * 7, '19000107'), 120)
I'm trying to convert a start + end date into multiple rows filled with weeknumbers and the year connected to it. My idea is to convert this start + end date in SSIS with a Script component, and create rows for each week by setting SynchronousInput to 'None'. I tried using a Timespan for this however this does not count through to the next year when reaching above week 53.
Does anyone know how to accomplish this?
Example:
Input:
StartDate EndDate
2014-12-29 2015-02-01
Wanted output:
Year WeekNr
2015 1
2015 2
2015 3
2015 4
2015 5
At first, I used the formular from here, but then remembered that, there is a better way using sqlserver 2008+ and changed the script to be more readable:
;WITH N(N)AS
(SELECT 1 FROM(VALUES(1),(1),(1),(1),(1),(1),(1),(1),(1),(1))M(N)),
tally(N)AS(SELECT ROW_NUMBER()OVER(ORDER BY N.N)FROM N,N a,N b,N c,N d,N e,N f)
SELECT
datepart(iso_week, dateadd(week, n-1, #from)) weeknr,
datepart(year, dateadd(week, datediff(day, -7*n, #from)/7, -4)) Year
FROM TALLY
WHERE
datediff(ww, dateadd(d, -1, #from), dateadd(d, -1, #to))+1 >= n
Result:
weeknr Year
1 2015
2 2015
3 2015
4 2015
5 2015