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
Related
I have a rolling date with associated hours. I want to sum the hours over my defined 7-day week range of Saturday to Friday.
So I need to define any date as a week beginning Saturday, week ending Friday and sum over this range.
The table is in the form of:
Date
Day
Hours
2021-06-12
Saturday
3
2021-06-18
Friday
3
2021-06-21
Monday
1
2021-06-22
Tuesday
2
Grouping the above table into Saturday-Friday Week sums.
Saturday
Friday
Total_Hours
2021-06-12
2021-06-18
6
2021-06-19
2021-06-25
3
You can use date functions to get for each date (other than Saturday) the previous Saturday which is the start of the week it belongs too and group by that:
WITH cte AS (
SELECT *, DATEADD(day, -DATEPART(dw, Date) % 7, Date) Saturday
FROM tablename
)
SELECT Saturday,
DATEADD(day, 6, Saturday) Friday,
SUM(Hours) Total_Hours
FROM cte
GROUP BY Saturday
See the demo.
This code works in SQL Server but similar logic and functions can be used for other databases.
You don't have to tie this to datepart() or datename(), which in turn depends on DATEFIRST (alas!).
Instead, you can use date arithmetic. This in turn depends on the fact that 0 date is a Monday. But that is not configurable (as far as I know).
So:
select v.week as saturday, dateadd(day, 6, v.week) as friday,
sum(hours)
from t cross apply
(values (dateadd(day, 7*datediff(week, 0, dateadd(day, 1, t.date)) - 2, 0))
) v(week)
group by v.week
order by v.week;
Here is a db<>fiddle.
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 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.
I am having a problem with week numbers. The customers week starts on a Tuesday, so ends on a Monday. So I have done:
Set DateFirst 2
When I then use
DateAdd(ww,#WeeksToShow, Date)
It occasionally gives me 8 weeks of information. I think it is because it goes over to the previous year, but I am not sure how to fix it.
If I do:
(DatePart(dy,Date) / 7) - #WeeksToShow
Then it works better, but obviously doesn't work going through to previous years as it just goes to minus figures.
Edit:
My currently SQL (If it helps at all without any data)
Set DateFirst 2
Select
DATEPART(yyyy,SessionDate) as YearNo,
DATEPART(ww,SessionDate) as WeekNo,
DATEADD(DAY, 1 - DATEPART(WEEKDAY, SessionDate + SessionTime), CAST(SessionDate +SessionTime AS DATE)) [WeekStart],
DATEADD(DAY, 7 - DATEPART(WEEKDAY, SessionDate + SessionTime), CAST(SessionDate + SessionTime AS DATE)) [WeekEnd],
DateName(dw,DATEADD(DAY, 7 - DATEPART(WEEKDAY, SessionDate + SessionTime), CAST(SessionDate + SessionTime AS DATE))) as WeekEndName,
Case when #ConsolidateSites = 1 then 0 else SiteNo end as SiteNo,
Case when #ConsolidateSites = 1 then 'All' else CfgSites.Name end as SiteName,
GroupNo,
GroupName,
DeptNo,
DeptName,
SDeptNo,
SDeptName,
PluNo,
PluDescription,
SUM(Qty) as SalesQty,
SUM(Value) as SalesValue
From
PluSalesExtended
Left Join
CfgSites on PluSalesExtended.SiteNo = CfgSites.No
Where
Exists (Select Descendant from DescendantSites where Parent in (#SiteNo) and Descendant = PluSalesExtended.SiteNo)
AND (DATEPART(WW,SessionDate + SessionTime) !=DATEPART(WW,GETDATE()))
AND SessionDate + SessionTime between DATEADD(ww,#NumberOfWeeks * -1,#StartingDate) and #StartingDate
AND TermNo = 0
AND PluEntryType <> 4
Group by
DATEPART(yyyy,SessionDate),
DATEPART(ww,SessionDate),
DATEADD(DAY, 1 - DATEPART(WEEKDAY, SessionDate + SessionTime), CAST(SessionDate +SessionTime AS DATE)),
DATEADD(DAY, 7 - DATEPART(WEEKDAY, SessionDate + SessionTime), CAST(SessionDate + SessionTime AS DATE)),
Case when #ConsolidateSites = 1 then 0 else SiteNo end,
Case when #ConsolidateSites = 1 then 'All' else CfgSites.Name end,
GroupNo,
GroupName,
DeptNo,
DeptName,
SDeptNo,
SDeptName,
PluNo,
PluDescription
order by WeekEnd
There are two issues here, the first is that I suspect you are defining 8 weeks of data as having 8 different values for DATEPART(WEEK, in which case you can replicate the root cause of the issue by looking at what ISO would define as the first week of 2015:
SET DATEFIRST 2;
SELECT Date, Week = DATEPART(WEEK, Date)
FROM (VALUES
('20141229'), ('20141230'), ('20141231'), ('20150101'),
('20150102'), ('20150103'), ('20150104')
) d (Date);
Which gives:
Date Week
-----------------
2014-12-29 52
2014-12-30 53
2014-12-31 53
2015-01-01 1
2015-01-02 1
2015-01-03 1
2015-01-04 1
So although you only have 7 days, you have 3 different week numbers. The problem is that DATEPART(WEEK is quite a simplistic function, and will simply return the number of week boundaries passed since the first day of the year, a better function would be ISO_WEEK since this takes into account year boundaries nicely:
SET DATEFIRST 2;
SELECT Date, Week = DATEPART(ISO_WEEK, Date)
FROM (VALUES
('20141229'), ('20141230'), ('20141231'), ('20150101'),
('20150102'), ('20150103'), ('20150104')
) d (Date);
Which gives:
Date Week
-----------------
2014-12-29 1
2014-12-30 1
2014-12-31 1
2015-01-01 1
2015-01-02 1
2015-01-03 1
2015-01-04 1
The problem is, that this does not take into account that the week starts on Tuesday, since the ISO week runs Monday to Sunday, you could adapt your usage slightly to get the week number of the day before:
SET DATEFIRST 2;
SELECT Date, Week = DATEPART(ISO_WEEK, DATEADD(DAY, -1, Date))
FROM (VALUES
('20141229'), ('20141230'), ('20141231'), ('20150101'),
('20150102'), ('20150103'), ('20150104')
) d (Date);
Which would give:
Date Week
-----------------
2014-12-29 52
2014-12-30 1
2014-12-31 1
2015-01-01 1
2015-01-02 1
2015-01-03 1
2015-01-04 1
So Monday the 29th December is now recognized as the previous week. The problem is that there is no ISO_YEAR built in function, so you will need to define your own. This is a fairly trivial function, even so I almost never create scalar functions because they perform terribly, instead I use an inline table valued function, so for this I would use:
CREATE FUNCTION dbo.ISOYear (#Date DATETIME)
RETURNS TABLE
AS
RETURN
( SELECT IsoYear = DATEPART(YEAR, #Date) +
CASE
-- Special cases: Jan 1-3 may belong to the previous year
WHEN (DATEPART(MONTH, #Date) = 1 AND DATEPART(ISO_WEEK, #Date) > 50) THEN -1
-- Special case: Dec 29-31 may belong to the next year
WHEN (DATEPART(MONTH, #Date) = 12 AND DATEPART(ISO_WEEK, #Date) < 45) THEN 1
ELSE 0
END
);
Which just requires a subquery to be used, but the extra typing is worth it in terms of performance:
SET DATEFIRST 2;
SELECT Date,
Week = DATEPART(ISO_WEEK, DATEADD(DAY, -1, Date)),
Year = (SELECT ISOYear FROM dbo.ISOYear(DATEADD(DAY, -1, Date)))
FROM (VALUES
('20141229'), ('20141230'), ('20141231'), ('20150101'),
('20150102'), ('20150103'), ('20150104')
) d (Date);
Or you can use CROSS APPLY:
SET DATEFIRST 2;
SELECT Date,
Week = DATEPART(ISO_WEEK, DATEADD(DAY, -1, Date)),
Year = y.ISOYear
FROM (VALUES
('20141229'), ('20141230'), ('20141231'), ('20150101'),
('20150102'), ('20150103'), ('20150104')
) d (Date)
CROSS APPLY dbo.ISOYear(d.Date) y;
Which gives:
Date Week Year
---------------------------
2014-12-29 52 2014
2014-12-30 1 2015
2014-12-31 1 2015
2015-01-01 1 2015
2015-01-02 1 2015
2015-01-03 1 2015
2015-01-04 1 2015
Even with this method, by simply getting a date 6 weeks ago you sill still end up with 7 weeks if the date you are using is not a Tuesday, because you will have 5 full weeks, and a part week at the start and a part week at the end, this is the second issue. So you need to make sure your start date is a Tuesday. The following will get you Tuesday of 7 weeks ago:
SELECT CAST(DATEADD(DAY, 1 - DATEPART(WEEKDAY, GETDATE()), DATEADD(WEEK, -6, GETDATE())) AS DATE);
The logic of this is explained better in this answer, the following is the part that will get the start of the week (based on your datefirst settings):
SELECT DATEADD(DAY, 1 - DATEPART(WEEKDAY, GETDATE()), GETDATE());
Then all I have done is substitute the second GETDATE() with DATEADD(WEEK, -6, GETDATE()) so that it is getting the start of the week 6 weeks ago, then there is just a cast to date to remove the time element from it.
This will get you current week + 5 previous weeks starting tuesday:
WHERE dateadd(week, datediff(d, 0, getdate()-1)/7 - 4, 1) <= yourdatecolumn
This will show examples:
DECLARE #wks int = 6 -- Weeks To Show
SELECT
dateadd(week, datediff(d, 0, getdate()-1)/7 - 4, 1) tuesday5weeksago,
dateadd(week, datediff(d, 0, getdate()-1)/7 - 5, 1) tuesday6weeksago,
dateadd(week, datediff(d, 0, getdate()-1)/7 - 6, 1) tuesday7weeksago,
dateadd(week, datediff(d, 0, getdate()-1)/7 - #wks + 1, 1) tuesdaydynamicweeksago
Result:
tuesday5weeksago tuesday6weeksago tuesday7weeksago tuesdaydynamicweeksago
2015-01-27 2015-01-20 2015-01-13 2015-01-20