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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 have a SQL query that to return the number of items per week. I have a query that returns so far this:
Number of Items | Week Number
-------------------------------
100 | 18
80 | 19
120 | 20
And would like to return the following:
Number of Items | Week Beginning
-------------------------------
100 | 1st May 2017
80 | 8th May 2017
120 | 15th May 2017
What I have so far is:
SELECT COUNT(*) AS 'Number of Items', DATEPART(WEEK, Date) FROM table
where DATEPART(Year, Date) = '2017' and DATEPART(MONTH, Date) = 5
group by DATEPART(WEEK, Date)
You are talking about the 1st day of the current week:
example: select FORMAT(dateadd(ww,datediff(ww,0,getdate()),0),'dd MMM yyyy')--if you are using SQL 2012+
answer:
SELECT COUNT(*) AS 'Number of Items', FORMAT(dateadd(ww,datediff(ww,0,date_column),0),'dd MMM yyyy')
FROM table
where DATEPART(Year, Date) = '2017' and DATEPART(MONTH, Date) = 5
group by DATEPART(WEEK, Date)
As you need Monday to be the first day of the week
select DATEPART(WEEK, MyDate),DATEADD(DAY,1,(DATEADD(DAY, 1-DATEPART(WEEKDAY, MyDate), MyDate)))
from (
select '5/3/2017' MyDate
union all select '5/10/2017'
union all select '5/14/2017')A
SELECT DATEADD(DAY, DATEDIFF(DAY, 0, Date) /7*7, 0) AS StartDateOfWeek
check this if it solves
DECLARE #WeekNum INT
, #YearNum char(4);
SELECT #WeekNum = 20
, #YearNum = 2017
-- once you have the #WeekNum and #YearNum set, the following calculates the date range.
SELECT DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + #YearNum) + (#WeekNum-1), 6) AS StartOfWeek;
SELECT DATEADD(wk, DATEDIFF(wk, 5, '1/1/' + #YearNum) + (#WeekNum-1), 5) AS EndOfWeek;
thanks to http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=185440
I have some dates and week numbers and first day of week is 5 (Friday) in SQL Server. Now I want to calculate previous week and week start date from this below query.
DECLARE #LocationID tinyint = 1,
#FromDate date = '2016-05-20',
#ToDate date = '2016-05-29';
--Step1: ==================================================================
--SET first day of the week of the year
DECLARE #CurrentDayOfWeek INT
SET DATEFIRST 1 -- normalize current dkoray of week to Monday
SET #CurrentDayOfWeek = DATEPART(DW, DATEADD(yy, DATEDIFF(yy, 0, GETDATE()), 0))
SET DATEFIRST #CurrentDayOfWeek -- first day of week is now 5 (Friday)
--SELECT #CurrentDayOfWeek
-----Get Weekend number--------------
DECLARE #WeekStart int, #WeekEnd int
SET #WeekStart=(SELECT { fn WEEK( #FromDate) })
SET #WeekEnd=(SELECT { fn WEEK( #ToDate) })
;WITH AllDate AS
(
SELECT #FromDate as TheDate
UNION ALL
SELECT DATEADD(DAY, 1, TheDate)
FROM AllDate
WHERE DATEADD(DAY, 1, TheDate) <= #ToDate
),
AllDateDetail AS
(
SELECT TheDate,
DATEPART(WEEK, TheDate) AS WeekNumber,
YEAR(TheDate) AS [Year],
UPPER(CONVERT(NVARCHAR(3),DATENAME(WEEKDAY, TheDate))) AS [DayName]
FROM AllDate
)
SELECT
*
, CONVERT(DATE,DATEADD(WK, DATEDIFF(WK, 0, TheDate), 0)) AS LastWeekStart
, { fn WEEK( CONVERT(DATE,DATEADD(WK, DATEDIFF(WK, 0, TheDate) - 1, 0))) } AS WeekNumber
FROM AllDateDetail
--Select dateadd(wk, datediff(wk, 0, getdate()) - 1, 0) as LastWeekStart
--Select dateadd(wk, datediff(wk, 0, getdate()), 0) as ThisWeekStart
--Select dateadd(wk, datediff(wk, 0, getdate()) + 1, 0) as NextWeekStart
The current result of this query:
Expected result:
TheDate WeekNumber [Year] [DayName] LastWeekStart LastWeekNumber
20-05-2016 21 2016 FRI 13-05-2016 20
21-05-2016 21 2016 SAT 13-05-2016 20
22-05-2016 21 2016 SUN 13-05-2016 20
23-05-2016 21 2016 MON 13-05-2016 20
24-05-2016 21 2016 TUE 13-05-2016 20
25-05-2016 21 2016 WED 13-05-2016 20
26-05-2016 21 2016 THU 13-05-2016 20
27-05-2016 22 2016 FRI 20-05-2016 21
28-05-2016 22 2016 SAT 20-05-2016 21
29-05-2016 22 2016 SUN 20-05-2016 21
Can you help me? Thanks
Here's some handy ways of calculating First/Last day of This/Last week respecting DATEFIRST that I blogged a couple of weeks ago.
SET DATEFIRST 5 --Friday
-- Start/End of Weeks respecting DATEFIRST
SELECT DATEADD(DAY, 1-DATEPART(WEEKDAY, GETDATE()), GETDATE()) 'First Day of Current Week (DATEFIRST)';
SELECT DATEADD(DAY, 7-DATEPART(WEEKDAY, GETDATE()), GETDATE()) 'Last Day of Current Week (DATEFIRST)';
SELECT DATEADD(DAY, -6-DATEPART(WEEKDAY, GETDATE()), GETDATE()) 'First Day of Last Week (DATEFIRST)';
SELECT DATEADD(DAY, 0-DATEPART(WEEKDAY, GETDATE()), GETDATE()) 'Last Day of Last Week (DATEFIRST)';
(Gratuitous blog promo https://www.rednotebluenote.com/2016/04/first-day-of-the-week-and-datefirst/ )
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
basically I want to be able to select monday and friday for every week in the year.
So for example this week coming i want 9/29/2014 and 10/3/2014, but i want this for every week in the year.
Here's one way (you might need to check which day of the week is setup to be the first, here I have Sunday as the first day of the week)
You can use a table with many rows (more than 365) to CROSS JOIN to in order to get a run of dates (a tally table).
My sys columns has over 800 rows in, you could use any other table or even CROSS JOIN a table onto itself to multiply up the number of rows
Here I used the row_number function to get a running count of rows and incremented the date by 1 day for each row:
select
dateadd(d, row_number() over (order by name), cast('31 Dec 2013' as datetime)) as dt
from sys.columns a
With the result set of dates now, it's trivial to check the day of week using datepart()
SELECT
dt,
datename(dw, dt)
FROM
(
select
dateadd(d, row_number() over (order by name), cast('31 Dec 2013' as datetime)) as dt
from
sys.columns a
) as dates
WHERE
(datepart(dw, dates.dt) = 2 OR datepart(dw, dates.dt) = 6)
AND dt >= '01 Jan 2014' AND dt < '01 Jan 2015'
Edit:
Here's an example SqlFiddle
http://sqlfiddle.com/#!6/d41d8/21757
Edit 2:
If you want them on the same row, days of the week at least are constant, you know Friday is always 4 days after Monday so do the same but only look for Mondays, then just add 4 days to the Monday...
SELECT
dt as MonDate,
datename(dw, dt) as MonDateName,
dateadd(d, 4, dt) as FriDate,
datename(dw, dateadd(d, 4, dt)) as FriDateName
FROM
(
select
dateadd(d, row_number() over (order by name), cast('31 Dec 2013' as datetime)) as dt
from
sys.columns a
) as dates
WHERE
datepart(dw, dates.dt) = 2
AND dt >= '01 Jan 2014' AND dt < '01 Jan 2015'
AND dt >= '01 Jan 2014' AND dt < '01 Jan 2015'
Example SqlFiddle for this:
http://sqlfiddle.com/#!6/d41d8/21764
(note that only a few rows come back because sys.columns is quite small on the SqlFiddle server, try another system table if this is a problem)
You can use a suitable table with numbers, like the master..spt_values table as basis for the range generation:
;WITH dates AS (
SELECT DATEADD(DAY,number,CAST('2014-01-01' AS DATE)) d
FROM master..spt_values WHERE TYPE = 'p'
AND number < 366
)
SELECT
Week = DATEPART(WEEK, d),
DayOfWeek = DATENAME(dw, d),
Date = d
FROM dates
WHERE DATENAME(dw, d) IN ('Monday', 'Friday')
-- or use datepart instead as datename might be specific to language
-- WHERE DATEPART(dw, d) IN (2,6)
Sample output:
Week DayOfWeek Date
----------- ------------------------------ ----------
1 Friday 2014-01-03
2 Monday 2014-01-06
2 Friday 2014-01-10
3 Monday 2014-01-13
3 Friday 2014-01-17
4 Monday 2014-01-20
4 Friday 2014-01-24
5 Monday 2014-01-27
5 Friday 2014-01-31
SELECT extract(isodow from "date_source") as datee 1-monday 7-sunday
and then in WHERE write datee IN ('1','2'....)