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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
How get first day of week (Monday) where week = 6 and year = 2020 I need get 10.02.2020
eg. week 1 in 2020 is date from 06.01.2020 - 12.01.2020
week 6 in 2020 is date from 10.02.2020 - 16.02.2020
DECLARE #YEAR int = 2020;
DECLARE #WEEKSTOADD int = 6;
SET DATEFIRST 1;
SELECT
DATEADD(day,
1 - DATEPART(dw,DATEADD(week,#WEEKSTOADD,cast(cast(#YEAR as varchar(4)) + '0101' as date))),
DATEADD(week,#WEEKSTOADD,cast(cast(#YEAR as varchar(4)) + '0101' as date)))
The following code will get the date of Monday in the week of a given date regardless of the setting of DateFirst or Language:
Cast( DateAdd( day, - ( ##DateFirst + DatePart( weekday, Datum ) - 2 ) % 7, Datum ) as Date )
An example with sample data:
with SampleData as (
select GetDate() - 30 as Datum
union all
select DateAdd( day, 1, Datum )
from SampleData
where Datum < GetDate() )
select Datum,
-- 1 = Monday through 7 = Sunday.
( ##DateFirst + DatePart( weekday, Datum ) - 2 ) % 7 + 1 as WeekDay,
-- Date of Monday in the week of the supplied date.
Cast( DateAdd( day, - ( ##DateFirst + DatePart( weekday, Datum ) - 2 ) % 7, Datum ) as Date ) as Monday
from SampleData;
As per sample data you need substring() :
select t.*, substring(datnum, charindex(' ', datnum) + 1, 10) as dt
from table t
where t.week = 6;
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 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
User will select a date in frontend and flexibledays, say for example if they have selected '2014-07-17' as date and flexibledays as 2, then we need to display both 2 previous and next 2 working days as like below,
2014-07-15
2014-07-16
2014-07-17
2014-07-20
2014-07-21
excluding weekends (friday and saturday), for use weekends is friday and saturday.
I have used the below query
DECLARE #MinDate DATE, #MaxDate DATE;
SELECT #MinDate = DATEADD(Day, -#inyDays ,#dtDate), #MaxDate = DATEADD(Day,#inyDays ,#dtDate)
DECLARE #DayExclusionValue VARCHAR(20)
SELECT #DayExclusionValue = dbo.UDF_GetConfigSettingValue('DaysToExclude')
DECLARE #NumOfWeekends INT
SELECT #NumOfWeekends= (DATEDIFF(wk, #MinDate, #MaxDate) * 2) +(CASE WHEN DATENAME(dw, #MinDate) = 'Friday' THEN 1 ELSE 0 END) +(CASE WHEN DATENAME(dw, #MaxDate) = 'Saturday' THEN 1 ELSE 0 END)
SET #MaxDate = DATEADD(Day,#inyDays + #NumOfWeekends ,#dtDate)
;WITH CalculatedDates AS
(
SELECT dates = #MinDate
UNION ALL
SELECT DATEADD(day, 1, dates)
FROM CalculatedDates
WHERE DATEADD(day, 1, dates) <= #MaxDate
)
SELECT dates FROM CalculatedDates
WHERE dates >= CAST(GETDATE() AS DATE)
AND DATENAME(DW, dates) NOT IN (SELECT Value FROM UDF_GetTableFromString(#DayExclusionValue))
OPTION (MAXRECURSION 0);
but the above query is not working properly.
Can you pls suggest me any other solution.
This example will work for Oracle, you did not say what DB you were using. If you have a list of vacations you need to join that in as indicated. It would need to be a outerjoin, and you need to add a case or something so that the vacation tables 'exclude' days override the generated days.
Also I chose a multiplier on random. When only dealing with weekend 8 was more than enough, but if your vacation table includes a lot of consecutive vacation days it might no longer be.
select d from(
select rownum nn, d, sysdate - d, first_value (rownum) over (order by abs(sysdate-d)) zero_valu
from (
select sysdate+n d, to_char(sysdate+n,'DAY'), CASE to_char(sysdate+n,'D') WHEN '6' THEN 'exclude' WHEN '7' THEN 'exclude' ELSE 'include' END e_or_i from
(SELECT ROWNUM-9 n -- 9=flexibleday*8/2 +1
FROM ( SELECT 1 just_a_column
FROM dual
CONNECT BY LEVEL <= 16 -- 8=flexibleday * 8
)
)
) where e_or_i = 'include' -- in this step you need to join in a table of holidays or such if you need that.
) where abs(nn-7) <= 2 -- 2=flexiday
order by d
DECLARE #StartDate DATE = '2014-07-17';
SELECT *
FROM
(
--Show Closest Previous 2 Days Not In Friday or Saturday
SELECT TOP 2
DATEADD(DAY, -nr, #StartDate) CheckDate,
DATENAME(DW, DATEADD(DAY, -nr, #StartDate)) CheckName,
-nr CheckCount
FROM (VALUES(1),(2),(3),(4)) AS Numbers(nr)
WHERE DATENAME(DW, DATEADD(DAY, -nr, #StartDate)) NOT IN ('Friday','Saturday')
UNION ALL
--Show Todays Date If Not Friday or Saturday
SELECT TOP 1
DATEADD(DAY, +nr, #StartDate) CheckDate,
DATENAME(DW, DATEADD(DAY, +nr, #StartDate)) CheckName,
nr CheckCount
FROM (VALUES(0)) AS Numbers(nr)
WHERE DATENAME(DW, DATEADD(DAY, +nr, #StartDate)) NOT IN ('Friday','Saturday')
UNION ALL
--Show Closest Next 2 Days Not In Friday or Saturday
SELECT TOP 2
DATEADD(DAY, +nr, #StartDate) CheckDate,
DATENAME(DW, DATEADD(DAY, +nr, #StartDate)) CheckName,
nr CheckCount
FROM (VALUES(1),(2),(3),(4)) AS Numbers(nr)
WHERE DATENAME(DW, DATEADD(DAY, +nr, #StartDate)) NOT IN ('Friday','Saturday')
) d
ORDER BY d.CheckDate
I break it into 3 parts, previous 2 days, today (if applicable) and next 2 days
Here is the output:
CheckDate CheckName CheckCount
2014-07-15 Tuesday -2
2014-07-16 Wednesday -1
2014-07-17 Thursday 0
2014-07-20 Sunday 3
2014-07-21 Monday 4
I use the datename since not sure what ##datefirst your server is set to. the values() section is just a numbers table (you should create a numbers table as big as the amount of records you want to return plus any weekends you are crossing over) and then the TOP 2 in the first and last sections would be replaced with the number of days you wanted to return before and after.
**** Update with generic numbers table functionality added:
Here we declare the starting date and the number of previous and next days we would like to pull:
DECLARE #StartDate DATE = '2014-07-20';
DECLARE #MaxBusDays INT = 5
This next section creates a numbers table (can be easily found via google)
DECLARE #number_of_numbers INT = 100000;
;WITH
a AS (SELECT 1 AS i UNION ALL SELECT 1),
b AS (SELECT 1 AS i FROM a AS x, a AS y),
c AS (SELECT 1 AS i FROM b AS x, b AS y),
d AS (SELECT 1 AS i FROM c AS x, c AS y),
e AS (SELECT 1 AS i FROM d AS x, d AS y),
f AS (SELECT 1 AS i FROM e AS x, e AS y),
numbers AS
(
SELECT TOP(#number_of_numbers)
ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) AS number
FROM f
)
Now we use the numbers table and a row_number setting to pull only the number of rows before and after (plus the date of, if it's not fri/sat as wanted) that are working days (not fri/sat)
SELECT *
FROM
(
--Show Closest Previous x Working Days (Not Friday or Saturday)
SELECT * FROM
(
SELECT
DATEADD(DAY, -number, #StartDate) CheckDate,
DATENAME(DW, DATEADD(DAY, -number, #StartDate)) CheckName,
-number CheckCount,
ROW_NUMBER() OVER (ORDER BY number ASC) AS RowCounter
FROM Numbers
WHERE DATENAME(DW, DATEADD(DAY, -number, #StartDate)) NOT IN ('Friday','Saturday')
) a
WHERE a.RowCounter <= #MaxBusDays
UNION ALL
--Show Todays Date If Working Day (Not Friday or Saturday)
SELECT TOP 1
#StartDate CheckDate,
DATENAME(DW, #StartDate) CheckName,
0 CheckCount,
0 RowCounter
WHERE DATENAME(DW, #StartDate) NOT IN ('Friday','Saturday')
UNION ALL
--Show Closest Next x Working Days (Not Friday or Saturday)
SELECT * FROM
(
SELECT
DATEADD(DAY, +number, #StartDate) CheckDate,
DATENAME(DW, DATEADD(DAY, +number, #StartDate)) CheckName,
number CheckCount,
ROW_NUMBER() OVER (ORDER BY number ASC) AS RowCounter
FROM Numbers
WHERE DATENAME(DW, DATEADD(DAY, +number, #StartDate)) NOT IN ('Friday','Saturday')
) b
WHERE b.RowCounter <= #MaxBusDays
) c
ORDER BY c.CheckDate
Here is the output: (2014-07-20 is the middle row)
CheckDate CheckName CheckCount RowCounter
2014-07-13 Sunday -7 5
2014-07-14 Monday -6 4
2014-07-15 Tuesday -5 3
2014-07-16 Wednesday -4 2
2014-07-17 Thursday -3 1
2014-07-20 Sunday 0 0
2014-07-21 Monday 1 1
2014-07-22 Tuesday 2 2
2014-07-23 Wednesday 3 3
2014-07-24 Thursday 4 4
2014-07-27 Sunday 7 5