The query below grabs the max date from column Time_Stamp as StartDate from Survey. Adding 90 days to it for EndDate. Then creating 2 more ranges with 90 day intervals by just adding more to the original EndDate.
I'm trying to have the StartDate fall within 4 buckets of either Jan1st, April1st, July1st, or Oct1. If the Max Time_Stamp is before 1 of these dates then that will be the first StartDate of my ranges...So for my example below, the max Time_Stamp is June4th so the StartDate needs to be July 1st. Is this doable within sql server?
Time_Stamp for Hospital1
Time_Stamp
-----------
2014-06-04 16:01:14.000
2014-06-04 15:55:33.000
2014-06-04 15:45:05.000
2014-06-04 15:36:15.000
2014-06-04 15:00:34.000
2014-06-04 14:35:24.000
2014-06-04 14:04:50.000
2014-06-04 13:46:55.000
2014-06-04 13:23:57.000
2014-06-04 11:27:51.000
Current output:
StartDate EndDate
----------- -----------
Jun 4 2014 Sep 2 2014
Sep 3 2014 Dec 2 2014
Dec 3 2014 Mar 3 2015
query
WITH Start AS
(
SELECT
MAX(Time_Stamp) as StartDate,
DATEADD(day, 90, MAX(Time_Stamp)) as EndDate
FROM Survey
WHERE MainHospital = 'Hospital1'
),
Results AS
(
SELECT StartDate, EndDate from Start
UNION
SELECT DATEADD(DAY, 1, EndDate), DATEADD(day, 91, EndDate) FROM Start
UNION
SELECT DATEADD(DAY, 92, EndDate), DATEADD(day, 182, EndDate) FROM Start
)
SELECT LEFT(StartDate,11) AS StartDate, LEFT(EndDate,11) AS EndDate FROM Results
Just an update, this gives me what I need for the 1st StartDate...
--Return first day of next quarter
SELECT DATEADD(qq, DATEDIFF(qq, 0, MAX(Time_Stamp)) + 1, 0)
FROM Survey
WHERE MainHospital = 'Hospital1'
DECLARE #Year DATE = '2013-01-01'
DECLARE #Quarter INT = 1;
SELECT DATEADD(QUARTER, #Quarter - 1, #Year) ,
DATEADD(DAY, -1, DATEADD(QUARTER, #Quarter, #Year))
You can use DATEPART(QUARTER, #Date) to figure out which quarter the record being selected and then you can use this query to find the begin/end dates of that quarter.
Use it like this:
SELECT DATEADD(QUARTER, DATEPART(QUARTER, Time_Stamp) - 2, DATEADD(YEAR, DATEDIFF(YEAR, 0, Time_Stamp), 0)) AS StartDate,
DATEADD(SECOND, -1, DATEADD(QUARTER, DATEPART(QUARTER, Time_Stamp) - 1, DATEADD(YEAR, DATEDIFF(YEAR, 0, Time_Stamp), 0))) AS EndDate
FROM Hospital1
It seems that you are just trying to calculate the quarter of your date, Am i correct?
If so, you can just use:
SELECT DATENAME(Quarter, CAST(CONVERT(VARCHAR(8), GETDATE()) AS DATETIME)) as Quarter
to calculate a quarters date ranges:
select DATEADD(qq, datediff(qq, 0, getdate()),0) as first
select dateadd(dd, -1, DATEADD(qq, datediff(qq, 0, getdate()) +1, 0)) as last
Related
For example consider the April 2023. In this year the Monday dates are 3, 10, 17, and 24. Now I want to get the date 10 position of April.
My target result is: April 10 = second or 2
SELECT (DAY('2023-04-10') - 1) / 7 + 1
Position outputs the searched number for a date
WITH CTE_DATES
AS(
SELECT CAST(DATEADD(mm, DATEDIFF(mm, 0, GETDATE()) -1, 0) AS DATE) AS [DATE] -- start previous month
UNION ALL
SELECT CAST(DATEADD(DD, 1, [DATE]) AS DATE)
FROM [CTE_DATES]
WHERE [DATE] < CAST(DATEADD(ms, -3, DATEADD(mm, DATEDIFF(mm, 0, GETDATE()) + 0, 0)) AS DATE) -- end previous month
)
SELECT
[date]
,DATENAME(WEEKDAY, [date]) AS 'Weekday'
,DATEPART(WEEKDAY, [date]) AS 'Day'
,ROW_NUMBER() OVER (ORDER BY [date] ASC) AS 'POSITION'
FROM [CTE_DATES]
WHERE DATEPART(WEEKDAY, [date]) IN (1)
How can I get the date of specific day ? Like if I have Thursday or month number ?
If I give 12 for instance I want to get the date of 12th day of this month. Or if I give 'Sun' or 'Sat' is it possible to get the dates of these days ?
DATEFROMPARTS function can construct a date from day, month and year.
DATEPARTS does the opposite - gives you the day, month, year, hour, etc. of a date. Or you can use functions like YEAR, MONTH and DAY.
You can deconstruct the value returned by GETDATE function and construct whatever date you want. Here is for example how to get the date for 12th day of the current month:
select DATEFROMPARTS(YEAR(GETDATE()), MONTH(GETDATE()), 12)
Converting 'Sun' or 'Sat' to date is a bit more difficult. First, they aren't quite deterministic. If today is Friday, "Sunday this week" means "next Sunday" in some parts of the world and "last Sunday" in others. You should implement your own logic based on the value returned by DATEPART(dw, GETDATE()) (which will give you the day of the week).
To find the weekday of the current month
DECLARE #daynumber INT = 12
SELECT datename(weekday, dateadd(d, #daynumber - 1, getdate()))
To find the dates of the current month of a given weekday
DECLARE #dayname char(3) = 'sat'
;WITH CTE as
(
SELECt TOP
(datediff(D, eomonth(getdate(), -1),eomonth(getdate())))
dateadd(d,row_number()over(ORDER BY 1/0),
eomonth(getdate(),-1))date
FROM
(values(1),(2),(3),(4),(5),(6))x(x),
(values(1),(2),(3),(4),(5),(6))y(x)
)
SELECT day(date) monthday, date
FROM CTE
WHERE left(datename(weekday, date),3) = #dayname
select sysdatetime(); --2018-12-13 16:29:56.0560574
---If I give 12 for instance I want to get the date of 12th day of this month.
declare #numDate int = 12;
select dateadd(m, datediff(m,0,getdate()),#numDate - 1 ); --2018-12-12 00:00:00.000
--Or if I give 'Sun' or 'Sat' is it possible to get the dates of these days ?
declare #text nvarchar(20) = 'Sunday';
declare #dateStart date = dateadd(month, datediff(month, 0, sysdatetime()), 0),
#days int =( select (DAY(dateadd(dd,-1,DATEADD(m,1,cast(2018 as varchar(4)) + '-' + cast(12 as varchar(2)) +'-01')))));
declare #dateEnd date = DATEADD(day,#days-1,#dateStart);
;WITH CTE (Dates,EndDate) AS
(
SELECT #dateStart AS Dates,#dateEnd AS EndDate
UNION ALL
SELECT DATEADD(day,1,Dates),EndDate
FROM CTE
WHERE DATEADD(day,1,Dates) <= EndDate
)
SELECT CTE.Dates, DATENAME(DW, CTE.Dates)
FROM CTE
where DATENAME(DW, CTE.Dates) = #text;
Result:
Dates,Day
2018/12/2,Sunday
2018/12/9,Sunday
2018/12/16,Sunday
2018/12/23,Sunday
2018/12/30,Sunday
-- Here is how to get week day name to week day number
DECLARE #T TABLE (Dow INT, NameOfDay VARCHAR(15), ShortName CHAR(3));
WITH Days AS
(
SELECT TOP 7
ROW_NUMBER() OVER(PARTITION BY object_id ORDER BY object_id) AS RowNo
FROM
sys.all_columns
)
INSERT INTO #T
SELECT
RowNo,
DATENAME(WEEKDAY, RowNo - 1),
LEFT(DATENAME(WEEKDAY, RowNo - 1), 3)
FROM
Days
SELECT
*
FROM
#T;
-- Here is how to get start of period
SELECT
DATEADD(DAY, DATEDIFF(DAY, 0, GETDATE()), 0) AS StartOfDay,
DATEADD(WEEK, DATEDIFF(WEEK, 0, GETDATE()), 0) AS StartOfWeek,
DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0) AS StartOfMonth,
DATEADD(YEAR, DATEDIFF(YEAR, 0, GETDATE()), 0) AS StartOfYear;
-- An example
WITH
StartPeriods AS
(
SELECT DATEADD(WEEK, DATEDIFF(WEEK, 0, GETDATE()), 0) AS StartOfWeek
),
SelectedDay AS
(
SELECT
Dow - 1 AS Dow,
(SELECT StartOfWeek FROM StartPeriods) AS StartOfWeek
FROM
#T
WHERE
ShortName = 'Wed'
)
SELECT
DATEADD(DAY, Dow, StartOfWeek)
FROM
SelectedDay;
One of my clients defines (for strange financial reasons) a financial month as a period of time that begins the Wednesday immediately after the last Tuesday of a Month (inclusive) and lasts until the last tuesday of the following month (inclusive).
I need to find the start of the last and the current financial month.
Some examples:
if today is September 23rd 2015 i need to get July 29th and August 26th because the current financial month goes from August 26th to September 29th.
If today is September 30th 2015 I need to get August 26th to September 30th.
I have different clients with different definitions and this means that some of them are using Wednesday and others are using Monday so i need this day to be a parameter, like Monday = 1 and Wednesday = 3. I call it FDOM, FirstDayOfMonth.
My work so far focused on using the formulas i found around with first and last days of current and last month, modified to take into account FDOM. I managed to get last Wednesday of Last Month but this sometimes is not correct because I am considering a day of the month that belongs to a solar month but also to the next financial month, like September 30th belongs to solar September but to Financial October, as financial October begins September 30th.
DECLARE #BASE AS DateTime = '19000101 00:00'
DECLARE #FDOM AS INT = 3 --Wednesday
DECLARE #Datevalue AS DATE = GETDATE()
SET DATEFIRST #FDOM
select DATEADD(D,1-(DATEPART(dw,DATEADD(D,-1,DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #Datevalue) , #BASE)))),DATEADD(D,-1,DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #Datevalue) , #BASE)))
This gives me the first wednesday after the last tuesday of last month and this would be correct from September 1st to September 29th (it gives August 26th) as "the beginning of the current financial month". But it would be wrong on September 30th as it should give September 30th and also wrong from August 26th till the end of August as it should give August 26th but instead gives July 29th.
I think this answers your requirements. Its quite long but hopefully, by breaking things out and naming things, I'm making it clear how we get to the final answer, and so if it's not quite right, it can be adapted:
declare #FDOM int
set #FDOM = 3 --Wednesday. 0 = Sunday, 6 = Saturday
declare #KnownDay datetime
set #KnownDay = DATEADD(day,#FDOM - 1,'20150301') --Offset from a "known good" Sunday to the day before FDOM
declare #EOLastDec datetime
set #EOLastDec = DATEADD(year,DATEDIFF(year,'20010101',GETDATE()),'20001231')
declare #Today datetime
set #Today = DATEADD(day,DATEDIFF(day,0,GETDATE()),0) --You can change this to test other key dates
;With Numbers(n) as (--If you have a numbers table, you can skip this CTE
select ROW_NUMBER() OVER (ORDER BY so1.object_id) - 1
from sys.objects so1 cross join sys.objects so2
), LastOfMonths as (
select DATEADD(month,n,#EOLastDec) as LOM
from Numbers
where n between 0 and 13
), LastImportant as (
select DATEADD(day,-n,LOM) as EOFMonth
from LastOfMonths cross join Numbers
where n between 0 and 6 and
DATEPART(weekday,DATEADD(day,-n,LOM)) = DATEPART(weekday,#KnownDay)
)
select DATEADD(day,1,li0.EOFMonth) as StartOfMonth,DATEADD(day,1,li1.EOFMonth) as EndOfMonth
from
LastImportant li1
cross join
LastImportant li2
left join
LastImportant li1_anti
on
li1.EOFMonth < li1_anti.EOFMonth and
li1_anti.EOFMonth <= #Today
left join
LastImportant li2_anti
on
li2.EOFMonth > li2_anti.EOFMonth and
li2_anti.EOFMonth >= #Today
inner join
LastImportant li0
on
li0.EOFMonth < li1.EOFMonth
left join
LastImportant li0_anti
on
li0_anti.EOFMonth < li1.EOFMonth and
li0.EOFMonth < li0_anti.EOFMonth
where
li1.EOFMonth <= #Today and
li2.EOFMonth >= #Today and
li1_anti.EOFMonth is null and
li2_anti.EOFMonth is null and
li0_anti.EOFMonth is null
Hopefully, the CTEs are reasonably explanatory. We generate a numbers table, and then we calculate the last day of each month, and from there, we step up to 6 days backwards to locate a day of the right type (i.e. a Tuesday, if #FDOM is 3)
I originally had a simpler final query using just li1 and li2 (and li1_anti and li2_anti), but realised that the query was just finding the current financial month - so I've added another couple of joins (using li0 and li0_anti) to find the start of the previous financial month.
Calculate the start date from previous month and the last day from current month and used a CTE to generate all dates between them. Later, get the MAX weekday from both months.
DECLARE #CurrentDate DATE = '2015-08-23'
DECLARE #StartDate DATE,
#EndDate DATE,
#MonthEnd INT = 3
-- Get the first day from previous month and last day from current month
SELECT #StartDate = DATEADD(MONTH , DATEDIFF(MONTH, 0, #CurrentDate)-1, 0),
#EndDate = DATEADD(SECOND,-1,
DATEADD(MONTH , DATEDIFF(MONTH, 0, #CurrentDate)+1,0))
;WITH Calendar AS
( -- Generate all dates between #StartDate and #EndDate
SELECT #StartDate [Date]
UNION ALL
SELECT DATEADD(D, +1, Calendar.[Date])
FROM Calendar
WHERE Calendar.[Date] < #EndDate
)
SELECT DATEADD(DAY, +1, MAX(StartDate.[Date])) StartDate,
DATEADD(DAY, +1, MAX(EndDate .[Date])) EndDate
FROM Calendar StartDate,
Calendar EndDate
WHERE -- Get the max weekday from previous month
DATEPART(MONTH , StartDate.[Date]) = DATEPART(MONTH, #StartDate) AND
DATEPART(WEEKDAY, StartDate.[Date]) = #MonthEnd AND
-- Get the max weekday from current month
DATEPART(MONTH , EndDate .[Date]) = DATEPART(MONTH, #EndDate) AND
DATEPART(WEEKDAY, EndDate .[Date]) = #MonthEnd
Try this. You can use EOMONTH function to get the end of month on Sql Server 2012 or above.
Click to see the fiddle demo.
DECLARE #date DATETIME = GETDATE()
DECLARE #LastMonthEnd DATETIME = DATEADD(day, -1, DATEADD(month, DATEDIFF(month, 0, #date), 0))
DECLARE #CurrentMonthEnd DATETIME = DATEADD(day, -1, DATEADD(month, DATEDIFF(month, 0, #date) + 1, 0))
SET DATEFIRST 1
;WITH CTE1 AS
(
SELECT 1 number, DATEPART(WEEKDAY, #LastMonthEnd) FirstDay,
DATEPART(WEEKDAY, #CurrentMonthEnd) LastDay
UNION ALL
SELECT 1+number, DATEPART(WEEKDAY, DATEADD(DAY, -number, #LastMonthEnd)),
DATEPART(WEEKDAY, DATEADD(DAY, -number, #CurrentMonthEnd))
FROM CTE1
WHERE number < 7
)
SELECT DATEADD(DAY, -(SELECT Number FROM CTE1 WHERE FirstDay = 3), #LastMonthEnd) StartDate,
DATEADD(DAY, -(SELECT Number FROM CTE1 WHERE LastDay = 3), #CurrentMonthEnd) EndDate
After a lot of effort I managed to find an expression that DOES NOT use CTE as i am not sure i can't use CTE in all the places i will have to employ this.
So basically first i understand with a CASE if the date i am considering is before or after the last Wednesday of the Month it belongs to. Then i return the last wednseday of two and of one month ago OR the last wednesday of one month ago and of this month.
This works also changing FDOM and i tested it for several months of this year. It seems to always work.
Probably the use of EOMonth would shorten it but i have to verify i can use it in my server.
I am sorry i just specified as a requirement that i can't use a CTE only in the comments but thank you for your help
DECLARE #datevalue AS Datetime = getdate()
DECLARE #BASE AS DateTime = '19000101 00:00'
DECLARE #FDOM AS INT = 3 --1 is for Monday
SET DATEFIRST #FDOM
SELECT CASE WHEN (#datevalue < DATEADD(D, 1-(DATEPART(dw,DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) + 1, #BASE))),DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) + 1, #BASE)))
THEN(DATEADD(D, 1-(DATEPART(dw,DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) - 1, #BASE))),DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) - 1, #BASE)))
ELSE(DATEADD(D, 1-(DATEPART(dw,DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) - 0, #BASE))),DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) - 0, #BASE)))
END AS [Start of Last Financial Month]
,CASE WHEN (#datevalue < DATEADD(D, 1-(DATEPART(dw,DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) + 1, #BASE))),DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) + 1, #BASE)))
THEN(DATEADD(D, 1-(DATEPART(dw,DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) + 0, #BASE))),DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) + 0, #BASE)))
ELSE(DATEADD(D, 1-(DATEPART(dw,DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) + 1, #BASE))),DATEADD(MONTH, DATEDIFF(MONTH, #BASE, #datevalue) + 1, #BASE)))
END AS [Start of Current Financial Month]
I want records from table which stores the current date when a record is inserted with in current week only.
I have tried:
SELECT PId
,WorkDate
,Hours
,EmpId
FROM Acb
WHERE EmpId=#EmpId AND WorkDate BETWEEN DATEADD(DAY, -7, GETDATE()) AND GETDATE()
Do it like this:
SET DATEFIRST 1 -- Define beginning of week as Monday
SELECT [...]
AND WorkDate >= dateadd(day, 1-datepart(dw, getdate()), CONVERT(date,getdate()))
AND WorkDate < dateadd(day, 8-datepart(dw, getdate()), CONVERT(date,getdate()))
Explanation:
datepart(dw, getdate()) will return the number of the day in the current week, from 1 to 7, starting with whatever you specified using SET DATEFIRST.
dateadd(day, 1-datepart(dw, getdate()), getdate()) subtracts the necessary number of days to reach the beginning of the current week
CONVERT(date,getdate()) is used to remove the time portion of GETDATE(), because you want data beginning at midnight.
A better way would be
select datepart(ww, getdate()) as CurrentWeek
You can also use wk instead of ww.
Datepart Documentation
Its Working For Me.
Select * From Acb Where WorkDate BETWEEN DATEADD(DAY, -7, GETDATE()) AND DATEADD(DAY, 1, GETDATE())
You have to put this line After the AND Clause AND DATEADD(DAY, 1, GETDATE())
datepart(dw, getdate()) is the current day of the week, dateadd(day, 1-datepart(dw, getdate()), getdate()) should be the first day of the week, add 7 to it to get the last day of the week
You can use following query to extract current week:
select datepart(dw, getdate()) as CurrentWeek
SET DATEFIRST 1;
;With CTE
AS
(
SELECT
FORMAT(CreatedDate, 'MMMM-yyyy') as Months,
CASE
WHEN YEAR(DATEADD(DAY, 1-DATEPART(WEEKDAY, Min(CreatedDate)), Min(CreatedDate))) < YEAR(Min(CreatedDate))
THEN FORMAT(DATEADD(YEAR, DATEDIFF(YEAR, 0,DATEADD(YEAR, 0 ,GETDATE())), 0) ,'MMM dd') + ' - ' + FORMAT(DATEADD(dd, 7-(DATEPART(dw, Min(CreatedDate))), Min(CreatedDate)) ,'MMM dd')
ELSE
FORMAT(DATEADD(DAY, 1-DATEPART(WEEKDAY, Min(CreatedDate)), Min(CreatedDate)) ,'MMM dd') + ' - ' + FORMAT(DATEADD(dd, 7-(DATEPART(dw, Min(CreatedDate))), Min(CreatedDate)) ,'MMM dd')
END DateRange,
Sum(ISNULL(Total,0)) AS Total,
sum(cast(Duration as int)) as Duration
FROM TL_VriandOPI_Vendorbilling where VendorId=#userID and CompanyId=#CompanyID
Group By DATEPART(wk, CreatedDate) ,FORMAT(CreatedDate, 'MMMM-yyyy')
)
SELECT Months,DateRange,Total,Duration,
case when DateRange=(select FORMAT(DATEADD(DAY, 1-DATEPART(WEEKDAY, Min(getdate())), Min(getdate())) ,'MMM dd') + ' - ' +
FORMAT(DATEADD(dd, 7-(DATEPART(dw, Min(getdate()))), Min(getdate())) ,'MMM dd'))
then 1 else 0 end as Thisweek
FROM CTE order by Months desc
Using DATEDIFF works as well, however a bit hacky since it doesn't care about datefirst:
set datefirst 1; -- set monday as first day of week
declare #Now datetime = '2020-09-28 11:00';
select *
into #Temp
from
(select 1 as Nbr, '2020-09-22 10:00' as Created
union
select 2 as Nbr, '2020-09-25 10:00' as Created
union
select 2 as Nbr, '2020-09-28 10:00' as Created) t
select * from #Temp where DATEDIFF(ww, dateadd(dd, -##datefirst, Created), dateadd(dd, -##datefirst, #Now)) = 0 -- returns 1 result
select * from #Temp where DATEDIFF(ww, dateadd(dd, -##datefirst, Created), dateadd(dd, -##datefirst, #Now)) = 1 -- returns 2 results
drop table #Temp
Here is a great way of how to get day of week for a date, Deterministic scalar function to get day of week for a date.
Now, could anyone help me to create a deterministic scalar function to get week of year for a date please? Thanks.
This works deterministically, I can use it as a computed column.
datediff(week, dateadd(year, datediff(year, 0, #DateValue), 0), #DateValue) + 1
Test code:
;
with
Dates(DateValue) as
(
select cast('2000-01-01' as date)
union all
select dateadd(day, 1, DateValue) from Dates where DateValue < '2050-01-01'
)
select
year(DateValue) * 10000 + month(DateValue) * 100 + day(DateValue) as DateKey, DateValue,
datediff(day, dateadd(week, datediff(week, 0, DateValue), 0), DateValue) + 2 as DayOfWeek,
datediff(week, dateadd(month, datediff(month, 0, DateValue), 0), DateValue) + 1 as WeekOfMonth,
datediff(week, dateadd(year, datediff(year, 0, DateValue), 0), DateValue) + 1 as WeekOfYear
from Dates option (maxrecursion 0)