Related
Here is what my table looks like
ID | Date
------------
13 2013-03-21 00:00:00
13 2013-03-23 00:00:00
13 2013-03-24 00:00:00
25 2013-03-21 00:00:00
25 2013-03-22 00:00:00
25 2013-03-21 00:00:00
25 2013-03-23 00:00:00
25 2013-03-28 00:00:00
25 2013-03-21 00:00:00
82 2013-03-22 00:00:00
82 2013-03-22 00:00:00
I want it to output (combine sat & sun into 'weekend'). It'll combine all the Mondays, Tuesdays, Wednesdays, etc. and combine the values.
day | total
--------------
Friday 2
Thursday 3
Weekend 3
This is my query so far
SELECT
CASE
WHEN DATENAME(weekday, [date]) = 'Saturday' OR DATENAME(weekday, [date]) = 'Sunday' THEN 'Weekend'
ELSE DATENAME(weekday, [date])
END AS Day,
COUNT(DISTINCT [id]) AS score
FROM [table]
GROUP BY
CASE
WHEN DATENAME(weekday, [date]) = 'Saturday' OR DATENAME(weekday, [date]) = 'Sunday' THEN 'Weekend'
ELSE DATENAME(weekday, [date])
END
Currently I can get my query to output this, but it seems like it isn't adding both thursdays or sat + sun.
Day | total
--------------
Friday 2
Thursday 2
Weekend 2
Appears like what you are actually after is a distinct count of the ID and date combined. Seems like the easiest way, therefore, would be to concatenate the 2 values to make a new unique value and COUNT those DISTINCT values:
SELECT CASE WHEN DATENAME(WEEKDAY, V.[Date]) IN ('Saturday','Sunday') THEN 'Weekend' ELSE DATENAME(WEEKDAY, V.[Date]) END AS [Weekday],
COUNT(DISTINCT CONCAT(ID,[date]))
FROM (VALUES (13, CONVERT(datetime2(0),'2013-03-21T00:00:00')),
(13, CONVERT(datetime2(0),'2013-03-23T00:00:00')),
(13, CONVERT(datetime2(0),'2013-03-24T00:00:00')),
(25, CONVERT(datetime2(0),'2013-03-21T00:00:00')),
(25, CONVERT(datetime2(0),'2013-03-22T00:00:00')),
(25, CONVERT(datetime2(0),'2013-03-21T00:00:00')),
(25, CONVERT(datetime2(0),'2013-03-23T00:00:00')),
(25, CONVERT(datetime2(0),'2013-03-28T00:00:00')),
(25, CONVERT(datetime2(0),'2013-03-21T00:00:00')),
(82, CONVERT(datetime2(0),'2013-03-22T00:00:00')),
(82, CONVERT(datetime2(0),'2013-03-22T00:00:00'))) V (ID, [Date])
GROUP BY CASE WHEN DATENAME(WEEKDAY, V.[Date]) IN ('Saturday','Sunday') THEN 'Weekend' ELSE DATENAME(WEEKDAY, V.[Date]) END;
Try this!
select count(id)as Total,Day from
(
select id, case when Datename(weekday,dt1) IN('Saturday','Sunday') then 'Weekend' else Datename(weekday,dt1) end as 'Day' from tab
)x
group by Day
Demo here
select Day,Count(day) cnt from (
SELECT
CASE
WHEN DATENAME(weekday, [date]) = 'Saturday' OR DATENAME(weekday, [date]) = 'Sunday' THEN 'Weekend'
ELSE DATENAME(weekday, [date])
END AS Day
,[date]
FROM [Infinite_campus].[dbo].[AttendanceRecord]
) as t
group by Day
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;
My company's reporting week is Monday through Sunday. On Monday morning's I run several queries using Microsoft SQL Server Management Studio to report on business activity for the previous week. I currently use declaration statements to pull the desired date range. This works quite well as long as I'm running reports on Monday. However, if Monday was a holiday and I'm not running reports until Tuesday, I need to manually modify my date range for each query. How can I modify my date filter to retrieve records through the previous "Sunday" so it doesn't matter what day I actually run the report.
Here's a sample query;
Declare #DATEFROM SMALLDATETIME = (CONVERT(datetime, getdate() + cast('00:00' as datetime)) - 8), #DATETO smalldatetime = (CONVERT(datetime, getdate() + cast('23:59' as datetime))-2);
Create Table #SALES ([PartNumber] CHAR(5), [DateSold] SMALLDATETIME)
Insert Into #SALES Select '10190', '6/3/2018 11:00'
Insert Into #SALES Select '10213', '6/8/2018 8:00:00 AM'
Insert Into #SALES Select '10214', '6/5/2018 9:30:00 AM'
Insert Into #SALES Select '10215', '6/4/2018 1:00:00 PM'
Insert Into #SALES Select '10217', '6/6/2018 1:00:00 PM'
Insert Into #SALES Select '10219', '6/7/2018 12:00:00 PM'
Insert Into #SALES Select '10220', '6/9/2018 3:30:00 PM'
Insert Into #SALES Select '10221', '6/11/2018 8:30:00 AM'
Insert Into #SALES Select '10222', '6/11/2018 2:30:00 PM'
Insert Into #SALES Select '10225', '6/8/2018 8:00:00 AM'
Insert Into #SALES Select '10227', '6/10/2018 9:00:00 AM'
Insert Into #SALES Select '10231', '6/10/2018 1:00:00 PM'
Insert Into #SALES Select '10233', '6/2/2018 8:00:00 AM';
SELECT S.PartNumber, S.DateSold
FROM #SALES S
WHERE DATESOLD BETWEEN #DATEFROM AND #DATETO
ORDER BY DateSold;
DROP TABLE #SALES
First, a few items of interest:
SQL Server's SET DATEFIRST command allows you to choose any weekday as the first day of the week, from 1 (Monday) through 7 (Sunday).
You can query the current value of this setting using the expression ##datefirst.
DATEPART(weekday, getdate()) will return a number for the current weekday where 1 indicates the day set by SET DATEFIRST, 2 indicates the day after, etc.
So suppose I want to answer the question: how many days prior to some arbitrary date #TestDate was the most recent Monday? The number of days that have passed since the most recent beginning of a week is DATEPART(weekday, #TestDate) - 1, and the number of days that pass between a Monday and the beginning of a week is ##datefirst - 1, so the number of days that have passed since the most recent Monday is the sum of those quantities modulo 7:
declare #TestDate date = convert(date, getdate());
declare #DaysPastMonday int = (##datefirst + datepart(weekday, #TestDate) - 2) % 7;
With this information you can get your date range very easily:
declare #DateTo date = dateadd(day, -#DaysPastMonday, #TestDate);
declare #DateFrom date = dateadd(day, -7, #DateTo);
But note that both of the dates I've selected here are Mondays, whereas you want a Monday through a Sunday. The reason I've done it this way is that if you're going to be looking at fields that potentially have a time component as well as a date (e.g., datetime or datetime2), and you want to use BETWEEN, then you need to make sure that your end date has the latest time of day that SQL Server is capable of representing. I find it cleaner to use a date range that's inclusive on the start date and exclusive on the end date. So in place of an expression like this:
x BETWEEN #DateFrom AND #DateTo
You'd write your queries to use an expression like this:
x >= #DateFrom AND x < #DateTo
Declare #DATEFROM datetime = DATEADD(wk, DATEDIFF(wk, 6, GETDATE()), 0),
#DATETO datetime = DATEADD(ms, -3, (select DATEADD(wk, DATEDIFF(wk, 6, GETDATE()), 7)));
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
EmployeeID RecordID DateRecord
1 1 2/19/2013 12:00:00 AM
1 2 2/21/2013 12:00:00 AM
1 3 2/23/2013 12:00:00 AM
1 4 2/27/2013 12:00:00 AM
1 5 3/3/2013 12:00:00 AM
2 11 3/10/2013 12:00:00 AM
2 12 3/14/2013 12:00:00 AM
1 14 3/16/2013 12:00:00 AM
How can I count the number of days?
Example in February 2013 which has "19, 21, 23, 27" that should be count to "4" days .. ??
I found this method ..
SELECT DATEPART(yy, Daterecord),
DATEPART(mm, Daterecord),
DATEPART(dd, Daterecord),
COUNT(*)
FROM Records
GROUP BY DATEPART(yy, Daterecord),
DATEPART(mm, Daterecord),
DATEPART(dd, Daterecord)
and resulted to ..
2013 2 19 1
2013 2 21 1
2013 2 23 1
2013 2 27 1
2013 3 3 1
2013 3 10 1
2013 3 14 1
2013 3 16 1
it just get the specific dates but didm't count the total number of days in each month .. help me .. pls
I have change few names hopr you won't mind
WITH Emp_CTE AS (
SELECT EmployeeID ,DATEPART(yy, Daterecord) AS years,
DATEPART(mm, Daterecord) AS months
-- DATEPART(dd, Daterecord) AS days
FROM testTrial
)
SELECT COUNT(months) AS noOfMonths ,* FROM Emp_CTE GROUP BY months,EmployeeID,years
SqlFiddle
Let you try this:-
1: Find the number of days in whatever month we're currently in
DECLARE #dt datetime
SET #dt = getdate()
SELECT #dt AS [DateTime],
DAY(DATEADD(mm, DATEDIFF(mm, -1, #dt), -1)) AS [Days in Month]Solution
2: Find the number of days in a given month-year combo
DECLARE #y int, #m int
SET #y = 2012
SET #m = 2
SELECT #y AS [Year],
#m AS [Month],
DATEDIFF(DAY,
DATEADD(DAY, 0, DATEADD(m, ((#y - 1900) * 12) + #m - 1, 0)),
DATEADD(DAY, 0, DATEADD(m, ((#y - 1900) * 12) + #m, 0))
) AS [Days in Month]
If your table is called Employee then this will do the trick:
select convert(varchar, DateRecord, 112)/ 100, count(*)
from Employee
group by convert(varchar, DateRecord, 112)/ 100
Your initial query was almost right, just needed to remove the DATEPART(dd, Daterecord) from the grouping and it would work. Add in a HAVING clause to find the records from the month of February:
SELECT
DATEPART(yy, Daterecord),
DATEPART(mm, Daterecord),
COUNT(1)
FROM
Records
GROUP BY
DATEPART(yy, Daterecord),
DATEPART(mm, Daterecord)
HAVING
DATEPART(yy, eCreationTime) = 2013
AND DATEPART(mm, Daterecord) = 2
there is no 'yearmonth' in the suggested code ??
try this perhaps
select
datename(month,daterecord) as [Month]
, year(DateRecord) as [Year]
, count(distinct DateRecord ) as day_count
, count(distinct dateadd(day, datediff(day,0, DateRecord ), 0)) as daytime_count
from your_table
where ( DateRecord >= '20130201' and DateRecord < '20130301' )
group by
datename(month,daterecord)
, year(DateRecord)
note the column [daytime_count] is only required if the field [DateRecord] has times othe than 12:00 AM (i.e. it "trims off" times so you deal with dates at 12:AM)
Regarding date range selections: many people will feel that using 'between' is the solution however that isn't true and the safest most reliable method is as I shown above. Note that the higher date is 1st March, but we are asking for information that is less than the 1st March, so we don't need to worry about leap years and we don't have to worry about hours and minutes either.
see: What do BETWEEN and the devil have in common?
try this...
declare #date2 nvarchar(max)
set #date2 = (select getdate())
select DateDiff(Day,#date2,DateAdd(month,1,#date2))