Is it possible to do in SQL: for example I have period where #s_date = '20130101' and #e_date = '20130601' and I want to select all last days of months in this period.
This is example of result:
20130131
20130228
20130331
20130430
20130531
Thanks.
The easiest option is to have a calendar table, with a last day of the month flag, so your query would simply be:
SELECT *
FROM dbo.Calendar
WHERE Date >= #StartDate
AND Date <= #EndDate
AND EndOfMonth = 1;
Assuming of course that you don't have a calendar table you can generate a list of dates on the fly:'
DECLARE #s_date DATE = '20130101',
#e_date DATE = '20130601';
SELECT Date = DATEADD(DAY, ROW_NUMBER() OVER(ORDER BY Object_ID) - 1, #s_date)
FROM sys.all_objects;
Then once you have your dates you can limit them to where the date is the last day of the month (where adding one day makes it the first of the month):
DECLARE #s_date DATE = '20130101',
#e_date DATE = '20130601';
WITH Dates AS
( SELECT Date = DATEADD(DAY, ROW_NUMBER() OVER(ORDER BY Object_ID) - 1, #s_date)
FROM sys.all_objects
)
SELECT *
FROM Dates
WHERE Date <= #e_Date
AND DATEPART(DAY, DATEADD(DAY, 1, Date)) = 1;
Example on SQL Fiddle
You can run the following query and then adjust it by using your table details:
declare #s_date as datetime= '20130101'
declare #e_date as datetime= '20131020'
SELECT DateAdd(m, number, '1990-01-31')
FROM master.dbo.spt_values
WHERE 'P' = type
AND DateAdd(m, number, #s_date) < #e_date
example for 20130101 :
select CONVERT(VARCHAR(8),
dateadd(day, -1, dateadd(month, 1,
convert(datetime, '20130101',112))), 112)
result :
20130131
Try this query
WITH sample
AS (SELECT Cast('2013-04-01' AS DATETIME) Date
UNION ALL
SELECT Dateadd(day, 1, date) dt
FROM sample
WHERE date < Cast('2013-05-05' AS DATETIME))
SELECT *
FROM sample
Fiddle
EOMONTH(#date) is the function you need.
Here is the help page https://learn.microsoft.com/en-us/sql/t-sql/functions/eomonth-transact-sql?view=sql-server-2017
This query gets the las 50 End Of Months.
The original query used as an example is from here.
https://dba.stackexchange.com/a/186829
WITH cte AS
(
SELECT ROW_NUMBER() OVER (ORDER BY (SELECT 1)) - 1 AS [Incrementor]
FROM [master].[sys].[columns] sc1
CROSS JOIN [master].[sys].[columns] sc2
)
SELECT top 50 EOMONTH(DATEADD(Month, -1 * cte.[Incrementor], GETDATE()))
FROM cte
WHERE EOMONTH(DATEADD(Month, -1 * cte.[Incrementor], GETDATE())) < GETDATE();
Related
My end goal is to see end of month data for previous month.
Our processing is a day behind so if today is 7/28/2021 our Process date is 7/27/2021
So, I want my data to be grouped.
DECLARE
#ProcessDate INT
SET #ProcessDate = (SELECT [PrevMonthEnddatekey] FROM dbo.dimdate WHERE datekey = (SELECT [datekey] FROM sometable [vwProcessDate]))
SELECT
ProcessDate
, LoanOrigRiskGrade
,SUM(LoanOriginalBalance) AS LoanOrigBalance
,Count(LoanID) as CountofLoanID
FROM SomeTable
WHERE
ProcessDate in (20210131, 20210228,20210331, 20210430, 20210531, 20210630)
I do not want to hard code these dates into my WHERE statement. I have attached a sample of my results.
I am GROUPING BY ProcessDate, LoanOrigRiskGrade
Then ORDERING BY ProcessDate, LoanOrigIRskGrade
It looks like you want the last day of the month for months within a specified range. You can parameterize that.
For SQL Server:
DECLARE #ProcessDate INT
SET #ProcessDate = (
SELECT [PrevMonthEnddatekey]
FROM dbo.dimdate
WHERE datekey = (
SELECT [datekey]
FROM sometable [vwProcessDate]
)
)
DECLARE #startDate DATE
DECLARE #endDate DATE
SET #startDate = '2021-01-01'
SET #endDate = '2021-06-30'
;
with d (dt, eom) as (
select #startDate
, convert(int, replace(convert(varchar(10), eomonth(#startDate), 102), '.', ''))
union all
select dateadd(month, 1, dt)
, eomonth(dateadd(month, 1, dt))
from d
where dateadd(month, 1, dt) < #endDate
)
SELECT ProcessDate
, LoanOrigRiskGrade
, SUM(LoanOriginalBalance) AS LoanOrigBalance
, Count(LoanID) as CountofLoanID
FROM SomeTable
inner join d on d.eom = SomeTable.ProcessDate
Difficult to check without sample data.
I need to calculate using SQL Query, how many days within a given range fall into each calendar month.
I have given 2 dates, which define a date range; for example 2020-01-01 to 2020-08-03. I need to find how many days in that range fall in to each month i.e. how many fall into July, and how many into August.
In the example given, the expected result is 31 days in July and 3 days in August.
One approach uses a recusive query. Using date artithmetics, we can build the query so it performs one iteration per month rather than one per day, so this should be a rather efficient approach:
with cte as (
select
datefromparts(year(#dt_start), month(#dt_start), 1) month_start,
1 - day(#dt_start) + day(
case when #dt_end > eomonth(#dt_start)
then eomonth(#dt_start)
else #dt_end
end
) as no_days
union all
select
dateadd(month, 1, month_start),
case when #dt_end > dateadd(month, 2, month_start)
then day(eomonth(dateadd(month, 1, month_start)))
else day(#dt_end)
end
from cte
where dateadd(month, 1, month_start) <= #dt_end
)
select * from cte
Demo on DB Fiddle.
If we set the boundaries as follows:
declare #dt_start date = '2020-07-10';
declare #dt_end date = '2020-09-10';
Then the query returns:
month_start | no_days
:---------- | ------:
2020-07-01 | 22
2020-08-01 | 31
2020-09-01 | 10
You can refer this
;with dates(thedate) as (
select dateadd(yy,years.number,0)+days.number
from master..spt_values years
join master..spt_values days
on days.type='p' and days.number < datepart(dy,dateadd(yy,years.number+1,0)-1)
where years.type='p' and years.number between 100 and 150
-- note: 100-150 creates dates in the year range 2000-2050
-- adjust as required
)
select dateadd(m,datediff(m, 0, d.thedate),0) themonth, count(1)
from dates d
where d.thedate between '2020-01-01' and '2020-08-03'
group by datediff(m, 0, d.thedate)
order by themonth;
Please refer the link below where RichardTheKiwi user given a clear example for your scenario.
SQL Server query for total number of days for a month between date ranges
You can do all the work at the month level rather than the day level -- which should be a bit faster. Here is a method using a recursive CTE:
with cte as (
select #startdate as startdate, #enddate as enddate,
datefromparts(year(#startdate), month(#startdate), 1) as month
union all
select startdate, enddate, dateadd(month, 1, month)
from cte
where dateadd(month, 1, month) < #enddate
)
select month,
(case when month <= startdate and dateadd(month, 1, month) >= enddate
then day(enddate) - day(startdate) + 1
when month <= startdate
then day(eomonth(month)) - day(startdate) + 1
when dateadd(month, 1, month) < enddate
then day(eomonth(month))
when dateadd(month, 1, month) >= enddate
then day(enddate)
end)
from cte;
And the db<>fiddle.
The logic is simpler at the day level:
with cte as (
select #startdate as dte, #enddate as enddate
union all
select dateadd(day, 1, dte), enddate
from cte
where dte < enddate
)
select datefromparts(year(dte), month(dte), 1) as yyyymm, count(*)
from cte
group by datefromparts(year(dte), month(dte), 1)
order by yyyymm
option (maxrecursion 0)
Here is a solution with recursive CTE.
declare #startDate date = '2020-07-01'
declare #endDate date = '2020-08-03'
; WITH cte (n, year, month, daycnt)
AS (
SELECT
0
, DATEPART(year, #startDate)
, DATENAME(MONTH, #startDate)
, DATEPART(day, EOMONTH( #startDate ) ) - DATEPART(day, #startDate ) + 1
UNION ALL
SELECT
n + 1
, DATEPART(year, DATEADD(month, n + 1, #startDate) )
, DATENAME(MONTH, DATEADD(month, n + 1, #startDate) )
, IIF(
n = ( DATEPART(month, #endDate) - DATEPART(month, #startDate) ) + ( DATEPART(year, #endDate) - DATEPART(year, #startDate) ) * 12 - 1
, DATEPART(day, #endDate )
, DATEPART(day, EOMONTH( DATEADD(month, n + 1, #startDate) ) )
)
FROM
cte
WHERE
n <= ( DATEPART(month, #endDate) - DATEPART(month, #startDate) ) + ( DATEPART(year, #endDate) - DATEPART(year, #startDate) ) * 12 - 1
)
SELECT *
FROM cte
ORDER BY n
OPTION (maxrecursion 0)
This could be further simplified with a number function but that would also be essentially be a recursive CTE, though it would definitely look cleaner. But it requires defining a function on top of this SELECT statement.
This question already has answers here:
months between two dates in sql server with starting and end date of each of them in sql server
(6 answers)
Closed 4 years ago.
Given two #START_DATE and #END_DATE parameters, I'd like to write a query that generates pairs of (month_start_date, month_end_date) for every month that exists between those two days, including the ones those dates are in.
E.g. If #START_DATE = '2018-01-14' and #END_DATE = '2018-05-04' (yyyy-MM-dd), I'd like the output to be
month_start_date, month_end_date
2018-01-01, 2018-01-31
2018-02-01, 2018-02-28
2018-03-01, 2018-03-31
2018-04-01, 2018-04-30
2018-05-01, 2018-05-31
I tend to go for recursive CTEs for this purpose:
with dates as (
select datefromparts(year(#start_date), month(#start_date), 1) as dte
union all
select dateadd(month, 1, dte)
from dates
where dateadd(month, 1, dte) <= #end_date
)
select dte as start_date, eomonth(dte) as end_date
from dates;
This works as-is for up to 100 months. For more than that, you need to use set the max recursion option.
Following query returns the required result.
declare #StartDate date = '2018-01-14'
, #EndDate date = '2018-05-04';
;with Months as (
select top (datediff(month,#StartDate,#EndDate)+1)
[month_start_date] = dateadd(month
, datediff(month, 0, #StartDate) + row_number() over (order by number) -1
, 0)
, month_end_date = dateadd(day,-1,dateadd(month
, datediff(month, 0, #StartDate) + row_number() over (order by number)
,0))
from master.dbo.spt_values
order by [month_start_date]
)
select * from Months;
You need recursive cte :
with t as (
select dateadd(day, 1, eomonth(#START_DATE, -1)) as start_dt, #END_DATE as end_dt
union all
select dateadd(mm, 1, start_dt), end_dt
from t
where dateadd(mm, 1, start_dt) < #END_DATE
)
select start_dt as month_start_date, eomonth(start_dt) as month_end_date
from t
option (maxrecursion 0);
Is there is a way to group dates by week of month in SQL Server?
For example
Week 2: 05/07/2012 - 05/13/2012
Week 3: 05/14/2012 - 05/20/2012
but with Sql server statement
I tried
SELECT SOMETHING,
datediff(wk, convert(varchar(6), getdate(), 112) + '01', getdate()) + 1 AS TIME_
FROM STATISTICS_
GROUP BY something, TIME_
ORDER BY TIME_
but it returns the week number of month. (means 3)
How to get the pair of days for current week ?
For example, now we are in third (3rd) week and I want to show 05/14/2012 - 05/20/2012
I solved somehow:
SELECT DATEADD(ww, DATEDIFF(ww,0,<my_column_name>), 0)
select DATEADD(ww, DATEDIFF(ww,0,<my_column_name>), 0)+6
Then I will get two days and I will concatenate them later.
All right, bear with me here. We're going to build a temporary calendar table that represents this month, including the days from before and after the month that fall into your definition of a week (Monday - Sunday). I do this in a lot of steps to try to make the process clear, but I probably haven't excelled at that in this case.
We can then generate the ranges for the different weeks, and you can join against your other tables using that.
SET DATEFIRST 7;
SET NOCOUNT ON;
DECLARE #today SMALLDATETIME, #fd SMALLDATETIME, #rc INT;
SELECT #today = DATEADD(DAY, DATEDIFF(DAY, 0, GETDATE()), 0), -- today
#fd = DATEADD(DAY, 1-DAY(#today), #today), -- first day of this month
#rc = DATEPART(DAY, DATEADD(DAY, -1, DATEADD(MONTH, 1, #fd)));-- days in month
DECLARE #thismonth TABLE (
[date] SMALLDATETIME,
[weekday] TINYINT,
[weeknumber] TINYINT
);
;WITH n(d) AS (
SELECT TOP (#rc+12) DATEADD(DAY, ROW_NUMBER() OVER
(ORDER BY [object_id]) - 7, #fd) FROM sys.all_objects
)
INSERT #thismonth([date], [weekday]) SELECT d, DATEPART(WEEKDAY, d) FROM n;
DELETE #thismonth WHERE [date] < (SELECT MIN([date]) FROM #thismonth WHERE [weekday] = 2)
OR [date] > (SELECT MAX([date]) FROM #thismonth WHERE [weekday] = 1);
;WITH x AS ( SELECT [date], weeknumber, rn = ((ROW_NUMBER() OVER
(ORDER BY [date])-1) / 7) + 1 FROM #thismonth ) UPDATE x SET weeknumber = rn;
-- now, the final query given all that (I've only broken this up to get rid of the vertical scrollbars):
;WITH ranges(w,s,e) AS (
SELECT weeknumber, MIN([date]), MAX([date]) FROM #thismonth GROUP BY weeknumber
)
SELECT [week] = CONVERT(CHAR(10), r.s, 120) + ' - ' + CONVERT(CHAR(10), r.e, 120)
--, SOMETHING , other columns from STATISTICS_?
FROM ranges AS r
-- LEFT OUTER JOIN dbo.STATISTICS_ AS s
-- ON s.TIME_ >= r.s AND s.TIME_ < DATEADD(DAY, 1, r.e)
-- comment this out if you want all the weeks from this month:
WHERE w = (SELECT weeknumber FROM #thismonth WHERE [date] = #today)
GROUP BY r.s, r.e --, SOMETHING
ORDER BY [week];
Results with WHERE clause:
week
-----------------------
2012-05-14 - 2012-05-20
Results without WHERE clause:
week
-----------------------
2012-04-30 - 2012-05-06
2012-05-07 - 2012-05-13
2012-05-14 - 2012-05-20
2012-05-21 - 2012-05-27
2012-05-28 - 2012-06-03
Note that I chose YYYY-MM-DD on purpose. You should avoid regional formatting like M/D/Y especially for input but also for display. No matter how targeted you think your audience is, you're always going to have someone who thinks 05/07/2012 is July 5th, not May 7th. With YYYY-MM-DD there is no ambiguity whatsoever.
Create a calendar table, then you can query week numbers, first/last days of specific weeks and months etc. You can also join on it queries to get a date range etc.
How about a case statement?
case when datepart(day, mydatetime) between 1 and 7 then 1
when datepart(day, mydatetime) between 8 and 14 then 2
...
You'll also have to include the year & month unless you want all the week 1s in the same group.
It's not clear of you want to "group dates by week of month", or alternately "select data from a given week"
If you mean "group" this little snippet should get you 'week of month':
SELECT <stuff>
FROM CP_STATISTICS
WHERE Month(<YOUR DATE COL>) = 5 --april
GROUP BY Year(<YOUR DATE COL>),
Month(<YOUR DATE COL>),
DATEDIFF(week, DATEADD(MONTH, DATEDIFF(MONTH, 0, <YOUR DATE COL>), 0)
, <YOUR DATE COL>) +1
Alternately, if you want "sales for week 1 of April, ordered by date" You could do something like..
DECLARE #targetDate datetime2 = '5/3/2012'
DECLARE #targetWeek int = DATEDIFF(week, DATEADD(MONTH,
DATEDIFF(MONTH, 0, #targetDate), 0), #targetDate) +1
SELECT <stuff>
FROM CP_STATISTICS
WHERE MONTH(#targetDate) = Month(myDateCol) AND
YEAR(#targetDate) = Year (myDateCol) AND
#targetWeek = DATEDIFF(week, DATEADD(MONTH,
DATEDIFF(MONTH, 0, myDateCol), 0), myDateCol) +1
ORDER BY myDateCol
Note, things would get more complicated if you use non-standard weeks, or want to reach a few days into an earlier month for weeks that straddle a month boundary.
EDIT 2
From looking at your 'solved now' section. I think your question is "how do I get data out of a table for a given week?"
Your solution appears to be:
DECLARE #targetDate datetime2 = '5/1/2012'
DECLARE #startDate datetime2 = DATEADD(ww, DATEDIFF(ww,0,targetDate), 0)
DECLARE #endDate datetime2 = DATEADD(ww, DATEDIFF(ww,0,#now), 0)+6
SELECT <stuff>
FROM STATISTICS_
WHERE dateStamp >= #startDate AND dateStamp <= #endDate
Notice how if the date is 5/1 this solution results in a start date of '4/30/2012'. I point this out because your solution crosses month boundaries. This may or may not be desirable.
I have a table called FcData and the data looks like:
Op_Date
2011-02-14 11:53:40.000
2011-02-17 16:02:19.000
2010-02-14 12:53:40.000
2010-02-17 14:02:19.000
I am looking to get the Number of weeks in That Month from Op_Date. So I am looking for output like:
Op_Date Number of Weeks
2011-02-14 11:53:40.000 5
2011-02-17 16:02:19.000 5
2010-02-14 12:53:40.000 5
2010-02-17 14:02:19.000 5
This page has some good functions to figure out the last day of any given month: http://www.sql-server-helper.com/functions/get-last-day-of-month.aspx
Just wrap the output of that function with a DATEPART(wk, last_day_of_month) call. Combining it with an equivalent call for the 1st-day-of-week will let you get the number of weeks in that month.
Use this to get the number of week for ONE specific date. Replace GetDate() by your date:
declare #dt date = cast(GetDate() as date);
declare #dtstart date = DATEADD(day, -DATEPART(day, #dt) + 1, #dt);
declare #dtend date = dateadd(DAY, -1, DATEADD(MONTH, 1, #dtstart));
WITH dates AS (
SELECT #dtstart ADate
UNION ALL
SELECT DATEADD(day, 1, t.ADate)
FROM dates t
WHERE DATEADD(day, 1, t.ADate) <= #dtend
)
SELECT top 1 DatePart(WEEKDAY, ADate) weekday, COUNT(*) weeks
FROM dates d
group by DatePart(WEEKDAY, ADate)
order by 2 desc
Explained: the CTE creates a result set with all dates for the month of the given date. Then we query the result set, grouping by week day and count the number of occurrences. The max number will give us how many weeks the month overlaps (premise: if the month has 5 Mondays, it will cover five weeks of the year).
Update
Now, if you have multiple dates, you should tweak accordingly, joining your query with the dates CTE.
Here is my take on it, might have missed something.
In Linq:
from u in TblUsers
let date = u.CreateDate.Value
let firstDay = new DateTime(date.Year, date.Month, 1)
let lastDay = firstDay.AddMonths(1)
where u.CreateDate.HasValue
select Math.Ceiling((lastDay - firstDay).TotalDays / 7)
And generated SQL:
-- Region Parameters
DECLARE #p0 Int = 1
DECLARE #p1 Int = 1
DECLARE #p2 Float = 7
-- EndRegion
SELECT CEILING(((CONVERT(Float,CONVERT(BigInt,(((CONVERT(BigInt,DATEDIFF(DAY, [t3].[value], [t3].[value2]))) * 86400000) + DATEDIFF(MILLISECOND, DATEADD(DAY, DATEDIFF(DAY, [t3].[value], [t3].[value2]), [t3].[value]), [t3].[value2])) * 10000))) / 864000000000) / #p2) AS [value]
FROM (
SELECT [t2].[createDate], [t2].[value], DATEADD(MONTH, #p1, [t2].[value]) AS [value2]
FROM (
SELECT [t1].[createDate], CONVERT(DATETIME, CONVERT(NCHAR(2), DATEPART(Month, [t1].[value])) + ('/' + (CONVERT(NCHAR(2), #p0) + ('/' + CONVERT(NCHAR(4), DATEPART(Year, [t1].[value]))))), 101) AS [value]
FROM (
SELECT [t0].[createDate], [t0].[createDate] AS [value]
FROM [tblUser] AS [t0]
) AS [t1]
) AS [t2]
) AS [t3]
WHERE [t3].[createDate] IS NOT NULL
According to this MSDN article: http://msdn.microsoft.com/en-us/library/ms174420.aspx you can only get the current week in the year, not what that month returns.
There may be various approaches to implementing the idea suggested by #Marc B. Here's one, where no UDFs are used but the first and the last days of month are calculated directly:
WITH SampleData AS (
SELECT CAST('20110214' AS datetime) AS Op_Date
UNION ALL SELECT '20110217'
UNION ALL SELECT '20100214'
UNION ALL SELECT '20100217'
UNION ALL SELECT '20090214'
UNION ALL SELECT '20090217'
),
MonthStarts AS (
SELECT
Op_Date,
MonthStart = DATEADD(DAY, 1 - DAY(Op_Date), Op_Date)
/* alternatively: DATEADD(MONTH, DATEDIFF(MONTH, 0, Op_Date), 0) */
FROM FcData
),
Months AS (
SELECT
Op_Date,
MonthStart,
MonthEnd = DATEADD(DAY, -1, DATEADD(MONTH, 1, MonthStart))
FROM FcData
)
Weeks AS (
SELECT
Op_Date,
StartWeek = DATEPART(WEEK, MonthStart),
EndWeek = DATEPART(WEEK, MonthEnd)
FROM MonthStarts
)
SELECT
Op_Date,
NumberOfWeeks = EndWeek - StartWeek + 1
FROM Weeks
All calculations could be done in one SELECT, but I chose to split them into steps and place every step in a separate CTE so it could be seen better how the end result was obtained.
You can get number of weeks per month using the following method.
Datepart(WEEK,
DATEADD(DAY,
-1,
DATEADD(MONTH,
1,
DATEADD(DAY,
1 - DAY(GETDATE()),
GETDATE())))
-
DATEADD(DAY,
1 - DAY(GETDATE()),
GETDATE())
+1
)
Here how you can get accurate amount of weeks:
DECLARE #date DATETIME
SET #date = GETDATE()
SELECT ROUND(cast(datediff(day, dateadd(day, 1-day(#date), #date), dateadd(month, 1, dateadd(day, 1-day(#date), #date))) AS FLOAT) / 7, 2)
With this code for Sep 2014 you'll get 4.29 which is actually true since there're 4 full weeks and 2 more days.