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In SQL Server 2008 i want to set a default date value of every Friday to show up in the column when i insert a new record?
ALTER TABLE myTable ADD CONSTRAINT_NAME DEFAULT GETDATE() FOR myColumn
Whats the best way to show every Friday?
I want the default value to be based on the now date then knowing that the next available date is 05-07/2013
I have the following:
dateadd(d, -((datepart(weekday, getdate()) + 1 + ##DATEFIRST) % 7), getdate())
But when passing todays date, it gave me: 2013-06-28 which is actually LAST Friday!, it should be the up and coming Friday!
SELECT DATEADD(day,-3, DATEADD(week, DATEDIFF(week, 0, current_timestamp)+1, 0)) AS LastFridayDateOfWeek
Gets the last date of current week (sunday) then subtracts 3 from that to get Friday.
Replace current_timestamp if you need a different dates friday.
EDIT:
I thought about this a bit, and if the above (Friday THIS WEEK, so for Saturday it gives the previous date) does not work, you could easily use a reference date set like so:
DATEADD(DAY,7 + DATEDIFF(day,'20100109',#checkDateTime)/7*7,'20100108') as FridayRefDate
Same thing but with no hard coded Friday/Saturday in it:
DATEADD(DAY,7 + DATEDIFF(day,DATEADD(wk, DATEDIFF(wk,0,#checkDateTime),5),#checkDateTime)/7*7,DATEADD(wk, DATEDIFF(wk,0,#checkDateTime), 4))
So for 20100109 is a Friday.
SET #checkDateTime = '2012-01-14 3:34:00.000'
SELECT DATEADD(DAY,7 + DATEDIFF(day,'20100109',#checkDateTime)/7*7,'20100108') as FridayRefDate
it returns "2012/1/20"
But for SET #checkDateTime = '2012-01-13 3:34:00.000' it returns "2012/1/13"
If your current query gives you last Friday, the easiest thing to do is simply to add 7 to it:
select dateadd(d, 7-((datepart(weekday, getdate()) + 1 + ##DATEFIRST) % 7), getdate())
------------------^
SELECT CONVERT(DATE, ( CASE WHEN DATEPART(dw, GETDATE()) - 6 <= 0
THEN DATEADD(dd,
( DATEPART(dw, GETDATE()) - 6 ) * -1,
GETDATE())
ELSE DATEADD(dd, ( DATEPART(dw, GETDATE()) ) - 1,
GETDATE())
END )) AS NearestFriday
Just add 7 to the formula
SELECT DATEADD(dd,CAST(5-GETDATE() AS int)%7,GETDATE()+7)
To verify the formula:
WITH test AS (
SELECT GETDATE() AS d UNION ALL
SELECT DATEADD(dd,1,d)
FROM test WHERE d < GETDATE() + 30
)
SELECT
d AS [input],
DATEADD(dd,CAST(5-d AS int)%7,d+7) AS [output]
FROM test
To tweak the the formula, adjust the 5- and the +7
How do I get yesterdays date and if the current date is Monday I would need Sunday, Saturday and Friday. This is already asked here for ms access. I now need this for SQL server. How would I go about this?
Will create an inline view that will return the previous date in SQL It will return between 1 and 3 rows depending on the current date.
If current date is Monday Return:
Sundays date
Saturdays date
Fridays date
If current date is Tuesday then Return: Mondays date
If current date is Wednesday then Return: Tuesdays date
If current date is Thursday then Return: Wednesdays date
If current date is Friday then Return: Thursdays date
If current date is Saturday then Return: Fridays date
If current date is Sunday then Return: Saturdays date
I hope this helps explain what I am trying to do more clearly.
sample select query
--get previous date
select * from [Purchase Orders] where MyDate in (previous date(s))
Not entirely clear about the question though. But you can do something on the similar lines.
Using DATEPART and DATEADD
DECLARE #TodaysDate DATETIME
SET #TodaysDate = '2013-04-10'
SELECT CASE
WHEN DATEPART (DW, DATEADD (DD, -1, #TodaysDate )) IN (1, 6, 7)
THEN 'WeekEnd' ELSE 'WeekDay'
END D
Using the query you linked to, you can try the code below:
SELECT *
FROM [Purchase Order]
WHERE MyDate >= CASE
WHEN DATENAME(dw, CONVERT(CHAR(8), GETDATE() , 112)) LIKE 'Monday' THEN CONVERT(CHAR(8), DATEADD(dd, -3, GETDATE()), 112)
ELSE CONVERT(CHAR(8), DATEADD(dd, -1, GETDATE()), 112)
END
AND MyDate < CONVERT(CHAR(8),GETDATE(),112)
You can use the following table function:
CREATE FUNCTION dbo.ufnPreviousDay (#suppliedDate DATE)
RETURNS #PreviousDates TABLE (PreviousDate DATE)
AS
BEGIN
INSERT INTO #PreviousDates (
PreviousDate)
SELECT
PreviousDate = DATEADD(DAY, -1, #suppliedDate)
IF DATEPART(WEEKDAY, DATEADD(DAY, ##DATEFIRST - 1, #suppliedDate)) = 1 -- If Monday
BEGIN
INSERT INTO #PreviousDates (
PreviousDate)
SELECT
PreviousDate = DATEADD(DAY, -2, #suppliedDate)
UNION ALL
SELECT
PreviousDate = DATEADD(DAY, -3, #suppliedDate)
END
RETURN
END
You can use it like this:
SELECT
T.*
FROM
dbo.ufnPreviousDay('2018-03-19') AS T
/*
Result:
2018-03-18
2018-03-17
2018-03-16
*/
If you want today's previous dates:
SELECT
T.*
FROM
YourTable AS T
INNER JOIN dbo.ufnPreviousDay(GETDATE()) AS M ON T.MyDate = M.PreviousDate
I'm trying to get the correct SQL code to obtain last Friday's date. A few days ago, I thought I had my code correct. But just noticed that it's getting last week's Friday date, not the last Friday. The day I'm writing this question is Saturday, 8/11/2012 # 12:23am. In SQL Server, this code is returning Friday, 8/3/2012. However, I want this to return Friday, 8/10/2012 instead. How can I fix this code? Since we're getting to specifics here, if the current day is Friday, then I want to return today's date. So if it were yesterday (8/10/2012) and I ran this code yesterday, then I would want this code to return 8/10/2012, not 8/3/2012.
SELECT DATEADD(DAY, -3, DATEADD(WEEK, DATEDIFF(WEEK, 0, GETDATE()), 0))
try this:
declare #date datetime;
set #date='2012-08-09'
SELECT case when datepart(weekday, #date) >5 then
DATEADD(DAY, +4, DATEADD(WEEK, DATEDIFF(WEEK, 0, #date), 0))
else DATEADD(DAY, -3, DATEADD(WEEK, DATEDIFF(WEEK, 0, #date), 0)) end
result:
2012-08-03
Example2:
declare #date datetime;
set #date='2012-08-10'
SELECT case when datepart(weekday, #date) >5 then
DATEADD(DAY, +4, DATEADD(WEEK, DATEDIFF(WEEK, 0, #date), 0))
else DATEADD(DAY, -3, DATEADD(WEEK, DATEDIFF(WEEK, 0, #date), 0)) end
result:
2012-08-10
Modular arithmetic is the most direct approach, and order of operations decides how Fridays are treated:
DECLARE #test_date DATETIME = '2012-09-28'
SELECT DATEADD(d,-1-(DATEPART(dw,#test_date) % 7),#test_date) AS Last_Friday
,DATEADD(d,-(DATEPART(dw,#test_date+1) % 7),#test_date) AS This_Friday
Use this :
SELECT DATEADD(day, (DATEDIFF (day, '19800104', CURRENT_TIMESTAMP) / 7) * 7, '19800104') as Last_Friday
None of that? Try this:
DECLARE #D DATE = GETDATE()
SELECT DATEADD(D,-(DATEPART(W,#D)+1)%7,#D)
A tested function which works no matter what ##DATEFIRST is set to.
-- ==============
-- fn_Get_Week_Ending_forDate
-- Author: Shawn C. Teague
-- Create date: 2017
-- Modified date:
-- Description: Returns the Week Ending Date on DayOfWeek for a given stop date
-- Parameters: DayOfWeek varchar(10) i.e. Monday,Tues,Wed,Friday,Sat,Su,1-7
-- DateInWeek DATE
-- ==============
CREATE FUNCTION [dbo].[fn_Get_Week_Ending_forDate] (
#DayOfWeek VARCHAR(10),#DateInWeek DATE)
RETURNS DATE
AS
BEGIN
DECLARE #End_Date DATE
,#DoW TINYINT
SET #DoW = CASE WHEN ISNUMERIC(#DayOfWeek) = 1
THEN CAST(#DayOfWeek AS TINYINT)
WHEN #DayOfWeek like 'Su%' THEN 1
WHEN #DayOfWeek like 'M%' THEN 2
WHEN #DayOfWeek like 'Tu%' THEN 3
WHEN #DayOfWeek like 'W%' THEN 4
WHEN #DayOfWeek like 'Th%' THEN 5
WHEN #DayOfWeek like 'F%' THEN 6
ELSE 7
END
select #End_Date =
CAST(DATEADD(DAY,
CASE WHEN (#DoW - (((##datefirst) + datepart(weekday, #DateInWeek)) % 7)) = 7
THEN 0
WHEN (#DoW - (((##datefirst) + datepart(weekday, #DateInWeek)) % 7)) < 0
THEN 7 - ABS(#DoW - (((##datefirst) + datepart(weekday, #DateInWeek)) % 7))
ELSE (#DoW - (((##datefirst) + datepart(weekday, #DateInWeek)) % 7) )
END
,#DateInWeek) AS DATE)
RETURN #End_Date
END
This will give you the Friday of Last week.
SELECT DATEADD(day, -3 - (DATEPART(dw, GETDATE()) + ##DATEFIRST - 2) % 7, GETDATE()) AS LastWeekFriday
This will give you last Friday's Date.
SELECT DATEADD(day, +4 - (DATEPART(dw, GETDATE()) + ##DATEFIRST-2) % 7, GETDATE()) AS LastFriday
select convert(varchar(10),dateadd(d, -((datepart(weekday, getdate()) + 1 + ##DATEFIRST) % 7), getdate()),101)
Following code can be use to return any last day by replacing #dw_wk, test case below use friday as asked in original questions
DECLARE #date SMALLDATETIME
,#dw_wk INT --last day of week required - its integer representation
,#dw_day int --current day integer reprsentation
SELECT #date='8/11/2012'
SELECT #dw_day=DATEPART(dw,#date)
SELECT #dw_wk=DATEPART(dw,'1/2/2015') --Just trying not to hard code 5 for friday, here we can substitute with any date which is friday
SELECT case when #dw_day<#dw_wk then DATEADD(DAY, #dw_wk-7-#dw_day,#date) else DATEADD(DAY,#dw_wk-#dw_day, #date) END
Here's an answer I found here adapted from MySQL to T-SQL that is a one liner using all basic arithmetic (no division or modulos):
SELECT DATEADD(d, 1 - datepart(weekday, dateadd(d, 2, GETDATE())), GETDATE())
You can do all sorts of combinations of this, like get next Friday's date unless today is Friday, or get last Thursday's date unless today is Thursday by just changing the 1 and the 2 literals in the command:
Get next Friday's date unless today is Friday
SELECT DATEADD(d, 7 - datepart(weekday, dateadd(d, 1, GETDATE())), GETDATE())
Get last Thursday's date unless today is Thursday
SELECT DATEADD(d, 1 - datepart(weekday, dateadd(d, 3, GETDATE())), GETDATE())
I have had this same issue, and created the following example to show how to do this and to make it flexible to use whichever day of the week you want. I have different lines in the SELECT statement, just to show what this is doing, but you just need the [Results] line to get the answer. I also used variables for the current date and the target day of the week, to make it easier to see what needs to change.
Finally, there is an example of results when you want to include the current date as a possible example or when you always want to go back to the previous week.
DECLARE #GetDate AS DATETIME = GETDATE();
DECLARE #Target INT = 6 -- 6 = Friday
SELECT
#GetDate AS [Current Date] ,
DATEPART(dw, #GetDate) AS [Current Day of Week],
#Target AS [Target Day of Week] ,
IIF(#Target = DATEPART(dw, #GetDate), 'Yes' , 'No') AS [IsMatch] ,
IIF(#Target = DATEPART(dw, #GetDate), 0 , ((7 + #Target - DATEPART(dw, #GetDate)) % 7) - 7) AS [DateAdjust] ,
------------------------------------------------------------------------------------------------------------------------------------------------
CAST(IIF(#Target = DATEPART(dw, #GetDate), #GetDate, DATEADD(d, (((7 + #Target - DATEPART(dw, #GetDate)) % 7) - 7), #GetDate)) AS DATE) AS [Result]
------------------------------------------------------------------------------------------------------------------------------------------------
;
SELECT
#GetDate AS [Current Date] ,
DATEPART(dw, #GetDate) AS [Current Day of Week],
#Target AS [Target Day of Week] ,
((7 + #Target - DATEPART(dw, #GetDate)) % 7) - 7 AS [DateAdjust] ,
------------------------------------------------------------------------------------------------------------------------------------------------
CAST(DATEADD(d, (((7 + #Target - DATEPART(dw, #GetDate)) % 7) - 7), #GetDate) AS DATE) AS [NOTIncludeCurrent]
------------------------------------------------------------------------------------------------------------------------------------------------
;
SELECT DECODE(TO_CHAR(SYSDATE,'DY'),'FRI',SYSDATE,NEXT_DAY(SYSDATE, 'FRI')-7) FROM dual;
I may be using the wrong term (hence why I can't find it on google).
"Are there any functions or common code for Accounting Months Deliminations?"
For Example, this month started on a friday but on most accounting journals the weeks are measured by the first monday of the month so instead of having the 1st of July it would be the 4th of July. Same thing with the month end (29th instead of the 31st)
Again, I'm sure someone has created this 'wheel' before, and I can't seem to find it for the life of me.
The following query assumes a table, SalesTable, has a field called Amount (the value you want to sum) and a field called SaleDate (the date on which the sale occured.) It also assumes that accounting months begin the first Monday of the month and end on the Sunday prior to the beginning of the next accounting month.
Again, I highly recommend a table-based approach to this, but if you can't modify the schema, this should do the trick in T-SQL:
SELECT
CASE WHEN s.SaleDate < DATEADD(WEEK, DATEDIFF(WEEK, 0, DATEADD(DAY, 6 - DATEPART(DAY, s.SaleDate ),s.SaleDate )), 0)
THEN DATEADD(WEEK, DATEDIFF(WEEK, 0, DATEADD(DAY, 6 - DATEPART(DAY, DATEADD(day,-7,s.SaleDate) ),DATEADD(day,-7,s.SaleDate) )), 0)
ELSE DATEADD(WEEK, DATEDIFF(WEEK, 0, DATEADD(DAY, 6 - DATEPART(DAY, s.SaleDate ),s.SaleDate )), 0)
END AccountingMonth,
SUM(s.Amount) TotalSales
FROM SalesTable s
GROUP BY
CASE WHEN s.SaleDate < DATEADD(WEEK, DATEDIFF(WEEK, 0, DATEADD(DAY, 6 - DATEPART(DAY, s.SaleDate ),s.SaleDate )), 0)
THEN DATEADD(WEEK, DATEDIFF(WEEK, 0, DATEADD(DAY, 6 - DATEPART(DAY, DATEADD(day,-7,s.SaleDate) ),DATEADD(day,-7,s.SaleDate) )), 0)
ELSE DATEADD(WEEK, DATEDIFF(WEEK, 0, DATEADD(DAY, 6 - DATEPART(DAY, s.SaleDate ),s.SaleDate )), 0)
END
Note that the AccountingMonth return field actually contains the date of the first Monday of the month. In actual practice, you probably want to wrap this entire query in another query that reformats AccountingMonth to whatever you like... "2011-07", "2011-08", etc.
Here's how it works: This bit of code is the important part:
DATEADD(WEEK, DATEDIFF(WEEK, 0, DATEADD(DAY, 6 - DATEPART(DAY, s.SaleDate ),s.SaleDate )), 0)
It takes any date and returns the first Monday of the month in which that date occurred. In your case, however, you have to do a little more work because a sale might have occurred in the window between the first of the month and the first Monday of the month. The CASE statement detects that scenario and, if it's true, subtracts a week off of the date before calculating the first Monday.
Good luck!
-Michael
I have some code that takes in a year and month and returns the fiscal start and end dates. Perhaps this will give you something to go by:
DECLARE #yr int;
DECLARE #mo int;
SELECT #yr = 2011
SELECT #mo = 7
DECLARE #FiscalMonthStartDate datetime
DECLARE #FiscalMonthEndDate datetime
DECLARE #startOfMonth datetime
DECLARE #startOfNextMonth datetime
select #startOfMonth = CAST((CAST(#yr AS VARCHAR(4)) + '-' + CAST(#mo AS VARCHAR(2)) + '-' + '01') as DATE)
select #startOfNextMonth = CAST((CAST(#yr AS VARCHAR(4)) + '-' + CAST((#mo + 1) AS VARCHAR(2)) + '-' + '01') as DATE)
SELECT #FiscalMonthStartDate =
CASE
WHEN DATEPART(DW,#startOfMonth) = 0
THEN DATEADD(DD, 1, #startOfMonth)
ELSE
DATEADD(DD, 8 - DATEPART(DW,#startOfMonth), #startOfMonth)
END
SELECT #FiscalMonthEndDate =
CASE
WHEN DATEPART(DW,#startOfNextMonth) = 0
THEN DATEADD(DD, 1, #startOfNextMonth)
ELSE
DATEADD(DD, 8 - DATEPART(DW,#startOfNextMonth), #startOfNextMonth)
END
-- subtract one day to get end of fiscal month (not start of next fiscal month)
SELECT #FiscalMonthEndDate = DATEADD(DD, -1, #FiscalMonthEndDate)
SELECT #FiscalMonthStartDate, #FiscalMonthEndDate
I need to get the last day of the month given as a date in SQL. If I have the first day of the month, I can do something like this:
DATEADD(DAY, DATEADD(MONTH,'2009-05-01',1), -1)
But does anyone know how to generalize it so I can find the last day of the month for any given date?
From SQL Server 2012 you can use the EOMONTH function.
Returns the last day of the month that contains the specified date,
with an optional offset.
Syntax
EOMONTH ( start_date [, month_to_add ] )
How ... I can find the last day of the month for any given date?
SELECT EOMONTH(#SomeGivenDate)
Here's my version. No string manipulation or casting required, just one call each to the DATEADD, YEAR and MONTH functions:
DECLARE #test DATETIME
SET #test = GETDATE() -- or any other date
SELECT DATEADD(month, ((YEAR(#test) - 1900) * 12) + MONTH(#test), -1)
You could get the days in the date by using the DAY() function:
dateadd(day, -1, dateadd(month, 1, dateadd(day, 1 - day(date), date)))
Works in SQL server
Declare #GivenDate datetime
SET #GivenDate = GETDATE()
Select DATEADD(MM,DATEDIFF(MM, 0, #GivenDate),0) --First day of the month
Select DATEADD(MM,DATEDIFF(MM, -1, #GivenDate),-1) --Last day of the month
I know this is a old question but here is another solution that works for me
SET #dtDate = "your date"
DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#dtDate)+1,0))
And if some one is looking for different examples here is a link http://blog.sqlauthority.com/2007/08/18/sql-server-find-last-day-of-any-month-current-previous-next/
I hope this helps some one else.
stackoverflow Rocks!!!!
For SQL server 2012 or above use EOMONTH to get the last date of month
SQL query to display end date of current month
DECLARE #currentDate DATE = GETDATE()
SELECT EOMONTH (#currentDate) AS CurrentMonthED
SQL query to display end date of Next month
DECLARE #currentDate DATE = GETDATE()
SELECT EOMONTH (#currentDate, 1 ) AS NextMonthED
Based on the statements:
SELECT DATEADD(MONTH, 1, #x) -- Add a month to the supplied date #x
and
SELECT DATEADD(DAY, 0 - DAY(#x), #x) -- Get last day of month previous to the supplied date #x
how about adding a month to date #x and then retrieving the last day of the month previous to that (i.e. The last day of the month of the supplied date)
DECLARE #x DATE = '20-Feb-2012'
SELECT DAY(DATEADD(DAY, 0 - DAY(DATEADD(MONTH, 1, #x)), DATEADD(MONTH, 1, #x)))
Note: This was test using SQL Server 2008 R2
Just extend your formula out a little bit:
dateadd(day, -1,
dateadd(month, 1,
cast(month('5/15/2009') as varchar(2)) +
'/1/' +
cast(year('5/15/2009') as varchar(4)))
This works for me, using Microsoft SQL Server 2005:
DATEADD(d,-1,DATEADD(mm, DATEDIFF(m,0,'2009-05-01')+1,0))
WinSQL to get last day of last month (i.e today is 2017-02-09, returns 2017-01-31:
Select dateadd(day,-day(today()),today())
Try to run the following query, it will give you everything you want :)
Declare #a date =dateadd(mm, Datediff(mm,0,getdate()),0)
Print('First day of Current Month:')
Print(#a)
Print('')
set #a = dateadd(mm, Datediff(mm,0,getdate())+1,-1)
Print('Last day of Current Month:')
Print(#a)
Print('')
Print('First day of Last Month:')
set #a = dateadd(mm, Datediff(mm,0,getdate())-1,0)
Print(#a)
Print('')
Print('Last day of Last Month:')
set #a = dateadd(mm, Datediff(mm,0,getdate()),-1)
Print(#a)
Print('')
Print('First day of Current Week:')
set #a = dateadd(ww, Datediff(ww,0,getdate()),0)
Print(#a)
Print('')
Print('Last day of Current Week:')
set #a = dateadd(ww, Datediff(ww,0,getdate())+1,-1)
Print(#a)
Print('')
Print('First day of Last Week:')
set #a = dateadd(ww, Datediff(ww,0,getdate())-1,0)
Print(#a)
Print('')
Print('Last day of Last Week:')
set #a = dateadd(ww, Datediff(ww,0,getdate()),-1)
Print(#a)
WinSQL: I wanted to return all records for last month:
where DATE01 between dateadd(month,-1,dateadd(day,1,dateadd(day,-day(today()),today()))) and dateadd(day,-day(today()),today())
This does the same thing:
where month(DATE01) = month(dateadd(month,-1,today())) and year(DATE01) = year(dateadd(month,-1,today()))
This query can also be used.
DECLARE #SelectedDate DATE = GETDATE()
SELECT DATEADD(DAY, - DAY(#SelectedDate), DATEADD(MONTH, 1 , #SelectedDate)) EndOfMonth
--## Useful Date Functions
SELECT
GETDATE() AS [DateTime],
CAST(GETDATE() AS DATE) AS [Date],
DAY(GETDATE()) AS [Day of Month],
FORMAT(GETDATE(),'MMMM') AS [Month Name],
FORMAT(GETDATE(),'MMM') AS [Month Short Name],
FORMAT(GETDATE(),'MM') AS [Month No],
YEAR(GETDATE()) AS [Year],
CAST(DATEADD(DD,-(DAY(GETDATE())-1),GETDATE()) AS DATE) AS [Month Start Date],
EOMONTH(GETDATE()) AS [Month End Date],
CAST(DATEADD(M,-1,DATEADD(MM, DATEDIFF(M,0,GETDATE()),0)) AS DATE) AS [Previous Month Start Date],
CAST(DATEADD(S,-1,DATEADD(MM, DATEDIFF(M,0,GETDATE()),0)) AS DATE) AS [Previous Month End Date],
CAST(DATEADD(M,+1,DATEADD(MM, DATEDIFF(M,0,GETDATE()),0)) AS DATE) AS [Next Month Start Date],
CAST(DATEADD(D,-1,DATEADD(MM, DATEDIFF(M,0,GETDATE())+2,0)) AS DATE) AS [Next Month End Date],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE()),0) AS DATE) AS [First Day of Current Week],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE())+1,-1) AS DATE) AS [Last Day of Current Week],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE())-1,0) AS DATE) AS [First Day of Last Week],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE()),-1) AS DATE) AS [Last Day of Last Week],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE())+1,0) AS DATE) AS [First Day of Next Week],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE())+2,-1) AS DATE) AS [Last Day of Next Week]
My 2 cents:
select DATEADD(DAY,-1,DATEADD(MONTH,1,DATEADD(day,(0-(DATEPART(dd,'2008-02-12')-1)),'2008-02-12')))
Raj
using sql server 2005, this works for me:
select dateadd(dd,-1,dateadd(mm,datediff(mm,0,YOUR_DATE)+1,0))
Basically, you get the number of months from the beginning of (SQL Server) time for YOUR_DATE. Then add one to it to get the sequence number of the next month. Then you add this number of months to 0 to get a date that is the first day of the next month. From this you then subtract a day to get to the last day of YOUR_DATE.
Take some base date which is the 31st of some month e.g. '20011231'. Then use the
following procedure (I have given 3 identical examples below, only the #dt value differs).
declare #dt datetime;
set #dt = '20140312'
SELECT DATEADD(month, DATEDIFF(month, '20011231', #dt), '20011231');
set #dt = '20140208'
SELECT DATEADD(month, DATEDIFF(month, '20011231', #dt), '20011231');
set #dt = '20140405'
SELECT DATEADD(month, DATEDIFF(month, '20011231', #dt), '20011231');
Using SQL Server, here is another way to find last day of month :
SELECT DATEADD(MONTH,1,GETDATE())- day(DATEADD(MONTH,1,GETDATE()))
I wrote following function, it works.
It returns datetime data type. Zero hour, minute, second, miliseconds.
CREATE Function [dbo].[fn_GetLastDate]
(
#date datetime
)
returns datetime
as
begin
declare #result datetime
select #result = CHOOSE(month(#date),
DATEADD(DAY, 31 -day(#date), #date),
IIF(YEAR(#date) % 4 = 0, DATEADD(DAY, 29 -day(#date), #date), DATEADD(DAY, 28 -day(#date), #date)),
DATEADD(DAY, 31 -day(#date), #date) ,
DATEADD(DAY, 30 -day(#date), #date),
DATEADD(DAY, 31 -day(#date), #date),
DATEADD(DAY, 30 -day(#date), #date),
DATEADD(DAY, 31 -day(#date), #date),
DATEADD(DAY, 31 -day(#date), #date),
DATEADD(DAY, 30 -day(#date), #date),
DATEADD(DAY, 31 -day(#date), #date),
DATEADD(DAY, 30 -day(#date), #date),
DATEADD(DAY, 31 -day(#date), #date))
return convert(date, #result)
end
It's very easy to use.
2 example:
select [dbo].[fn_GetLastDate]('2016-02-03 12:34:12')
select [dbo].[fn_GetLastDate](GETDATE())
Based on the most voted answer at below link I came up with the following solution:
declare #mydate date= '2020-11-09';
SELECT DATEADD(month, DATEDIFF(month, 0, #mydate)+1, -1) AS lastOfMonth
link: How can I select the first day of a month in SQL?
I couldn't find an answer that worked in regular SQL, so I brute forced an answer:
SELECT *
FROM orders o
WHERE (MONTH(o.OrderDate) IN ('01','03','05','07','08','10','12') AND DAY(o.OrderDate) = '31')
OR (MONTH(o.OrderDate) IN ('04','06','09','11') AND DAY(o.OrderDate) = '30')
OR (MONTH(o.OrderDate) IN ('02') AND DAY(o.OrderDate) = '28')
---Start/End of previous Month
Declare #StartDate datetime, #EndDate datetime
set #StartDate = DATEADD(month, DATEDIFF(month, 0, GETDATE())-1,0)
set #EndDate = EOMONTH (DATEADD(month, DATEDIFF(month, 0, GETDATE())-1,0))
SELECT #StartDate,#EndDate