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I am extracting below from a date:
Week Number
Day Number
Where start day of my Week is Monday.
For Example the date is : '1st May 2019' or '2019-05-01'
Then
Week Number = 1
Day Number = 3
Because the day is Wednesday on 1st May 2019.
I am using MS SQL Server 2012.
and used below code from net to get Week Number.
I get the Week Number from this but i do not understand what actually it is doing.
select datediff(week, dateadd(week, datediff(week, 0, dateadd(month, datediff(month, 0, #date), 0)), 0), #date - 1) + 1
Can anyone explain it to me and tell me how to extract the Day Number.
This select will give you the #date day of week (1: Monday, 7: Sunday), regardless of your ##DATEFIRST set MS Doc:
select (DATEPART(DW, #date) + ##DATEFIRST + 5) % 7 + 1 as [Day of Week]
If you set datefirst, you can simply do:
set datefirst 1; -- "1" = "Monday"
select datepart(week, getdate()), datepart(weekday, getdate())
Others have mentioned how to get the day-of-week.
In terms of your code for getting week-of-month:
select datediff(week, dateadd(week, datediff(week, 0, dateadd(month, datediff(month, 0, #date), 0)), 0), #date - 1) + 1
To my eye it appears to get the number of full months elapsed between day zero and the current date, giving us a month-serial.
Then adds zero months to this figure (which is apparently being used to cast the month-serial back to a date representing the beginning of the current month), giving us a beginning-of-month date.
It then gets the number of full weeks elapsed between day zero and the beginning-of-month date, giving us a week-serial.
The week-serial is then cast back to a date representing the beginning-of-week date, representing the Monday on or prior to the beginning-of-month date.
It then gets the difference, in weeks, between the monday-on-or-prior date and the day before the current date, and adds 1 (so that the first week of the month is represented as Week 1).
It sounds like a plausible algorithm, but I wouldn't endorse it without testing and inspecting the intermediary values, however, or unless you have found it from an authoritative source.
Check This-
SELECT ISNULL(NULLIF(DATEPART(DW,GETDATE())-1,0),7)
I have constructed the query as per your need. It works fine.
NOTE : Sunday is considered as 1st day of the week and Saturday is considered as 7th day of the week
DECLARE #dt datetime='2019-05-01'
DECLARE #Weekday INT
if(((DATEPART(DW, #dt) + ##DATEFIRST + 5) % 7 +2) =8)
SET #Weekday =1
else
SET #Weekday = ((DATEPART(DW, #dt) + ##DATEFIRST + 5) % 7 +2)
select DATEPART(week, #dt)- DATEPART(week, CONVERT(CHAR(6), #dt, 112)+'01')+ 1 AS [Week Number], #Weekday AS [Day of Week]
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 wanted to get the week from the given date, for this I tried with the DATENAME function to get the WEEK like,
Select DateName(WEEK,'2012-03-09')
am getting the output as 10. I want to get the starting date and ending date of this week like, 2012-03-04 to 2012-03-10 Is it possible?
Do something like this:
DECLARE #MyDate Date = '2012-03-09';
-- This gets you the SUNDAY of the week your date falls in...
SELECT DATEADD(DAY, -(DATEPART(WEEKDAY, #MyDate) - 1), #MyDate);
-- This gets you the SATURDAY of the week your date falls in...
SELECT DATEADD(DAY, (7 - DATEPART(WEEKDAY, #MyDate)), #MyDate);
-- This will show the range as a single column
SELECT
CONVERT(NVarChar, DATEADD(DAY, -(DATEPART(WEEKDAY, #MyDate) - 1), #MyDate))
+ ' through ' +
CONVERT(NVarChar, DATEADD(DAY, (7 - DATEPART(WEEKDAY, #MyDate)), #MyDate));
try the following, change getdate to your date
Select
DateAdd(d, 1- DatePart(dw,GetDate()),GetDate()) FirstDayOfWeek,
DateAdd(d, 7- DatePart(dw,GetDate()),GetDate()) LastDayOfWeek
This does not rely on your default setting of datefirst.
set datefirst 4 -- this row does nothing in this query.
-- It will hurt the queries using datepart though
declare #t table(dt datetime)
insert #t values('2012-03-09 11:12'), ('2012-03-10 11:12'),('2012-03-11 11:12')
-- week from sunday to saturday
Select dateadd(week, datediff(week, 0, dt),-1),
dateadd(week, datediff(week, 0, dt),+5)
from #t
I'm trying to get the most recent Friday in SQL Server 2008.
I have this. It gets the beginning of the week (monday) then subtracts 3 days to get Friday.
declare #recentFriday datetime = DATEADD(ww, DATEDIFF(dd,0,GETDATE()), 0)-3
When this is run during the week, it gets last Friday's date which is correct. But when run on Friday (or Saturday), it still gets last week's date instead of the current week's Friday. I'm about to use if/else conditions but I'm sure there's an easier way.
This works for any input and any setting of DATEFIRST:
dateadd(d, -((datepart(weekday, getdate()) + 1 + ##DATEFIRST) % 7), getdate())
It works by adjusting the weekday value so that 0 = Friday, simulating Friday as the beginning of the week. Then subtract the weekday value if non-zero to get the most recent Friday.
Edit: Updated to work for any setting of DATEFIRST.
DECLARE #date DATETIME = '20110512' -- Thursday
SELECT DATEADD(DAY,-(DATEDIFF(DAY,'19000105',#date)%7),#date) --20110506
SET #date = '20110513' -- Friday
SELECT DATEADD(DAY,-(DATEDIFF(DAY,'19000105',#date)%7),#date) --20110513
SET #date = '20110514' -- Saturday
SELECT DATEADD(DAY,-(DATEDIFF(DAY,'19000105',#date)%7),#date) --20110513
Calculate the number of days between a known Friday (05 Jan 1900) and the given date
The remainder left from dividing the difference in 1. by 7 will be the days elapsed since the last Friday
Subtract the remainder in 2. from the given date
you can check if the current day of week is friday or greater DATEPART(dw,GETDATE()) and then call (SELECT DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 0)+4) or (SELECT DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 0)-3)
SELECT
CASE WHEN DATEPART(dw,GETDATE()) >= 5 THEN
(SELECT DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 0)+4)
ELSE
(SELECT DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 0)-3)
END
Using a known Friday date (I'll use Jan 7, 2011) as a starting point, you can do this:
DECLARE #d DATETIME
SET #d = '2011-05-13' /* Friday */
SELECT DATEADD(DAY, (DATEDIFF (DAY, '20110107', #d) / 7) * 7, '20110107')
/* Returns 2011-05-13 */
SET #d = '2011-05-12' /* Thursday */
SELECT DATEADD(DAY, (DATEDIFF (DAY, '20110107', #d) / 7) * 7, '20110107')
/* Returns 2011-05-06 */
Just choose a known Friday that's older than any dates you'll be using in your calculations.
SELECT CONVERT(VARCHAR(12),GETDATE()) AS Today,
CASE WHEN (DATEPART(DW,GETDATE())< 7)
THEN CONVERT(VARCHAR(12),(DATEADD(dd,-(DATEPART(DW,GETDATE())+1),GETDATE())))
ELSE CONVERT(VARCHAR(12),(DATEADD(d,- 1,GETDATE())))
END AS [Last Friday]
Here is a completly set oriented way to achive the last Friday:
select Friday from
(
select max(GetDate()) as Friday where datepart(dw, getdate()) = 6
union all
select max((GetDate() - 1)) where datepart(dw, (getdate() - 1)) = 6
union all
select max((GetDate() - 2)) where datepart(dw, (getdate() - 2)) = 6
union all
select max((GetDate() - 3)) where datepart(dw, (getdate() - 3)) = 6
union all
select max((GetDate() - 4)) where datepart(dw, (getdate() - 4)) = 6
union all
select max((GetDate() - 5)) where datepart(dw, (getdate() - 5)) = 6
) x where Friday is not null
The other solutions were not working for my use case.
This works for finding any previous day by replacing 'Sunday' with the day you`re looking for:
DECLARE #myDate DATE = GETDATE()
WHILE DATENAME(WEEKDAY, #myDate) <> 'Sunday'
BEGIN
SET #myDate = DATEADD(DAY, -1, #myDate)
END
This is what I got I hope it helps
DECLARE #UserDate DateTime
SET #UserDate = '2020-09-03'
SELECT DATEADD(day, (6 - datepart(weekday, #UserDate)) , #UserDate)
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