I'm trying to make a dynamic query which searches records from 9 days ago at 21:00 to 2 days ago at 23:00.
I've been asked to make a weekly report in Maximo which lists all records between Sundays at 21:00 and the following Fridays at 23:00. The report must generate each week on Tuesdays at 6:00. I have pieced something together from other queries but can't figure out where to go from here:
startdatetime >= dateadd(hour, 21, dateadd(dd, datediff(dd, 0, getdate()), 0))
and startdatetime <= dateadd(hour, 23, dateadd(dd, datediff(dd, 0, getdate()), 0))
I've used DATEADD before to show all records between certain times of the current day. What I can't figure out is how to get those times for previous dates. I tried doing "...dateadd(dd,datediff(dd,0,getdate()-9),0) and ...dateadd(dd,datediff(dd,0,getdate()-2),0)", but I got a SQL error code 207. I obviously don't understand how the second DATEADD segment works. Can anyone show me what I'm doing wrong?
You're really close, you just need to change the second zero to the days you want to move.
WHERE startdatetime >= dateadd(hour,21,dateadd(dd,datediff(dd,0,getdate()),-9))
AND startdatetime <= dateadd(hour,23,dateadd(dd,datediff(dd,0,getdate()),-3))
The error 207 refers to an invalid column name. Make sure that you don't have any typos on your query and that the columns actually exist.
Using 0 as a base (which equates to 1900-01-01) is not very intuitive or self-documenting. Also writing a query that assumes it will only ever run on a Tuesday is a bit dangerous; we can make it so that if it needs to be run again on Wednesday or Thursday it still yields the right results.
-- first, let's make sure we know what weekday is a Sunday
DECLARE #OriginalDateFirst int = ##DATEFIRST;
IF #OriginalDateFirst <> 7
BEGIN
SET DATEFIRST 7;
END
-- let's figure out today
DECLARE #today datetime = CONVERT(date, GETDATE()),
#weekday int = DATEPART(WEEKDAY, GETDATE());
-- if weekday is 3 (Tuesday), the previous Sunday is today -2,
-- and the Sunday before that is today -9
-- if weekday is 4 (Wednesday), the days are -3, -10
-- if weekday is 5 (Thursday), the days are -4, -11
-- and so on
-- so the range is -6-#weekday to 1-#weekday
DECLARE #MostRecentSunday datetime = DATEADD(HOUR, 23, DATEADD(DAY, 1-#weekday, #today)),
#PreviousSunday datetime = DATEADD(HOUR, 21, DATEADD(DAY, -6-#weekday, #today));
-- make sure this is what you want:
SELECT #PreviousSunday, #MostRecentSunday;
-- now your query just says:
... WHERE startdatetime >= #PreviousSunday
AND startdatetime <= #MostRecentSunday;
-- let's put the datefirst setting back
IF #OriginalDateFirst <> 7
BEGIN
SET DATEFIRST #OriginalDateFirst;
END
Yes, it's definitely more code (though you can condense it quite a bit if you like, and if you know nobody messes with DATEFIRST, you can clear out some of that logic, too).
But I'm a big fan of making code self-documenting vs. terse/cryptic.
I need to set a where condition to the last date of previous year.
I looking for the solution using dateadd function but i can not figure it out. This code gives me the last day of current month. But how to get the last day of december last year '2015-12-31' of have tried different ways but it all gives me odd results.
declare #today datetime
select #today=getdate()
select convert(varchar(10), dateadd(dd, -day(dateadd(mm, 1, #today)),dateadd(mm, 1, #today)),101)
Assuming Sybase ASE, not ASA or IQ:
declare #lastyear smallint, #dec31 datetime
select #lastyear = datepart(year,getdate()) - 1
select #dec31 = convert(datetime, convert(char(4), #lastyear) + "1231")
select #dec31
This produces
--------------------------
Dec 31 2015 12:00AM
#lastyear contains 2015, the last year, then text "2015" is concatenaded into "20151231", which is the desired date in AAAAMMDD char format. Next step converts it into date, datetime or smalldatetime. #dec31 stores that result.
select convert(datetime, convert(varchar, datepart(year, getdate()) - 1 ) + '/12' + '/31')
Dec 31 2015 12:00AM
SELECT DATEADD(DD,-1,DATEADD(YEAR, DATEDIFF(YEAR, '', GETDATE())+1, ''))
I am trying to write a bit of SQL that will give the last day of the month in 6 months time.
E.g. if I have of a date of inspection 15-07-2015 I want the next inspection date to be 31-01-2016
The date of inspection could be any day of any month
Any advise would be appreciated.
if i understand. that's it
declare #d date = '2015-07-15'
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#d)+7,0))
If you are using SQL Server 2012+ you can use the eomonth function to get the last day of a month:
declare #d date = '2015-07-15'
select eomonth(#d,6)
result: 2016-01-31
The function takes a date and an optional integer that specifies the number of months to add to the date parameter.
try this
declare #FindLastDay datetime
set #FindLastDay=CONVERT(varchar(10),DATEADD(M,6,'2015-07-15'),120)
SELECT CAST(DATEADD(ms, -3, DATEADD(mm, DATEDIFF(m, 0, #FindLastDay) + 1, 0)) as DATE)
output:
2016-01-31
Select DateAdd(D, Day(DateAdd(Month, 7, '2015-07-15')) * -1, DateAdd(M, 7, '2015-07-15'))
Yields the answer you expect.
Could you please help me select a date which is the beginning of a particular following month, e.g. April?
For example, if it is Jan 08 2013, it should select April 01 2013, but if it is June 08 2013, it should select April 01 2014.
Thanks.
I would create a calendar table, then you can simply do something like this:
select
min([Date])
from
dbo.Calendar
where
MonthNumber = 4 and
DayNumber = 1 and
[Date] > getdate()
Querying a calendar table is usually clearer, simpler and more flexible than using date functions. You might also want to consider what happens if today is April 1: do you want today's date, or next year's?
If you're interested in April 1 because it's the start of a financial year, you can add that information to your calendar table directly:
select
min([Date])
from
dbo.Calendar
where
IsStartOfFinancialYear = 0x1 and
[Date] > getdate()
Use the DATEADD and the DATEPART function, like this short example:
DECLARE #Date Datetime
SET #Date = '2013.01.08 00:00:00'
SELECT DATEADD(year,
CASE WHEN DATEPART(month, #Date) < 4
THEN 0
ELSE 1 END,
DATEADD(day, -DATEPART(day, #Date) + 1,
DATEADD(month, -DATEPART(month, #Date) + 4, #Date)))
I am trying to group records by week, storing the aggregated date as the first day of the week. However, the standard technique I use for rounding off dates does not appear to work correctly with weeks (though it does for days, months, years, quarters and any other timeframe I've applied it to).
Here is the SQL:
select "start_of_week" = dateadd(week, datediff(week, 0, getdate()), 0);
This returns 2011-08-22 00:00:00.000, which is a Monday, not a Sunday. Selecting ##datefirst returns 7, which is the code for Sunday, so the server is setup correctly in as far as I know.
I can bypass this easily enough by changing the above code to:
select "start_of_week" = dateadd(week, datediff(week, 0, getdate()), -1);
But the fact that I have to make such an exception makes me a little uneasy. Also, apologies if this is a duplicate question. I found some related questions but none that addressed this aspect specifically.
To answer why you're getting a Monday and not a Sunday:
You're adding a number of weeks to the date 0. What is date 0? 1900-01-01. What was the day on 1900-01-01? Monday. So in your code you're saying, how many weeks have passed since Monday, January 1, 1900? Let's call that [n]. Ok, now add [n] weeks to Monday, January 1, 1900. You should not be surprised that this ends up being a Monday. DATEADD has no idea that you want to add weeks but only until you get to a Sunday, it's just adding 7 days, then adding 7 more days, ... just like DATEDIFF only recognizes boundaries that have been crossed. For example, these both return 1, even though some folks complain that there should be some sensible logic built in to round up or down:
SELECT DATEDIFF(YEAR, '2010-01-01', '2011-12-31');
SELECT DATEDIFF(YEAR, '2010-12-31', '2011-01-01');
To answer how to get a Sunday:
If you want a Sunday, then pick a base date that's not a Monday but rather a Sunday. For example:
DECLARE #dt DATE = '1905-01-01';
SELECT [start_of_week] = DATEADD(WEEK, DATEDIFF(WEEK, #dt, CURRENT_TIMESTAMP), #dt);
This will not break if you change your DATEFIRST setting (or your code is running for a user with a different setting) - provided that you still want a Sunday regardless of the current setting. If you want those two answers to jive, then you should use a function that does depend on the DATEFIRST setting, e.g.
SELECT DATEADD(DAY, 1-DATEPART(WEEKDAY, CURRENT_TIMESTAMP), CURRENT_TIMESTAMP);
So if you change your DATEFIRST setting to Monday, Tuesday, what have you, the behavior will change. Depending on which behavior you want, you could use one of these functions:
CREATE FUNCTION dbo.StartOfWeek1 -- always a Sunday
(
#d DATE
)
RETURNS DATE
AS
BEGIN
RETURN (SELECT DATEADD(WEEK, DATEDIFF(WEEK, '19050101', #d), '19050101'));
END
GO
...or...
CREATE FUNCTION dbo.StartOfWeek2 -- always the DATEFIRST weekday
(
#d DATE
)
RETURNS DATE
AS
BEGIN
RETURN (SELECT DATEADD(DAY, 1-DATEPART(WEEKDAY, #d), #d));
END
GO
Now, you have plenty of alternatives, but which one performs best? I'd be surprised if there would be any major differences but I collected all the answers provided so far and ran them through two sets of tests - one cheap and one expensive. I measured client statistics because I don't see I/O or memory playing a part in the performance here (though those may come into play depending on how the function is used). In my tests the results are:
"Cheap" assignment query:
Function - client processing time / wait time on server replies / total exec time
Gandarez - 330/2029/2359 - 0:23.6
me datefirst - 329/2123/2452 - 0:24.5
me Sunday - 357/2158/2515 - 0:25.2
trailmax - 364/2160/2524 - 0:25.2
Curt - 424/2202/2626 - 0:26.3
"Expensive" assignment query:
Function - client processing time / wait time on server replies / total exec time
Curt - 1003/134158/135054 - 2:15
Gandarez - 957/142919/143876 - 2:24
me Sunday - 932/166817/165885 - 2:47
me datefirst - 939/171698/172637 - 2:53
trailmax - 958/173174/174132 - 2:54
I can relay the details of my tests if desired - stopping here as this is already getting quite long-winded. I was a bit surprised to see Curt's come out as the fastest at the high end, given the number of calculations and inline code. Maybe I'll run some more thorough tests and blog about it... if you guys don't have any objections to me publishing your functions elsewhere.
For these that need to get:
Monday = 1 and Sunday = 7:
SELECT 1 + ((5 + DATEPART(dw, GETDATE()) + ##DATEFIRST) % 7);
Sunday = 1 and Saturday = 7:
SELECT 1 + ((6 + DATEPART(dw, GETDATE()) + ##DATEFIRST) % 7);
Above there was a similar example, but thanks to double "%7" it would be much slower.
For those who need the answer at work and creating function is forbidden by your DBA, the following solution will work:
select *,
cast(DATEADD(day, -1*(DATEPART(WEEKDAY, YouDate)-1), YourDate) as DATE) as WeekStart
From.....
This gives the start of that week. Here I assume that Sundays are the start of weeks. If you think that Monday is the start, you should use:
select *,
cast(DATEADD(day, -1*(DATEPART(WEEKDAY, YouDate)-2), YourDate) as DATE) as WeekStart
From.....
This works wonderfully for me:
CREATE FUNCTION [dbo].[StartOfWeek]
(
#INPUTDATE DATETIME
)
RETURNS DATETIME
AS
BEGIN
-- THIS does not work in function.
-- SET DATEFIRST 1 -- set monday to be the first day of week.
DECLARE #DOW INT -- to store day of week
SET #INPUTDATE = CONVERT(VARCHAR(10), #INPUTDATE, 111)
SET #DOW = DATEPART(DW, #INPUTDATE)
-- Magic convertion of monday to 1, tuesday to 2, etc.
-- irrespect what SQL server thinks about start of the week.
-- But here we have sunday marked as 0, but we fix this later.
SET #DOW = (#DOW + ##DATEFIRST - 1) %7
IF #DOW = 0 SET #DOW = 7 -- fix for sunday
RETURN DATEADD(DD, 1 - #DOW,#INPUTDATE)
END
Maybe you need this:
SELECT DATEADD(DD, 1 - DATEPART(DW, GETDATE()), GETDATE())
Or
DECLARE #MYDATE DATETIME
SET #MYDATE = '2011-08-23'
SELECT DATEADD(DD, 1 - DATEPART(DW, #MYDATE), #MYDATE)
Function
CREATE FUNCTION [dbo].[GetFirstDayOfWeek]
( #pInputDate DATETIME )
RETURNS DATETIME
BEGIN
SET #pInputDate = CONVERT(VARCHAR(10), #pInputDate, 111)
RETURN DATEADD(DD, 1 - DATEPART(DW, #pInputDate),
#pInputDate)
END
GO
Googled this script:
create function dbo.F_START_OF_WEEK
(
#DATE datetime,
-- Sun = 1, Mon = 2, Tue = 3, Wed = 4
-- Thu = 5, Fri = 6, Sat = 7
-- Default to Sunday
#WEEK_START_DAY int = 1
)
/*
Find the fisrt date on or before #DATE that matches
day of week of #WEEK_START_DAY.
*/
returns datetime
as
begin
declare #START_OF_WEEK_DATE datetime
declare #FIRST_BOW datetime
-- Check for valid day of week
if #WEEK_START_DAY between 1 and 7
begin
-- Find first day on or after 1753/1/1 (-53690)
-- matching day of week of #WEEK_START_DAY
-- 1753/1/1 is earliest possible SQL Server date.
select #FIRST_BOW = convert(datetime,-53690+((#WEEK_START_DAY+5)%7))
-- Verify beginning of week not before 1753/1/1
if #DATE >= #FIRST_BOW
begin
select #START_OF_WEEK_DATE =
dateadd(dd,(datediff(dd,#FIRST_BOW,#DATE)/7)*7,#FIRST_BOW)
end
end
return #START_OF_WEEK_DATE
end
go
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=47307
CREATE FUNCTION dbo.fnFirstWorkingDayOfTheWeek
(
#currentDate date
)
RETURNS INT
AS
BEGIN
-- get DATEFIRST setting
DECLARE #ds int = ##DATEFIRST
-- get week day number under current DATEFIRST setting
DECLARE #dow int = DATEPART(dw,#currentDate)
DECLARE #wd int = 1+(((#dow+#ds) % 7)+5) % 7 -- this is always return Mon as 1,Tue as 2 ... Sun as 7
RETURN DATEADD(dd,1-#wd,#currentDate)
END
For the basic (the current week's Sunday)
select cast(dateadd(day,-(datepart(dw,getdate())-1),getdate()) as date)
If previous week:
select cast(dateadd(day,-(datepart(dw,getdate())-1),getdate()) -7 as date)
Internally, we built a function that does it but if you need quick and dirty, this will do it.
Since Julian date 0 is a Monday just add the number of weeks to Sunday
which is the day before -1 Eg. select dateadd(wk,datediff(wk,0,getdate()),-1)
I found some of the other answers long-winded or didn't actually work if you wanted Monday as the start of the week.
Sunday
SELECT DATEADD(week, DATEDIFF(week, -1, GETDATE()), -1) AS Sunday;
Monday
SELECT DATEADD(week, DATEDIFF(week, 0, GETDATE() - 1), 0) AS Monday;
Set DateFirst 1;
Select
Datepart(wk, TimeByDay) [Week]
,Dateadd(d,
CASE
WHEN Datepart(dw, TimeByDay) = 1 then 0
WHEN Datepart(dw, TimeByDay) = 2 then -1
WHEN Datepart(dw, TimeByDay) = 3 then -2
WHEN Datepart(dw, TimeByDay) = 4 then -3
WHEN Datepart(dw, TimeByDay) = 5 then -4
WHEN Datepart(dw, TimeByDay) = 6 then -5
WHEN Datepart(dw, TimeByDay) = 7 then -6
END
, TimeByDay) as StartOfWeek
from TimeByDay_Tbl
This is my logic. Set the first of the week to be Monday then calculate what is the day of the week a give day is, then using DateAdd and Case I calculate what the date would have been on the previous Monday of that week.
This is a useful function for me
/* MeRrais 211126
select [dbo].[SinceWeeks](0,NULL)
select [dbo].[SinceWeeks](5,'2021-08-31')
*/
alter Function [dbo].[SinceWeeks](#Weeks int, #From datetime=NULL)
Returns date
AS
Begin
if #From is null
set #From=getdate()
return cast(dateadd(day, -(#Weeks*7+datepart(dw,#From)-1), #From) as date)
END
I don't have any issues with any of the answers given here, however I do think mine is a lot simpler to implement, and understand. I have not run any performance tests on it, but it should be neglegable.
So I derived my answer from the fact that dates are stored in SQL server as integers, (I am talking about the date component only). If you don't believe me, try this SELECT CONVERT(INT, GETDATE()), and vice versa.
Now knowing this, you can do some cool math equations. You might be able to come up with a better one, but here is mine.
/*
TAKEN FROM http://msdn.microsoft.com/en-us/library/ms181598.aspx
First day of the week is
1 -- Monday
2 -- Tuesday
3 -- Wednesday
4 -- Thursday
5 -- Friday
6 -- Saturday
7 (default, U.S. English) -- Sunday
*/
--Offset is required to compensate for the fact that my ##DATEFIRST setting is 7, the default.
DECLARE #offSet int, #testDate datetime
SELECT #offSet = 1, #testDate = GETDATE()
SELECT CONVERT(DATETIME, CONVERT(INT, #testDate) - (DATEPART(WEEKDAY, #testDate) - #offSet))
I had a similar problem. Given a date, I wanted to get the date of the Monday of that week.
I used the following logic: Find the day number in the week in the range of 0-6, then subtract that from the originay date.
I used: DATEADD(day,-(DATEPART(weekday,)+5)%7,)
Since DATEPRRT(weekday,) returns 1 = Sundaye ... 7=Saturday,
DATEPART(weekday,)+5)%7 returns 0=Monday ... 6=Sunday.
Subtracting this number of days from the original date gives the previous Monday. The same technique could be used for any starting day of the week.
I found this simple and usefull. Works even if first day of week is Sunday or Monday.
DECLARE #BaseDate AS Date
SET #BaseDate = GETDATE()
DECLARE #FisrtDOW AS Date
SELECT #FirstDOW = DATEADD(d,DATEPART(WEEKDAY,#BaseDate) *-1 + 1, #BaseDate)
Maybe I'm over simplifying here, and that may be the case, but this seems to work for me. Haven't ran into any problems with it yet...
CAST('1/1/' + CAST(YEAR(GETDATE()) AS VARCHAR(30)) AS DATETIME) + (DATEPART(wk, YOUR_DATE) * 7 - 7) as 'FirstDayOfWeek'
CAST('1/1/' + CAST(YEAR(GETDATE()) AS VARCHAR(30)) AS DATETIME) + (DATEPART(wk, YOUR_DATE) * 7) as 'LastDayOfWeek'