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I'm trying to retrieve the month difference of two dates but it seems like I can't find a way to get the accurate months.
Here are the queries I tried so far :
SELECT DATEDIFF(month,convert(datetime, '11/05/2015'), convert(datetime, '12/06/2015')) - 1
This will result to 0 which is wrong and when I used another date :
SELECT DATEDIFF(month,convert(datetime, '12/31/2015'), convert(datetime, '01/01/2016')) - 1
This would yield to 0 which is correct.
Leap year must also be considered.
The TSQL is return the correct results as you have written them. As you have it, it is taking the difference of the months between the two dates specified.
SELECT DATEDIFF(month, convert(datetime, '11/05/2015'), convert(datetime, '12/06/2015'))
The difference, in months, between November and December is "1".
However, if you are wanting the difference in terms of every 30 days is 1 month, then you would need to rewrite your query:
declare #daysPerMonth int = 30
SELECT (DATEDIFF(day, convert(datetime, '11/05/2015'), convert(datetime, '12/06/2015')) / #daysPerMonth)
This works if you define the number of days in a month as 30.
Here is a way to visualize what DATEDIFF(month, ...) is doing. Think of a monthly calendar hanging on the wall. For the purposes of this particular SQL function "number" of months between two dates is the number of pages you have to flip to get from one date to the next. It doesn't matter how many days are in each month or whether it's a leap year.
You aren't the first person to be confused about the behavior of the built-in function. If you need to count months by a different method then you'll need to describe the specifics of what you want to accomplish. It's very likely that a solution will be easy to create once you define the problem that needs to be addressed.
In T-SQL you can use CASE:
SELECT DATEDIFF(month, convert(datetime, '11/05/2015'),
convert(datetime, '12/06/2015')) -
CASE
WHEN (MONTH('11/05/2015') > MONTH('12/06/2015')
OR MONTH('11/05/2015') = MONTH('12/06/2015')
AND DAY('11/05/2015') > DAY('12/06/2015'))
THEN 1 ELSE 0
END
Which returns 1 which is correct and:
SELECT DATEDIFF(month, convert(datetime, '12/31/2015'),
convert(datetime, '01/01/2016')) -
CASE
WHEN (MONTH('12/31/2015') > MONTH('01/01/2016')
OR MONTH('12/31/2015') = MONTH('01/01/2016')
AND DAY('12/31/2015') > DAY('01/01/2016'))
THEN 1 ELSE 0
END
Which returns 0 which is correct.
declare
#start date = '20220105',
#end date = '20220406'
select 'Date_Diff_In_3 months' =
case
when datediff(month, #start, #end) < 3
or (datediff(month, #start, #end) = 3 and day(#start) >= day(#end))
then 'yes'
else 'no'
end
I need to get the number of elapsed days between any two dates with respect to the current date. IE:
mm/dd/yyyy
Current day = 07/10/2015
07/08/2013 ... 07/11/2013 - 4 days elapsed
Current day = 07/10/2015
07/08/2015 ... 07/11/2015 - 2 days have elapsed
I've tried several combinations using DATEDIFF with day as the date part, however, I can't seem to get a clean way to get the days elapsed when the date could be past or present.
EDIT
I know the start date and the end date of a certain business process. They could be this year, last year, two years ago and so on. I need a way via SQL Server functions to figure out the days total elapsed. If it's not the current year, obviously the entire span/range would have elapsed. If it's the current year, perhaps the entire span/range hasn't elapsed and it needs to say how many days are "into the process" based on the respected start time, end time and current time.
Hopefully this makes more sense?
Please help.
I used #Sean Lange, with a small tweak:
DATEDIFF(DAY, #StartDate, case when #EndDate < GETDATE() then #EndDate + 1 else GETDATE() end)
Thanks all.
This is pretty similar to the answer provided by Stan but here is my take on this.
with Something as
(
select CAST('2013-07-08' as datetime) as StartDate
, CAST('2013-07-11' as datetime) as EndDate
union all
select '2015-07-08', '2015-07-11'
)
select *
, DATEDIFF(DAY, StartDate, case when EndDate < GETDATE() then EndDate else GETDATE() end)
from Something
How about this:
Given:
CREATE TABLE dbo.test ( ChildID INT Identity,
Start DateTime
, Finish DateTime
)
and your test data:
insert into dbo.test (start,finish) values('07/08/2013','07/11/2013')
insert into dbo.test (start,finish) values('07/08/2015','07/11/2015')
then
select start,finish
, DATEDIFF(DAY, start, CASE WHEN GETDATE() BETWEEN start and finish
THEN GETDATE() - 1 ELSE finish END) + 1 as elapsed
from dbo.test
gives the result from your example.
You might have to tweak if there are other adjustments for how the current date fits between the range.
We have a requirement to bill our customers per day. We bill for an asset's existence in our system on that day. So, I started with datediff...
select datediff(dd ,'2015-04-24 12:59:32.050' ,'2015-05-01 00:59:59.000');
Returns this:
7
But I need to count the following dates: 4/24,4/25,4/26,4/27,4/28,4/29, 4/30, 5/1, which are 8 days. So datediff isn't quite working right. I tried these variations below
--too simple, returns 7, i need it to return 8
select datediff(dd ,'2015-04-24 12:59:32.050', '2015-05-01 23:59:59.000');
--looking better, this returns the 8 i need
select ceiling(datediff(hh,'2015-04-24 12:59:32.050', '2015-05-01 23:59:59.000')/24.0);
-- returns 7, even though the answer still needs to be 8. (changed enddate)
select ceiling(datediff(hh,'2015-04-24 12:59:32.050', '2015-05-01 00:59:59.000')/24.0);
So, my question... How, in SQL, would I derive the date count like i described, since I believe datediff counts the number of day boundaries crossed.... My current best approach is loop through each day in a cursor and count. Ick.
Use CONVERT to get rid of the time part, add 1 to get the desired result:
SELECT DATEDIFF(dd,
CONVERT(DATE, '2015-04-24 12:59:32.050'),
CONVERT(DATE, '2015-05-01 00:59:59.000')) + 1;
It turns out the time part does not play any significant role in DATEDIFF when dd is used as the datepart argument. Hence, CONVERT is redundant. This:
SELECT DATEDIFF(dd, '2015-04-24 23:59:59.59','2015-05-01 00:00:00.000') + 1
will return 8 as well.
You could try this which would return 8 days.
select datediff(dd ,'2015-04-24 12:59:32.050' ,CASE DATEDIFF(Second,'2015-05-01 00:00:00.000','2015-05-01 23:59:59.000') WHEN 0 THEN '2015-05-01 23:59:59.000' ELSE DATEADD(dd,+1,'2015-05-01 23:59:59.000') END)
If you want to use variables for your dates then something like this would work.
BEGIN
DECLARE #StartDate DATETIME
DECLARE #EndDate DATETIME
DECLARE #EndDateOnly DATE
SET #StartDate = '2015-04-24 12:59:32.050'
SET #EndDate = '2015-05-01 23:59:59.000'
SET #EndDateOnly = CAST(#EndDate AS DATE)
SELECT datediff(dd ,#StartDate ,CASE DATEDIFF(Second,CAST(#EndDateOnly||' 00:00:00.000' AS DATETIME),#EndDate) WHEN 0 THEN #EndDate ELSE DATEADD(dd,+1,#EndDate) END)
END
I've been looking around for a chunk of code to find the first day of the current week, and everywhere I look I see this:
DATEADD(WK, DATEDIFF(WK,0,GETDATE()),0)
Every place says this is the code I'm looking for.
The problem with this piece of code is that if you run it for Sunday it chooses the following Monday.
If I run:
SELECT GetDate() , DATEADD(WK, DATEDIFF(WK,0,GETDATE()),0)
Results for today (Tuesday):
2013-05-14 09:36:39.650................2013-05-13 00:00:00.000
This is correct, it chooses Monday the 13th.
If I run:
SELECT GetDate()-1 , DATEADD(WK, DATEDIFF(WK,0,GETDATE()-1),0)
Results for yesterday (Monday):
2013-05-13 09:38:57.950................2013-05-13 00:00:00.000
This is correct, it chooses Monday the 13th.
If I run:
SELECT GetDate()-2 , DATEADD(WK, DATEDIFF(WK,0,GETDATE()-2),0)
Results for the 12th (Sunday):
2013-05-12 09:40:14.817................2013-05-13 00:00:00.000
This is NOT correct, it chooses Monday the 13th when it should choose the previous Monday, the 6th.
Can anyone illuminate me as to what's going in here? I find it hard to believe that no one has pointed out that this doesn't work, so I'm wondering what I'm missing.
It is DATEDIFF that returns the "incorrect" difference of weeks, which in the end results in the wrong Monday. And that is because DATEDIFF(WEEK, ...) doesn't respect the DATEFIRST setting, which I'm assuming you have set to 1 (Monday), and instead always considers the week crossing to be from Saturday to Sunday, or, in other words, it unconditionally considers Sunday to be the first day of the week in this context.
As for an explanation for that, so far I haven't been able to find an official one, but I believe this must have something to do with the DATEDIFF function being one of those SQL Server treats as always deterministic. Apparently, if DATEDIFF(WEEK, ...) relied on the DATEFIRST, it could no longer be considered always deterministic, which I can only guess wasn't how the developers of SQL Server wanted it.
To find the first day of the week's date, I would (and most often do actually) use the first suggestion in #Jasmina Shevchenko's answer:
DATEADD(DAY, 1 - DATEPART(WEEKDAY, #Date), #Date)
DATEPART does respect the DATEFIRST setting and (most likely as a result) it is absent from the list of always deterministic functions.
Try this one -
SET DATEFIRST 1
DECLARE #Date DATETIME
SELECT #Date = GETDATE()
SELECT CAST(DATEADD(DAY, 1 - DATEPART(WEEKDAY, #Date), #Date) AS DATE)
SELECT CAST(#Date - 2 AS DATE), CAST(DATEADD(WK, DATEDIFF(WK, 0, #Date-2), 0) AS DATE)
Results:
---------- ----------
2013-05-12 2013-05-13
SQL Server has a SET DATEFIRST function which allows you to tell it what the first day of the week should be. SET DATEFIRST = 1 tells it to consider Monday as the first day of the week. You should check what the server's default setting is via ##DATEFIRST. Or you could simply change it at the start of your query.
Some references:
MSDN
Similar Question
That worked for me like a charm:
Setting moday as first day of the week without changing DATEFIRST variable:
-- FirstDayWeek
select dateadd(dd,(datepart(dw, case datepart(dw, [yourDate]) when 1 then dateadd(dd,-1,[yourDate]) else [yourDate] end) * -1) + 2, case datepart(dw, [yourDate]) when 1 then dateadd(dd,-1,[yourDate]) else [yourDate] end) as FirstDayWeek;
-- LastDayWeek
select dateadd(dd, (case datepart(dw, [yourDate]) when 1 then datepart(dw, dateadd(dd,-1,[yourDate])) else datepart(dw, [yourDate]) end * -1) + 8, case datepart(dw, [yourDate]) when 1 then dateadd(dd,-1,[yourDate]) else [yourDate] end) as LastDayWeek;
Setting sunday as fist day of the week without changing DATEFIRST variable
select convert(varchar(50), dateadd(dd, (datepart(dw, [yourDate]) * -1) + 2, [yourDate]), 103) as FirstDayWeek, convert(varchar(50), dateadd(dd, (datepart(dw, [yourDate]) * -1) + 8, [yourDate]), 103) as LastDayWeek;
You can change [yourDate] by GETDATE() for testing
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'