SQL Server date calculations in stored procedures - sql-server

I wanted to ask 3 questions about below code (please excuse the long code listing, I am including these lines in the hopes that it provides enough context).
Note that the code here depends on the date when it is executed. For this reason my questions refer to a hypothetical situation with two different execution dates:
March 1st 2014
January 1st 2014
My questions are on whether my understanding on some parts of this is correct, i.e:
A. that the SELECT DATEADD expression (on line 1) would:
On March 1st 2014 create the datetime 2014-02-31 23:59:59
B. that the code on lines 6-9 would:
On March 1st 2014 create the datetime 2014-02-01 00:00:00
On January 1st 2014 create the datetime 2013-02-01 00:00:00
and
C. that the code on lines 11-14 would:
On March 1st 2014 create the datetime 2015-01-30 00:00:00
On January 1st 2014 create the datetime 2014-01-30 00:00:00
Is this understanding correct?
1. SET #ldpmth = (SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE()),0)) )
2. SET #yr = (SELECT YEAR(#ldpmth))
3. SET #mth = 0
4. SET #dy = 0
5.
6. SET #fysdate = (SELECT CASE WHEN MONTH(#ldpmth) >= 2 THEN
7. DATEADD(MM, 1,CAST(CAST(#yr+#mth+#dy AS NVARCHAR(50)) AS DATETIME))
8. ELSE DATEADD(MM, 1, CAST(CAST((#yr-1)+#mth+#dy AS NVARCHAR(50))
9. AS DATETIME)) END )
10.
11. SET #fyedate = (SELECT CASE WHEN MONTH(#ldpmth) >= 2 THEN
12. DATEADD(YY, 1, CAST(CAST(#yr+#mth+#dy AS NVARCHAR(50)) AS DATETIME)) + 30
13. ELSE DATEADD(YY, 1, CAST(CAST((#yr-1)+#mth+#dy AS NVARCHAR(50))
14. AS DATETIME)) + 30 END )
(Thank you for all who answered so far. This is actually code that was developed (but not documented) at my place of employment some years ago and I have been tasked with converting it to a form that works client-side with Crystal Reports.)

On March 1st 2014 create the datetime 2014-02-31 23:59:59
No. AFAIK, there isn't any dbms that's so dumb it thinks Feb 31 is an actual date. The SQL statement on line 1 will return the value 'Feb 28 2014 23:59:59'.
You can test all your statements by substituting actual dates for GETDATE() and guesswork. In the first query, for example, use
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,'2014-03-01'),0));
^^^^^^^^^^
You can probably run a version of SQL Server locally. SQL Server 2008 even installs on Windows XP. If you can't bear that, though, there's always sqlfiddle.com.

This could be easily tested in SQL Server Management Studio (SSMS). I'll answer the first, and you can test the next two yourself.
Question:
A. that the DATEADD (on line 1) would on March 1st 2014 create the datetime 2014-02-31 23:59:59
Answer: No, because February can't possibly have 31 days, so it can't return '2014-02-31`
Proof:
select DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,'2014-03-01'),0))
Results
(No column name)
2014-02-28 23:59:59.000

For (B) and (C), if you'd stated your question, instead, as "My financial years run from 1st February until 30th(sic) January - how do I get the start and end dates for the current financial year, I'd have offered:
SET #fysdate = DATEADD(year,DATEDIFF(month,'19000201',GETDATE())/12,'19000201')
and:
SET #fyedate = DATEADD(year,DATEDIFF(month,'19000201',GETDATE())/12,'19010131')
(Note that this second finds the 31st Jan, not the 30th. I assume that was carelessness in your original question).
But, as to (A), as I indicated, I thought your question is again poorly phrased - if you're trying to find the end point of a period and that period is defined on datetimes (as opposed to just dates), it's far easier to construct an exclusive upper bound:
SET #ldpmth = DATEADD(month, DATEDIFF(month,0,GETDATE()),0)
This bound (to be used with a < comparison rather than <= or BETWEEN) doesn't artificially exclude any values with a time such as 23:59:59.527. Note also, that if the financial year end date calculation is going to be used against datetime values which include times, I'd again counsel computing an exclusive upper bound, and substitute 19010201 instead of 19010131.
So, in total, I don't think I've actually answered your questions as posed, but I think I've provided the answers you should have been seeking.

Related

Why SQL datepart(ww,GETDATE()) returns current week + 1

I am trying to get current week of the calendar. Let's assume today is 2022-03-14 12:00:00.
I am using these 2:
select GETDATE() // returns 2022-03-14 12:00:00
select datepart(ww,GETDATE() //returns 12
But as per calendar 2022-03-14 12:00:00 should be 11st week.
From where is the difference coming from and how I can get basically current week?
GetDatePart(ww, ...) in SQL server is dependent on the set datefirst as documented here.
You can use GetDatePart(isoww, getdate()) instead to get ISO Week Number.

Query all items for the week in SQL

I have a query that selects all items for the given week and sets their date to be the start day of that week. In some cases the query sets an incorrect date value for the given item and I have traced the problem to be in either DATEDIFF or DATEADD functions.
The query is
SELECT DATEADD(WEEK, DATEDIFF(WEEK, 0, DateTimeValue), 0) AS NewDateTimeValue
Let's take a date March 27 2016 as an example. The DATEDIFF returns value of 6065 and the DATEADD with that value returns March 28 2016.
For the date March 26 2016, the DATEDIFF returns 6064 and the DATEADD March 21, 2016.
To me this sounds like a FirstDayOfWeek issue and that SQL Server thinks Sunday is the first day of a week and thus giving different values for Sunday and for Saturday (March 27, 2016 is Sunday).
I tried to set the first day of the week by
SET DATEFIRST 1
but that didn't make any difference and SQL returned the same results.
So what causes the SQL function to behave like this and any ideas how to fix it?
I found out that DATEDIFF doesn't respect the DATEFIRST setting and always assumes the week starts on Sunday. This is the reason why my sample case works as it works. This question has some suggested workaround: Is it possible to set start of week for T-SQL DATEDIFF function?
This is happening because you are finding the no. of weeks from date 0 to the supplied date. 0 is actually '1900-01-01' which was a Monday. Therefore,DATEDIFF finds the number of completed weeks from this date until the supplied data. Thats why DATEDIFF for March 27 2016 returns 6065 (as it is the end of a week) and March 26 2016 returns 6064 (as it is still not the end of a week).
This is already explained in this link - Get first day of the week

Excel incorrectly converts Date into Int

I'm pulling the data from SQL database. I have a couple columns with date which need to be converted into Int type, but when I do this the date changes (-2 days). I tried Cast and Convert and it's always the same.
Converting to other type works fine and returns the correct date, but doesn't work for me. I need only the date part from datetime and it needs to be recognised as a date by Excel.
Why is this happening? Any ideas how to get it sorted?
I'm using the following query:
SELECT wotype3, CONVERT(INT,wo_date2 ,103), CAST(duedate AS int) FROM Tasks WHERE
duedate > DATEADD(DAY,1, GETDATE())
AND wo_date2>0
AND wo_date2<DATEADD(WEEK,3,GETDATE())
ORDER BY wotype3
I've had big problems with this, checking my SQL Server's calculation results with "expected results" which a user had created using Excel.
We had discrepancies just because of this 2-day date difference.
Why does it happen ?
Two reasons:
SQL Server uses a zero-based date count from Jan 1 1900, but Excel uses a 1-based date count from Jan 1 1900.
Excel has a bug in it (gasp!) which makes it think that the year 1900 was a leap year. It wasn't. SQL Server correctly refuses to let you have a date value containing "29-Feb-1900".
Combine these two discrepancies, and this is why all dates, from March 1 1900 onwards, are always 2-days out.
Apparently, this Excel bug is a known issue, to keep it in line with Lotus 1-2-3.
The Intentional Date Bug
Microsoft's own explanation
From now on, I think I'll justify bugs in my code with the same excuse.
;-)
For SQL Server 2008 and above, you can use the DATE datatype.
declare #dt datetime = '12/24/2013 10:45 PM' -- some date for example
SELECT #dt as OriginalDateTime, CAST(#dt as DATE) as OnlyDate
For versions prior to SQL Server 2008, you would need to truncate the time part using one or the other functions. Here is one way to do that:
declare #dt datetime = '12/24/2013 10:45 PM' -- some date for example
SELECT #dt as OriginalDateTime, CAST(FLOOR(CAST(#dt AS FLOAT)) as DATETIME) as OnlyDate

What's the best way to perform math on a floored date in SQL Server?

This is related to Floor a date in SQL server, but I'm taking it one step further:
I'm writing a query on a SQL Server table and I want to get all records for the current calendar year and the previous calendar year. So, right now, I'd want a query to return all records between January 1st, 2008 and today. But come January 1st, 2010, I want it to return records no older than 2009.
Essentially, I want to floor the current date to the beginning of the year and then subtract 1.
After rummaging through some SQL Server documentation, I came up with this:
WHERE create_date >= CAST((CAST(YEAR(GETDATE()) AS int) -1) AS varchar)
but it feels kind of ugly. Is there a better way?
Why not just use the year function on create_date as well?
WHERE YEAR(create_date) >= (YEAR(GETDATE()) -1)
This assumes (as you did) that there are no records in the database greater than today's date
I would suggest assigning a variable with the date lastyear-01-01, either by making an UDF for it, or something like
DECLARE #startOfLastYear DATETIME
SET #startOfLastYear = CAST(YEAR(GETDATE()) - 1 AS VARCHAR) + '-01-01'
Then do the query:
WHERE create_date >= #startOfLastYear
Because of two reasons:
Using YEAR() or any other function
on data from tables (i.e.
YEAR(create_date)) makes indices
unusable and decreases the
performance of the query
The variable name tells exactly what it is, and reading the code is easier.

Getting week number off a date in MS SQL Server 2005?

Is it possible to create an sql statement that selects the week number (NOT the day of week - or the day number in a week). I'm creating a view to select this extra information along with a couple of other fields and thus can not use a stored procedure. I'm aware that it's possible to create a UDF to do the trick, but if at all possible i'd rather only have to add a view to this database, than both a view and a function.
Any ideas? Also where i come from, the week starts monday and week 1 is the first week of the year with atleast 4 days.
Related:
How do I calculate the week number given a date?
Be aware that there are differences in what is regarded the correct week number, depending on the culture. Week numbers depend on a couple of assumptions that differ from country to country, see Wikipedia article on the matter. There is an ISO standard (ISO 8601) that applies to week numbers.
The SQL server integrated DATEPART() function does not necessarily do The Right Thing. SQL Server assumes day 1 of week 1 would be January 1, for many applications that's wrong.
Calculating week numbers correctly is non-trivial, and different implementations can be found on the web. For example, there's an UDF that calculates the ISO week numbers from 1930-2030, being one among many others. You'll have to check what works for you.
This one is from Books Online (though you probably want to use the one from Jonas Lincoln's answer, the BOL version seems to be incorrect):
CREATE FUNCTION ISOweek (#DATE DATETIME)
RETURNS INT
AS
BEGIN
DECLARE #ISOweek INT
SET #ISOweek = DATEPART(wk,#DATE)
+1
-DATEPART(wk,CAST(DATEPART(yy,#DATE) AS CHAR(4))+'0104')
-- Special cases: Jan 1-3 may belong to the previous year
IF (#ISOweek=0)
SET #ISOweek = dbo.ISOweek(CAST(DATEPART(yy,#DATE) - 1
AS CHAR(4))+'12'+ CAST(24+DATEPART(DAY,#DATE) AS CHAR(2)))+1
-- Special case: Dec 29-31 may belong to the next year
IF ((DATEPART(mm,#DATE)=12) AND
((DATEPART(dd,#DATE)-DATEPART(dw,#DATE))>= 28))
SET #ISOweek=1
RETURN(#ISOweek)
END
GO
You need the ISO week. From http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=60510, here's an implementation:
drop function dbo.F_ISO_WEEK_OF_YEAR
go
create function dbo.F_ISO_WEEK_OF_YEAR
(
#Date datetime
)
returns int
as
/*
Function F_ISO_WEEK_OF_YEAR returns the
ISO 8601 week of the year for the date passed.
*/
begin
declare #WeekOfYear int
select
-- Compute week of year as (days since start of year/7)+1
-- Division by 7 gives whole weeks since start of year.
-- Adding 1 starts week number at 1, instead of zero.
#WeekOfYear =
(datediff(dd,
-- Case finds start of year
case
when NextYrStart <= #date
then NextYrStart
when CurrYrStart <= #date
then CurrYrStart
else PriorYrStart
end,#date)/7)+1
from
(
select
-- First day of first week of prior year
PriorYrStart =
dateadd(dd,(datediff(dd,-53690,dateadd(yy,-1,aa.Jan4))/7)*7,-53690),
-- First day of first week of current year
CurrYrStart =
dateadd(dd,(datediff(dd,-53690,aa.Jan4)/7)*7,-53690),
-- First day of first week of next year
NextYrStart =
dateadd(dd,(datediff(dd,-53690,dateadd(yy,1,aa.Jan4))/7)*7,-53690)
from
(
select
--Find Jan 4 for the year of the input date
Jan4 =
dateadd(dd,3,dateadd(yy,datediff(yy,0,#date),0))
) aa
) a
return #WeekOfYear
end
go
Looks like the DATEPART mssql function should help you out with ...
DATEPART(wk, ‘Jan 1, xxxx’) = 1
Well I'll be.. turns out there is a way to set the first day of the week, DATEFIRST
SET DATEFIRST 1 -- for monday
Update: Now I understand better, what the OP wants.. which is custom-logic for this. I don't think MSSQL would have functions with such rich level of customization. But I may be wrong... I think you'll have to roll your own UDF here...sorry
FORGET THE OTHER ANSWERS
The question specifies "the week starts monday and week 1 is the first week of the year with atleast 4 days." This is ISO 8601 standard and what this answer provides. This function is used in production on our site.
This is all you need:
CREATE FUNCTION ISOweek (#DATE DATETIME)
RETURNS INT
AS
BEGIN
RETURN (datepart(DY, datediff(d, 0, #DATE) / 7 * 7 + 3)+6) / 7
END
GO
This will return you the week number of date entered in quotes
SELECT DATEPART( wk, 'enter the date over here' )
Looks like datepart will get you part of the way there, but you'll have to adjust to get your correct week number, based on the day of week of Jan 1 of the given year. I'm not familiar enough with T-SQL to do that, but it should be possible. Pity there isn't a mode argument as in MySQL
have you considered using the WEEK function?
This will get you the week of the year for the specified date that you pass in.
SELECT { fn WEEK(GETDATE()) } AS WeekNumber, { fn WEEK(CONVERT(DATETIME, '2008-01-01 00:00:00', 102)) } AS FirstWeekOfYear, { fn WEEK(CONVERT(DATETIME, '2008-12-31 00:00:00', 102)) } AS LastWeekOfYear
This outputs the following SQL2000 and SQL2005:
WeekNumber: 50
FirstWeekOfYear: 1
LastWeekOfYear: 53
I Hope this helps :)
Why yet again, people make mountains out of mole-hills, it astounds me?
So simple...
select DATEPART(wk, GETDATE())

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