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())
Related
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.
I have a report that I have been asked to produce. The first column of data is total time entered from the first of the year to the end of the previous month. The next column of data is total time from beginning of "current" month to the end of "current" month.
For example. If this report was being run for March then the first column would be total time for Jan and Feb and the second column would be total time for March. If I were to run it in April then the first column would be total time for Jan/Feb/Mar and the second column total time for April etc.
I am using various expressions to get first date of the year, last date of previous month, first date of this month, last date of this month. All working fine and it runs like a dream if you run the report in the current month (i.e. March) but if you want to run the report in April for March's data it won't do it as it's reading the date on the computer and using that to calculate the prev month.
In Crystal reports you can set a report date. Is there something similar in SQL reporting? I'm assuming you DECLARE the report date in your initial query but I haven't yet found the right combination of functions.
This report is for an external bit of software that we run.
The only parameters I can use are #FromDate and #ToDate and these are set as text rather than date
I was planning on using the #ToDate to set the report date but would obviously have to convert it from text first
Any guidance very much appreciated
Try this. You could actually do it without all the declare statements, I just did those to break it down to be easy to read.
declare #date_entered datetime = '2/7/2017'
declare #start_of_year datetime = DATEADD(yy, DATEDIFF(yy, 0, #date_entered), 0)
declare #start_of_month datetime = DATEADD(month, DATEDIFF(month, 0, #date_entered), 0)
declare #end_of_month datetime = DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#date_entered)+1,0))
declare #end_of_last_month datetime = DATEADD(day,-1,#start_of_month)
-- Now remove weekends from days found
declare #amount_of_days_previous tinyint = datediff(day,#start_of_year,#end_of_last_month) - (datediff(wk, #start_of_year, #end_of_last_month) * 2)
declare #amount_of_days_this tinyint = DATEDIFF(day,#start_of_month,#end_of_month) - (datediff(wk, #start_of_month, #end_of_month) * 2)
select #amount_of_days_previous * 8 as WorkHoursinDaysPrevious,
#amount_of_days_this * 8 as WorkHoursinDaysThis
I had a think over the weekend and realised that I was approaching the problem all wrong. I was trying to set the report date so that my query could read that date and then go back and find the relevant date (first date in the year, last date of previous month etc).
What I needed to do was use my date criteria (FromDate and ToDate) in my DATEADD selections.
CONVERT(datetime,#FromDate,103) AS FIRSTDAYOFYEAR,
DATEADD(D, - 1, DATEADD(MONTH, DATEDIFF(MONTH, '19000101', CONVERT(datetime,#ToDate,103)), '19000101')) AS LASTDAYPREVMONTH,
DATEADD(MONTH,DATEDIFF(MONTH, '19000101',CONVERT(datetime,#ToDate,103)), '19000101') AS FIRSTDAYMONTH,
CONVERT(datetime,#ToDate,103) AS LASTDAYMONTH,
This is now much better because my "first date of year" can be anything the user wants, not just the first day of the year.
I need to calculate an expiration date. Should be pretty easy right? Well, my "lifespan" is in years, not days, and to make it more challenging, the years value can be fractional. Here is what I am starting with:
set #ExpirationDate = DATEADD(year, #LifeSpanYears, #BeginDate)
The problem is, DATEADD allows a decimal to be passed in, but if you read the documentation, the decimal is truncated to an int.
How do you calculate an expiration date with a fractional year?
This is the answer I came up with:
create function Reporting.CalcExpirationDate ( #LifeSpanYears decimal(9,2), #BeginDate date)
returns date
begin
return DATEADD(day, (#LifeSpanYears * 365.25), #BeginDate)
end
Then use it like this:
select Reporting.CalcExpirationDate (13 , '2015-09-01') as CalculatedDate, '2028-09-01' as ExpectedDate
union all select Reporting.CalcExpirationDate (13.5 , '2015-09-01') as CalculatedDate, '2029-03-01' as ExpectedDate
Notes: As I was working on this I realized that it is impossible to determine the exact expiration date using a fractional year as a lifespan. How do you know when to add leap days or not? If you really want the exact answer, you need lifespan in days.
In practice, the "lifespan in years" is probably a lifespan in months. If so, you might consider:
return dateadd(month, round(#LIfeSpanYears * 12, 0), #BeginDate)
This handles the issue with leap years. However, it comes at the cost of only resolving to months.
I am working with dates so I have created a function that generates a SQL Table Calendar which returns Day, Month, WeekOfMonth, WeekOfYear and so on.
Right now, for the Day of Month field I am using the following function:
-- [WkNo]=Week number
[WkNo] = DATEPART(week,dt.DT),
But the problem is that when I run this on a SQL installed with Language = US English, the week setting is wrong cause the week starts from Sunday.
I need to set the week starting from Monday, is it possible without the use of DATEPART?
Update: based on asker's comment that he cannot use the DATEFIRST approach I am updating answer.
Note:
This answer is generic in nature.
If instead of Monday you want the week to start from Tuesday, you can change the dateadd(dd,-1,dt.DT) to dateadd(dd,-2,dt.DT) and for Wednesday to dateadd(dd,-3,dt.DT).
Basically the formula becomes dateadd(dd,-n,dt.DT) for value of n ranging over 1(Monday) to 6 (Saturday).
SELECT [WkNo]=
ISNULL(DATEPART(week,
case
when Year(dt.DT)>YEAR(dateadd(dd,-1,dt.DT))
then null
else dateadd(dd,-1,dt.DT)
end
),1)
See working fiddle http://sqlfiddle.com/#!6/10a80/14
Old Answer:
See MSDN documentation: https://msdn.microsoft.com/en-us/library/ms174420.aspx
When datepart is week (wk, ww) or weekday (dw), the return value
depends on the value that is set by using SET DATEFIRST. January 1 of
any year defines the starting number for the week datepart, for
example: DATEPART (wk, 'Jan 1, xxxx') = 1, where xxxx is any year.
SET DATEFIRST 1
-- [WkNo]=Week number
[WkNo] = DATEPART(week,dt.DT),
try
SET DATEFIRST {1,2,3,4,5,6,7(default, U.S. English)}
for your Query:
set datefirst 1
select [WkNo]= DATEPART(week,dt.DT)
See Here
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.