How can I extract the year and week combination from a date in SQL Server using T-SQL and have it match the MySQL yearweek result?
For example these MySQL queries (Click here for MySQL SQL Fiddle):
SELECT yearweek('2013-12-31');
SELECT yearweek('2014-01-01');
This returns 201352 for both dates. That is the expected result.
But in SQL Server...
datepart works as expected for the year extract
sometimes datepart does not return the expected value for iso_week
The MySQL result cannot be achieved with this T-SQL query...
SELECT datepart(year, #dt) * 100 + datepart (iso_week, #dt);
T-SQL versions of the MySQL queries above (Click here for T-SQL SQL Fiddle):
SELECT datepart(year, '2013-12-31') * 100 + datepart (iso_week, '2013-12-31');
SELECT datepart(year, '2014-01-01') * 100 + datepart (iso_week, '2014-01-01');
The result is 201352 for the first date and 201401 for the second date.
However, this is not the expected result because...
2014-01-01 belongs to the last week of 2013
So the expected result is 201352
Do any of you more experienced T-SQL developers know how to extract the year/week of a given date and have this match what I see in MySQL?
I need to have the week start on Monday. This is why I am using iso_week. I have tested the results with week anyway and found the same issue. This query also produces 201401 when 201352 is expected.
SELECT datepart(year, '2014-01-01') * 100 + datepart (week, '2014-01-01');
If you look at the ISO_Week datepart definition at http://msdn.microsoft.com/en-us/library/ms174420.aspx, you'll see the following:
ISO 8601 includes the ISO week-date system, a numbering system for
weeks. Each week is associated with the year in which Thursday
occurs...
The numbering system in different countries/regions might not comply
with the ISO standard.
Since January 1, 2014 was a Wednesday; January 2, 2014 was the first Thursday of the year and thus week 1 of 2014 (according to ISO 8601).
Furthermore, looking at the MySQL definitions for yearweek and week, there are several mode options (http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_week)
So, I think you're going to have to write your own yearweek function for the week counting rule you want:
CREATE FUNCTION dbo.yearweek(#date date)
RETURNS INT
as
begin
set #date = dateadd(dd,-datepart(dw,#date)+1, #date)
return datepart(year,#date)*100 + datepart(week,#date)
end
go
select dbo.yearweek('2013-12-31'), dbo.yearweek('2014-01-01')
NOTE: I haven't fully tested this code, and I'm not sure exactly what your requirements are. This is just meant as an example of the type of process you need to follow.
I think it's because you're using "ISO_WEEK" as the datepart
Look at the description under the "ISO_WEEK datepart" here: http://msdn.microsoft.com/en-us/library/ms174420.aspx
using "WEEK" instead might get the desired result
TO extract YEAR, MONTH or DAY you can simply use the YEAR, MONTH and DAY function as shown in the followinf example. for any other part you have to use the DATEPART function , read here for a detailed list of arguements you can pass to DATEPART function.
SELECT YEAR(GETDATE()) AS [YEAR]
, DATEPART(WEEK, GETDATE()) [Week]
UPDATE
SELECT CAST(YEAR(GETDATE()) AS VARCHAR(4)) +'-'
+ CAST(DATEPART(WEEK, GETDATE()) AS VARCHAR(2))
It seems to me that what you want is a week number based on the previous Monday. Is this correct? If so, iso_week is no use to you as it is based on this week's Thursday.
To get the year and week based on the previous Monday, I think you need something like the following.
set datefirst 1 ;
select
SomeDate ,
datename( weekday, SomeDate ) as [WeekdayName] ,
datepart( weekday, SomeDate ) as [Weekday] ,
datepart( week, SomeDate ) as [Week] ,
datepart( iso_week, SomeDate ) as [ISO_week] ,
dateadd( day, -( datepart( weekday, SomeDate ) - 1 ), SomeDate ) as [PreviousMonday] ,
datepart( year, dateadd( day, -( datepart( weekday, SomeDate ) - 1 ), SomeDate ) ) as [MondayYear] ,
datepart( week, dateadd( day, -( datepart( weekday, SomeDate ) - 1 ), SomeDate ) ) as [MondayWeek]
from
dbo.DateDemo
order by
SomeDate ;
Abbreviated sample output is as follows.
SomeDate PreviousMonday MondayYear MondayWeek
2013-12-25 2013-12-23 2013 52
2013-12-26 2013-12-23 2013 52
2013-12-27 2013-12-23 2013 52
2013-12-28 2013-12-23 2013 52
2013-12-29 2013-12-23 2013 52
2013-12-30 2013-12-30 2013 53
2013-12-31 2013-12-30 2013 53
2014-01-01 2013-12-30 2013 53
2014-01-02 2013-12-30 2013 53
2014-01-03 2013-12-30 2013 53
2014-01-04 2013-12-30 2013 53
2014-01-05 2013-12-30 2013 53
2014-01-06 2014-01-06 2014 2
Related
Here is my query:
SELECT ID AS 'securityid'
, date
, DATEADD(DW, 1, adate) AS 'lagged_date_v2'
, DATEADD(DAY, 1, adate) AS 'lagged_date_v1'
, aclose AS 'previous_close'
FROM mytable
WHERE adate BETWEEN '20170101 09:00' AND '20170630 18:00' AND
ID = 100056;
Here are the output from the code above:
I compared the results to DATEADDby day and by weekday and it returns the same results.
Column adate is just typical date, we can tell Jan 28, 2017 is Saturday. However, on this row, both lagged_date_v1 and lagged_date_v2 return Jan 28. If I use DATEADD by weekday correctly, I should see Jan 30 instead Jan 28, right?
My sql server version is 2008.
If I use DATEADD by weekday correctly, I should see Jan 30 instead Jan
28, right?
Wrong. In SQL Server, weekday doesn't mean days that aren't weekend days. It means which day of the week (1-7) is the day in question based on the current datefirst setting.
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, ''))
Here is a weird one for you all.
I need to determine the number of days in a Month
;WITH cteNetProfit AS
(
---- NET PROFIT
SELECT DT.CreateDate
, SUM(DT.Revenue) as Revenue
, SUM(DT.Cost) as Cost
, SUM(DT.GROSSPROFIT) AS GROSSPROFIT
FROM
(
SELECT CAST([createDTG] AS DATE) as CreateDate
, SUM(Revenue) as Revenue
, SUM(Cost) as Cost
, SUM(REVENUE - COST) AS GROSSPROFIT
FROM [dbo].[CostRevenueSpecific]
WHERE CAST([createDTG] AS DATE) > CAST(GETDATE() - 91 AS DATE)
AND CAST([createDTG] AS DATE) <= CAST(GETDATE() - 1 AS DATE)
GROUP BY createDTG
UNION ALL
SELECT CAST([CallDate] AS DATE) AS CreateDate
, SUM(Revenue) as Revenue
, SUM(Cost) as Cost
, SUM(REVENUE - COST) AS GROSSPROFIT
FROM abc.PublisherCallByDay
WHERE CAST([CallDate] AS DATE) > CAST(GETDATE() - 91 AS DATE)
AND CAST([CallDate] AS DATE) <= CAST(GETDATE() - 1 AS DATE)
GROUP BY CALLDATE
) DT
GROUP BY DT.CreateDate
)
select distinct MONTH(CREATEDATE), DateDiff(Day,CreateDate,DateAdd(month,1,CreateDate))
FROM cteNetProfit
For some reason it is returning two different results for the month of March 2016 one result is 30 and the other 31(which of course is correct) I validate that the underlying data only has 31 days worth of data for the Month of March. Since Feb is a leap year can this affect the DATEDIFF function. The remaining months return the correct #.
2 29
3 31
3 30
4 30
5 31
Thanks for the input, however, I found the solution elsewhere
select Distinct MONTH(CREATEDATE), Day(EOMONTH(CreateDate))
FROM cteNetProfit
The difference comes when you hit the 2016-03-31 date. If you run the query below for 2016-03-30 and 2016-03-31, the results of adding 1 MONTH using DATEADD, in both instances, is 2016-04-30. It returns the last day of the next month.
SELECT DATEADD(MONTH,1,'2016-03-30') , DATEADD(MONTH,1,'2016-03-31')
This syntax seemed to work (courtesy of https://raresql.com/2013/01/06/sql-server-get-number-of-days-in-month/).
SELECT DAY(DATEADD(ms,-2,DATEADD(MONTH, DATEDIFF(MONTH,0,#DATE)+1,0))) AS [Current Month]
I have two dates say '2011-01-23' and '2015-11-29',
'2011-01-23' falls in first quarter of 2011 so 'Jan 2011'
'2015-11-29' falls in fourth quarter of 2015 so 'Oct 2015'
In SQL Server I want get all quarters in a select list.
e.g.
Input: #StartDate='2011-01-23' , #EndDate='2015-11-29'
Output:
Jan 2011
Apr 2011
Jul 2011
Oct 2011
Jan 2012
Apr 2012
Jul 2013
Oct 2013
Jan 2014
......
......
......
Jul 2015
Oct 2015
You can use a recursive CTE to generate the dates as follows:
declare #StartDate datetime
declare #EndDate datetime
select #StartDate='2011-01-23' , #EndDate='2015-11-29'
;With Quarters as (
select DATEADD(quarter,DATEDIFF(quarter,0,#StartDate),0) as dt
union all
select DATEADD(quarter,1,dt)
from Quarters
where dt < DATEADD(quarter,DATEDIFF(quarter,0,#EndDate),0)
)
select
--I'd usually keep them as dates at this point, but to match your requirement
CONVERT(varchar(3),dt,109) + ' ' + CONVERT(varchar(4),dt,120)
from Quarters order by dt
This also uses a couple of other tricks - it uses CONVERT with far too short target datatypes to quickly truncate the strings to just the parts that we want to keep - and it uses a DATEADD/DATEDIFF pair to quickly round a datetime value down to it's nearest interval boundary.
For SQL Server 2012, you could instead use FORMAT to produce the output string, but I've not experimented with that much so I'll leave that as an exercise...
Finally, I found the solution of my question..
WITH mycte AS
(
SELECT CAST('2011-01-01' AS DATE) DateValue
UNION ALL
SELECT DATEADD(Q,1,DateValue) FROM mycte WHERE DATEADD(Q,1,DateValue) < '2012-12-31'
)
SELECT CONVERT(varchar(3), DATENAME(MONTH,DateValue))+ ' ' +Convert(varchar(4),DATEPART(YYYY,DateValue)) FROM mycte OPTION (MAXRECURSION 0)
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)))