I have a query where I need to calculate the week number based on the fiscal start of the year occurring on 01.07.YEAR
The problem I have is that when it goes to 30 Days of a calendar year I don't get the correct result and so I need to amend my script. I want a query that returns the correct week number regardless of what month I am in.
I have a calendar table populated; I add 18 months to the current date which give me the period and so I tried to do the same to the week but I am going around in circles.
Below is the code I am using:
SET DATEFIRST 1;
SELECT
[date],
DATEPART(wk, DATEADD(MONTH, 18, DATEADD(dd, -2, [date]))) 'FWeek',
LEFT(CONVERT(varchar, DATEADD(MONTH, 18, [date]), 112), 6) AS 'Period',
DATENAME(dw, [date]) AS 'Day'
FROM [dbo].[DP_PERIOD_DATES]
WHERE [date] >= '01/11/2018'
AND [date] < '04-01-2019'
ORDER BY [date] DESC
Try datediff to get the number of weeks from 1/7/x.
The extra datediff is to get the first of the year and add 6 to get the 7th of the year.
SELECT [date],
DateDiff(wk,
--JANUARY 7th of Current Year
DATEADD(yy, DATEDIFF(yy, 0, DATEADD(m,18,[date])), 0)+6
, DATEADD(m,18,[date]))
FROM [dbo].[DP_PERIOD_DATES]
WHERE [date] >= '01/11/2018'
AND [date] < '04-01-2019'
ORDER BY [date] DESC
SET FWEEK = DATENAME(yy, (DATEADD(month, 18, [date]))) + RIGHT('00'+DATENAME(dy,(datepart(DAYOFYEAR,DATEDIFF(DY,0,[date])/7*7+547)+5)/7),2)
, period = LEFT(CONVERT(varchar, DATEADD(month, 18, [date]),112),6)
Thanks Paul for all your help; it really wasnt easy as trying to apply someone elses logic to a date. I got it working using the above code in case anyone is looking for the same type of answer in the future
I've got an SSRS report that returns unique login count to our VDI Pools broken down by week.
For example for January it would read:
Week # of Logins
2015_JANUARY_WEEK_NO_1 3
2015_JANUARY_WEEK_NO_2 49
I'm using the Week column to link to another report that'll display the selected week's logins so I have to get the first day of the selected week and the last day of the select week and pass them to the other report as parameters.
I know how to get pull the year and week # from the Week column.
=Mid(Fields!Week.Value,1,4)
gives me the year and
=Trim(Mid(Fields!Week.Value,Len(Fields!Week.Value),Len(Fields!Week.Value)))
gives me the week.
I found this T-SQL that works:
DECLARE #WeekNum INT
, #YearNum char(4);
SELECT #WeekNum = 2
, #YearNum = 2015
-- once you have the #WeekNum and #YearNum set, the following calculates the date range.
SELECT DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + #YearNum) + (#WeekNum-1), 6) AS StartOfWeek;
SELECT DATEADD(wk, DATEDIFF(wk, 5, '1/1/' + #YearNum) + (#WeekNum-1), 5) AS EndOfWeek;
but I cannot figure out how to turn that into an expression that doesn't throw an error.
This is what I've got so far:
=DateAdd("w", DateDiff("w", 6, '1/1/' + (Mid(Fields!Week.Value,1,4))) + (Trim(Mid(Fields!Week.Value,Len(Fields!Week.Value),Len(Fields!Week.Value)))
- 1), 6)
and when I try to run the report in design view it returns an Expression expected error.
Edit
Sorry, I guess I should've posted my original query that's populating the report. Here it is below:
SELECT Convert(varchar(20),UPPER(DATENAME(YEAR, Time)))
+'_'+CONVERT(varchar(20),UPPER(DATENAME(MONTH, Time)))
+'_WEEK_NO_'+CONVERT(varchar(10),(DAY(Time)
+ (DATEPART(DW, DATEADD (MONTH, DATEDIFF (MONTH, 0, Time), 0))-1) -1)/7 + 1) as 'Week'
, Count(DISTINCT SUBSTRING(ModuleAndEventText,LEN('User ') + 2
, CHARINDEX(' requested', ModuleAndEventText) - LEN('User ') - 2)) as WeekCount
FROM VE1_UserLogins
WHERE DesktopId = #Pool
AND ([Time] BETWEEN (#StartDate) and (DATEADD(ms, -1, #EndDate +1)))
GROUP BY Convert(varchar(20),UPPER(DATENAME(YEAR, Time)))
+'_'+CONVERT(varchar(20),UPPER(DATENAME(MONTH, Time)))
+'_WEEK_NO_'+CONVERT(varchar(10),(DAY(Time)
+ (DATEPART(DW, DATEADD (MONTH, DATEDIFF (MONTH, 0, Time), 0))-1) -1)/7 + 1),YEAR(Time),MONTH(Time)
ORDER BY YEAR(Time), MONTH(Time), Week
I am assuming your Fields!Week.Value is string in date format like MM/dd/yyyy or something similar so if you want to get the start of that week Sunday as first day and Saturday as last day and you want to get there date then you should use below expression,
For start of week,
=DateAdd("d",1- DatePart("w", CDate(Fields!WeekDate.Value)), CDate(Fields!WeekDate.Value))
For end of week,
=DateAdd("d", 7 - DatePart("w", CDate(Fields!WeekDate.Value)), CDate(Fields!WeekDate.Value))
----------
UPDATE
Now there are two ways to achieve what you want
1)Calculate start of week and end of week at the sql and get the result to directly display on repott
2) Get time field from the sql and set expression in the SSRS
For the first way to achieve you need to add these two lines in your select sql statement
DATEADD(dd, -(DATEPART(dw, MIN(Time))-1),MIN(Time)) AS 'StartOfWeek'
,DATEADD(dd, 7-(DATEPART(dw, MIN(Time))), MIN(Time)) AS 'EndOfWeek'
Second way to achieve is
Update your Sql statement to get the date part and then include that date into the above given expression.
So for you your query you need to add,
,MIN(Time) AS WeekDate
Then you can use the above expression with using the incoming field WeekDate as input(I have updated the expression).
But, If you have no actual need for the date other than calculating the start and end of week then use the first method to get the data from the sql server as formatted.
I'm trying to solve a task: I have a table containing information about ships' battles. Battle is made of name and date. The problem is to get the last friday of the month when the battle occurred.
WITH num(n) AS(
SELECT 0
UNION ALL
SELECT n+1 FROM num
WHERE n < 31),
dat AS (
SELECT DATEADD(dd, n, CAST(battles.date AS DATE)) AS day,
dateadd(dd, 0, cast(battles.date as date)) as fight,
name FROM num,battles)
SELECT name, fight, max(day) FROM dat WHERE DATENAME(dw, day) = 'friday'
I thought there must be a maximum of date or something, but my code is wrong.
The result should look like this:
Please, help!!
P.S. DATE_FORMAT is not available
Possible problem: as spencer7593 noticed - and as I should have done and didn't - your original query is not MySQL at all. If you're porting a query that's OK. Otherwise this answer will not be helpful, as it makes use of MySQL functions.
The day you want is number 4 (0 being Sunday in MySQL).
So you want the last day of the month if the last day of the month is a 4; if the day of the month is a 5 you want a date which is 1 day earlier; if the day of the month is a 3 you want a date which is 1 day later, but that's impossible (the month ends), so you really need a date six days earlier.
This means that if the daynumber difference is negative, you want it modulo seven.
You can then build this expression (#DATE is your date; I use a fake date for testing)
SET #DATE='2015-02-18';
DATE_SUB(LAST_DAY(#DATE), INTERVAL ((WEEKDAY(LAST_DAY(#DATE))+7-4))%7 DAY);
It takes the last day of the month (LASTDAY(#DATE)), then it computes its weekday, getting a number from 0 to 6. Adds seven to ensure positivity after subtracting; then subtract the desired daynumber, in this case 4 for Friday.
The result, modulo seven, is the difference (always positive) from the last day's daynumber to the wanted daynumber. Since DATE_SUB(date, 0) returns the argument date, we needn't use IF.
SET #DATE='1962-10-20';
SELECT DATE_SUB(LAST_DAY(#DATE), INTERVAL ((WEEKDAY(LAST_DAY(#DATE))+7-4))%7 DAY) AS friday;
+------------+
| friday |
+------------+
| 1962-10-26 |
+------------+
Your query then would become something like:
SELECT `name`, `date`,
DATE_SUB(LAST_DAY(`date`),
INTERVAL ((WEEKDAY(LAST_DAY(`date`))+7-4))%7 DAY) AS friday
FROM battles;
Hi I have a MySQL database which on I am setting up a table for a Study Calendar, fields are as follows:
SELECT
studycalendarpk,
studytopic,
`module`,
startdate,
enddate,
syllabusoutline
FROM studycalendar
What I am trying to do is to create a query so that for a dashboard php page it has a query that dispays the current weeks study programme. Can someone please tell me how to setup query to filter it so that it is selected if the current date is between the startdate and enddate, thank you
You have a startdate and an enddate for each row in your table, and if I understand your requirement correctly, you want to display all rows that meet these criteria.
WHERE enddate >= start of week
AND startdate < start of next week
You already have startdate and enddate in your table. This answer assumes that each row's enddate is constrained to be greater than or equal to the starttdate. If it isn't you'll get strange results.
You need a MySQL expression to compute the first day of the present week. Here's how you do that.
FROM_DAYS(TO_DAYS(CURDATE()) -MOD(TO_DAYS(CURDATE()) -1, 7))
This expression yields the Sunday immediately preceding CURDATE(). If your weeks are considered to start on Monday, use this instead (notice the -2).
FROM_DAYS(TO_DAYS(CURDATE()) -MOD(TO_DAYS(CURDATE()) -2, 7))
These are very useful expressions because they yield actual dates. Those dates can then be manipulated by date arithmetic such as somedate + INTERVAL 7 DAY which conveniently gives you the date a week later. This sort of arithmetic even works for the last week, and the first week, of a calendar year.
Putting it all together, here's what you do to select the records you want.
WHERE enddate >= FROM_DAYS(TO_DAYS(CURDATE())-MOD(TO_DAYS(CURDATE())-1,7))
AND startdate < FROM_DAYS(TO_DAYS(CURDATE())-MOD(TO_DAYS(CURDATE())-1,7))
+ INTERVAL 7 DAY
This will get the records from your table relevant to the current week.
SELECT *
FROM studycalendar
where curdate() between startdate and enddate
Can you try it? We can gett week no of use by this week() method
SELECT
`studycalendarpk`,
`studytopic`,
`module`,
`startdate`,
`enddate`,
CAST(week(now()) AS UNSIGNED)
syllabusoutline
FROM studycalendar WHERE CAST(week(now()) AS UNSIGNED) between CAST(week('2014-09-01') AS UNSIGNED) and CAST(week('2014-09-07') AS UNSIGNED)
In SQL Statement in microsoft sql server, there is a built-in function to get week number but it is the week of the year.
Select DatePart(week, '2012/11/30') // **returns 48**
The returned value 48 is the week number of the year.
Instead of 48, I want to get 1, 2, 3 or 4 (week number of the month). I think the week number of the month can be achieved by modules with Month Number of this week. For e.g.
Select DATEPART(week, '2012/11/30')%MONTH('2012/11/30')
But I want to know is there other built-in functions to get WeekNumber of the month in MS SQL SERVER.
Here are 2 different ways, both are assuming the week starts on monday
If you want weeks to be whole, so they belong to the month in which they start:
So saturday 2012-09-01 and sunday 2012-09-02 is week 4 and monday 2012-09-03 is week 1 use this:
DECLARE #date date = '2012-09-01'
SELECT (day(datediff(d,0,#date)/7*7)-1)/7+1
If your weeks cut on monthchange so saturday 2012-09-01 and sunday 2012-09-02 is week 1 and monday 2012-09-03 is week 2 use this:
DECLARE #date date = '2012-09-01'
SELECT
datediff(ww,datediff(d,0,dateadd(m,datediff(m,7,#date),0)
)/7*7,dateadd(d,-1,#date))+1
I received an email from Gerald. He pointed out a flaw in the second method. This should be fixed now
I received an email from Ben Wilkins. He pointed out a flaw in the first method. This should be fixed now
DECLARE #DATE DATETIME
SET #DATE = '2013-08-04'
SELECT DATEPART(WEEK, #DATE) -
DATEPART(WEEK, DATEADD(MM, DATEDIFF(MM,0,#DATE), 0))+ 1 AS WEEK_OF_MONTH
No built-in function. It depends what you mean by week of month. You might mean whether it's in the first 7 days (week 1), the second 7 days (week 2), etc. In that case it would just be
(DATEPART(day,#Date)-1)/7 + 1
If you want to use the same week numbering as is used with DATEPART(week,), you could use the difference between the week numbers of the first of the month and the date in question (+1):
(DATEPART(week,#Date)- DATEPART(week,DATEADD(m, DATEDIFF(m, 0, #Date), 0))) + 1
Or, you might need something else, depending on what you mean by the week number.
Just look at the date and see what range it falls in.
Range 1-7 is the 1st week, Range 8-14 is the 2nd week, etc.
SELECT
CASE WHEN DATEPART(day,yourdate) < 8 THEN '1'
ELSE CASE WHEN DATEPART(day,yourdate) < 15 then '2'
ELSE CASE WHEN DATEPART(day,yourdate) < 22 then '3'
ELSE CASE WHEN DATEPART(day,yourdate) < 29 then '4'
ELSE '5'
END
END
END
END
Similar to the second solution, less code:
declare #date datetime = '2014-03-31'
SELECT DATEDIFF(week,0,#date) - (DATEDIFF(week,0,DATEADD(dd, -DAY(#date)+1, #date))-1)
Check this out... its working fine.
declare #date as datetime = '2014-03-10'
select DATEPART(week,#date) - DATEPART(week,cast(cast(year(#date) as varchar(4))+'-' + cast(month(#date) as varchar(2)) + '-01' as datetime))+1
WeekMonth = CASE WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 22 THEN '5'
WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 15 THEN '4'
WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 8 THEN '3'
WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 1 THEN '2'
ELSE '1'
END
There is no inbuilt function to get you the week number. I dont think dividing will help you anyway as the number of weeks in a month is not constant.
http://msdn.microsoft.com/en-us/library/bb675168.aspx
I guess you can divide the number(48) by 4 and take the modules of the same and project that as the week number of that month, by adding one to the result.
Here's a suggestion for getting the first and last days of the week for a month:
-- Build a temp table with all the dates of the month
drop table #tmp_datesforMonth
go
declare #begDate datetime
declare #endDate datetime
set #begDate = '6/1/13'
set #endDate = '6/30/13';
WITH N(n) AS
( SELECT 0
UNION ALL
SELECT n+1
FROM N
WHERE n <= datepart(dd,#enddate)
)
SELECT DATEADD(dd,n,#BegDate) as dDate
into #tmp_datesforMonth
FROM N
WHERE MONTH(DATEADD(dd,n,#BegDate)) = MONTH(#BegDate)
--- pull results showing the weeks' dates and the week # for the month (not the week # for the current month)
select MIN(dDate) as BegOfWeek
, MAX(dDate) as EndOfWeek
, datediff(week, dateadd(week, datediff(week, 0, dateadd(month, datediff(month, 0, dDate), 0)), 0), dDate) as WeekNumForMonth
from #tmp_datesforMonth
group by datediff(week, dateadd(week, datediff(week, 0, dateadd(month, datediff(month, 0, dDate), 0)), 0), dDate)
order by 3, 1
A dirty but easy one liner using Dense_Rank function. Performance WILL suffer, but effective none the less.
DENSE_RANK()over(Partition by Month(yourdate),Year(yourdate) Order by Datepart(week,yourdate) asc) as Week
Here is the query that brings the week number on whatever the startday and endday of the week it may be.
SET DATEFIRST 2
DECLARE #FROMDATE DATE='12-JAN-2015'
-- Get the first day of month
DECLARE #ALLDATE DATE=DATEADD(month, DATEDIFF(month, 0, #FROMDATE), 0)
DECLARE #FIRSTDATE DATE
;WITH CTE as
(
-- Get all dates in that month
SELECT 1 RNO,CAST(#ALLDATE AS DATE) as DATES
UNION ALL
SELECT RNO+1, DATEADD(DAY,1,DATES )
FROM CTE
WHERE DATES < DATEADD(MONTH,1,#ALLDATE)
)
-- Retrieves the first day of week, ie, if first day of week is Tuesday, it selects first Tuesday
SELECT TOP 1 #FIRSTDATE = DATES
FROM CTE
WHERE DATEPART(W,DATES)=1
SELECT (DATEDIFF(DAY,#FIRSTDATE,#FROMDATE)/7)+1 WEEKNO
For more information I have answered for the below question. Can check that.
How do I find week number of a date according to DATEFIRST
floor((day(#DateValue)-1)/7)+1
Here you go....
Im using the code below..
DATEPART(WK,#DATE_INSERT) - DATEPART(WK,DATEADD(DAY,1,DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#DATE_INSERT),0)))) + 1
Try Below Code:
declare #dt datetime='2018-03-15 05:16:00.000'
IF (Select (DatePart(DAY,#dt)%7))>0
Select (DatePart(DAY,#dt)/7) +1
ELSE
Select (DatePart(DAY,#dt)/7)
There is an inbuilt option to get the week number of the year
**select datepart(week,getdate())**
You can simply get week number by getting minimum week number of month and deduct it from week number. Suppose you have a table with dates
select
emp_id, dt , datepart(wk,dt) - (select min(datepart(wk,dt))
from
workdates ) + 1 from workdates
Solution:
declare #dt datetime='2018-03-31 05:16:00.000'
IF (Select (DatePart(DAY,#dt)%7))>0
Select (DatePart(DAY,#dt)/7) +1
ELSE
Select (DatePart(DAY,#dt)/7)
declare #end_date datetime = '2019-02-28';
select datepart(week, #end_date) - datepart(week, convert(datetime, substring(convert(nvarchar, convert(datetime, #end_date), 127), 1, 8) + '01')) + 1 [Week of Month];
Here is the tried and tested solution for this query in any situation - like if 1st of the month is on Friday , then also this will work -
select (DATEPART(wk,#date_given)-DATEPART(wk,dateadd(d,1-day(#date_given),#date_given)))+1
above are some solutions which will fail if the month's first date is on Friday , then 4th will be 2nd week of the month
Logic here works as well 4.3 weeks in every month. Take that from the DATEPART(WEEK) on every month but January. Just another way of looking at things. This would also account for months where there is a 5th week
DECLARE #date VARCHAR(10)
SET #date = '7/27/2019'
SELECT CEILING(DATEPART(WEEK,#date)-((DATEPART(MONTH,#date)-1)*4.3333)) 'Week of Month'
Below will only work if you have every week of the month represented in the select list. Else the rank function will not work, but it is a good solution.
SELECT DENSE_RANK() OVER (PARTITION BY MONTH(DATEFIELD)
ORDER BY DATEPART(WEEK,DATEFIELD) ASC) AS WeekofMont
try this one
declare #date datetime = '20210928'
select convert(int,(((cast(datepart(day,#date) as decimal(4,2))/7)-(1.00/7.00))+1.00))
select datepart(week,#date)-datepart(week,dateadd(day,1,eomonth(dateadd(m,-1,#date))))+1
or
select datepart(week,#date)-datepart(week,dateadd(d,-datepart(d,#date)+1,#date))+1
steps:
1,get the first day of month
2,week of year of the date - week of year of the first day of the month
3,+1
2023-1-1 is sunday
SET DATEFIRST 1;
DECLARE #date date = '2023-1-02';
select datepart(week,#date)-datepart(week,dateadd(day,1,eomonth(dateadd(m,-1,#date))))+1
return 2
SET DATEFIRST 7;
DECLARE #date date = '2023-1-02';
select datepart(week,#date)-datepart(week,dateadd(day,1,eomonth(dateadd(m,-1,#date))))+1
return 1
Code is below:
set datefirst 7
declare #dt datetime='29/04/2016 00:00:00'
select (day(#dt)+datepart(WEEKDAY,dateadd(d,-day(#dt),#dt+1)))/7
select #DateCreated, DATEDIFF(WEEK, #DateCreated, GETDATE())