I am trying to find an SSRS expression that will return the date of the most recent Saturday. If the expression runs on a Saturday, I want it to display the Saturday of the previous week.
I toyed around with an example I found online, and got it to produce the value I need it to produce. Listed below:
=IIf(Parameters!O_endDate.Value = nothing, DateAdd("d", -7-(WeekDay(Today(),7))+8, Today()), Parameters!O_endDate.Value)
I apologize for the stupid post. I cannot understand why the above expression is producing the value I need, and I want to know if it will continue to work as I want it to work, when run on other days of the week in the future. Note: I ran this code on a Sunday.
For your purposes, I don't think you need the IIF statement - this just returns the value of the parameter if it is not nothing.
This leaves the key operation as DateAdd("d", -7-(WeekDay(Today(),7))+8, Today()).
What this does is add -7-(WeekDay(Today(),7))+8 days to today - this can be simplified to 1-(WeekDay(Today(),7)) since -7+8=1.
Looking at the documentation here, the return_type appears to be invalid since the function is only defined for values of 1, 2 & 3 - if its working now then it may not continue to work.
This will work all the time DateAdd("d", 1 - WeekDay(Today(), 1), Today()) if you want the most recent Sunday to be today when today is a Sunday or use DateAdd("d", WeekDay(Today(), 2)), Today()) if the most recent Sunday is last week when today is a Sunday.
Note that in both cases this may produce the wrong answer at around midnight due to the 2 calls to Today() a few miliseconds apart. To be absolutely correct it should be:
Declare #DateToCheck datetime=Today()
Select DateAdd('d', 1 - WeekDay(#DateToCheck, 1), #DateToCheck)
This is a more generalised approach that should be closer to ANSI SQL.
See the SQL Fiddle for SQL Server 2008.
Becuase the return value of DATEPART depends on a global database setting, the return value is adjusted for a known Saturday. By changing this to another day of the week you will get the last that day.
You could try this:
=DateAdd("d", -1 * (weekday(today()) mod 7 ), today())
Related
I need to calculate the difference between DFU.HISTSTART and the last Sunday which is for today is 2/27. It should be dynamic and change every Sunday.
For some reason for this calculation I am getting 3 and should get 4.
,ABS(DATEDIFF(wk,
DATEADD(wk,
DATEDIFF(wk,6,GETDATE()), 0), DFU.HISTSTART))
AS '#WKS of Hist'
Does someone have any ideas?
You have two problems... the first is that you're trying to do the old offset trick with the "6". That works on other date parts but not on week. From the Microsoft Documentation...
For a week (wk, ww) or weekday (dw) datepart, the DATEPART return
value depends on the value set by SET DATEFIRST.
If your DATEFIRST is set to 7 (you can verify by running SELECT ##DATEFIRST;) AND your weeks start on Sundays, the following will work just fine and return a "4".
--===== Setup just the dates in question for a demo
DECLARE #HistStart DATE = '01-30-22'
,#Today DATE = '02-27-22'
;
--===== Demo the "right" way to use "wk".
-- I say "right" way because I don't trust DATEFIRST.
SELECT DATEDIFF(wk,#HistStart,#Today)
;
GO
The second thing is that it's generally a really bad practice to depend on the DATEFIRST setting in this global computing environment. Instead, do the much more universal/bullet-proof method of using Integer math to calculate the number of weeks it's been since date-serial 6, which you correctly identified as a Sunday.
--===== Setup just the dates in question for a demo
DECLARE #HistStart DATE = '01-30-22'
,#Today DATE = '02-27-22'
;
--===== Demo "Bulletproof" Way to calculate the difference in Weeks starting on Sunday
SELECT DATEDIFF(dd,6,#Today)/7 - DATEDIFF(dd,6,#HistStart)/7
;
If you need to calculate week differences in weeks a lot, you might want to turn that into a function so that if the company decides to change the day of the week that is the start of the week, you'll only need to change it in one place. In fact, you might want to have the function read it (the date-serial for the starting day of the week) from a "general settings table".
Another way of doing this:
with last_sunday as (
SELECT
case DAYNAME(current_date())
when 'Sun' then current_date()
when 'Mon'then current_date()-1
when 'Tue'then current_date()-2
when 'Wed'then current_date()-3
when 'Thu'then current_date()-4
when 'Fri'then current_date()-5
when 'Sat'then current_date()-6
else '2020-01-01' end "SUNDAY_DATE"
)
SELECT
DFU.HIST_START_DATE
,LAST_SUNDAY.SUNDAY_DATE
,datediff(week,DFU.HIST_START_DATE,LAST_SUNDAY.SUNDAY_DATE) weeks_diff
FROM DFU
JOIN LAST_SUNDAY
;
I need a WHERE statement where the date of the record is the previous day. I have the below code which will do this
WHERE DOC_DATE = dateadd(day,datediff(day,1,GETDATE()),0)
However I need this statement to get Friday's record when the current day is Monday. I have the below code but it will not work for me. No errors come back on SQL although no records results come back either. I have the below code for this
WHERE DOC_DATE = DATEADD(day, CASE WHEN datepart(dw,(GETDATE())) IN (2) then -3 ELSE -1 END ,0)
Important to add that this needs to be in a WHERE clause. This is for a Docuware administrative view I am creating. I have no control on how to write the SELECT statement, I only have access to edit the WHERE clause:
Here's a slightly "magical" way to compute the value that doesn't depend on any particular server settings such as datefirst. It's probably not immediately obvious how it works:
WHERE DOC_DATE = dateadd(day,datediff(day,'20150316',getdate()),
CASE WHEN DATEPART(weekday,getdate()) = DATEPART(weekday,'20150330')
THEN '20150313'
ELSE '20150315' END)
In the first line, we're computing the number of days which have elapsed since some arbitrary date. I picked a day in March 2015 to use as my base date.
The second line asks what today's day of the week is and if it's the same as some arbitrary "Known good" Monday. Just taking one value and comparing it to 2 depends on what your DATEFIRST setting is so I prefer not to write that.
In the third line, we decide what to do if it's a monday - I give a date that is 3 days before my arbitrary date above. If it wasn't a monday, we pick the day before.
Adding it all together, when we add the days difference from the arbitrary date back to one of these two dates from lines 3 and 4, it has the effect of shifting the date backwards 1 or 3 days.
It's can be an odd structure to see if you're not familiar with it - but combining dateadd/datediff and exploiting relationships between an arbitrary date and other dates computed from it can be useful for performing all kinds of calculations. A similar structure can be used for computing e.g. the last day of the month 15 months ago using just dateadd/datediff, an arbitrary date and another date with the right offset from the first:
SELECT DATEADD(month,DATEDIFF(month,'20010101',GETDATE()),'19991031')
As I said in a comment though, usually doing this sort of thing is only a short step away from needing to properly model your organisation's business days, at which point you'd typically want to introduce a calendar table. At one row per day, 20 years worth of pre-calculated calendar (adjusted as necessary as the business changes) is still less than 10000 rows.
You can try this.
WHERE DOC_DATE = DATEADD(DAY, CASE WHEN datepart(dw, GETDATE()) = 2 THEN -3 ELSE -1 END, CAST(GETDATE() AS DATE))
I have 70.000 rows of data, including a date time column (YYYY-MM-DD HH24-MM-SS.).
I want to split this data into 3 separate columns; Hour, day and Week number.
The date time column name is 'REGISTRATIONDATE' from the table 'CONTRACTS'.
This is what I have so far for the day and hour columns:
SELECT substr(REGISTRATIONDATE, 0, 10) AS "Date",
substr(REGISTRATIONDATE, 11, 9) AS "Hour"
FROM CONTRACTS;
I have seen the options to get a week number for specific dates, this assignment concerns 70.000 dates so this is not an option.
You (the OP) still have to explain what week number to assign to the first few days in a year, until the first Monday of the year. Do you assign a week number for the prior calendar year? In a Comment I asked about January 1, 2017, as an example; that was a Sunday. The week from January 2 to January 8 of 2017 is "week 1" according to your definition; what week number do you assign to Sunday, January 1, 2017?
The straightforward calculation below assigns to it week number 0. Other than that, the computation is trivial.
Notes: To find the Monday of the week for any given date dt, we can use trunc(dt, 'iw'). iw stands for ISO Week, standard week which starts on Monday and ends on Sunday.
Then: To find the first Monday of the year, we can start with the date January 7 and ask for the Monday of the week in which January 7 falls. (I won't explain that one - it's easy logic and it has nothing to do with programming.)
To input a fixed date, the best way is with the date literal syntax: date '2017-01-07' for January 7. Please check the Oracle documentation for "date literals" if you are not familiar with it.
So: to find the week number for any date dt, compute
1 + ( trunc(dt, 'iw') - trunc(date '2017-01-07', 'iw') ) / 7
This formula finds the Monday of the ISO Week of dt and subtracts the first Monday of the year - using Oracle date arithmetic, where the difference between two dates is the number of days between them. So to find the number of weeks we divide by 7; and to have the first Monday be assigned the number 1, instead of 0, we need to add 1 to the result of dividing by 7.
The other issue you will have to address is to convert your strings into dates. The best solution would be to fix the data model itself (change the data type of the column so that it is DATE instead of VARCHAR2); then all the bits of data you need could be extracted more easily, you would make sure you don't have dates like '2017-02-29 12:30:00' in your data (currently, if you do, you will have a very hard time making any date calculations work), queries will be a lot faster, etc. Anyway, that's an entirely different issue so I'll leave it out of this discussion.
Assuming your REGISTRATIONDATE if formatted as 'MM/DD/YYYY'
the simples (and the faster ) query is based ond to to_char(to_date(REGISTRATIONDATE,'MM/DD/YYYY'),'WW')
(otherwise convert you column in a proper date and perform the conversio to week number)
SELECT substr(REGISTRATIONDATE, 0, 10) AS "Date",
substr(REGISTRATIONDATE, 11, 9) AS "Hour",
to_char(to_date(REGISTRATIONDATE,'MM/DD/YYYY'),'WW') as "Week"
FROM CONTRACTS;
This is messy, but it looks like it works:
to_char(
to_date(RegistrationDate,'YYYY-MM-DD HH24-MI-SS') +
to_number(to_char(trunc(to_date(RegistrationDate,'YYYY-MM-DD HH24-MI-SS'),'YEAR'),'D'))
- 2,
'WW')
On the outside you have the solution previous given by others but using the correct date format. In the middle there is an adjustment of a certain number of days to adjust for where the 1st Jan falls. The trunc part gets the first of Jan from the date, the 'D' gets the weekday of 1st Jan. Since 1 represents Sunday, we have to use -2 to get what we need.
EDIT: I may delete this answer later, but it looks to me that the one from #mathguy is the best. See also the comments on that answer for how to extend to a general solution.
But first you need to:
Decide what to do dates in Jan before the first Monday, and
Resolve the underlying problems in the date which prevent it being converted to dates.
On point 1, if assigning week 0 is not acceptable (you want week 52/53) it gets a bit more complicated, but we'll still be able to help.
As I see it, on point 2, either there is something systematically wrong (perhaps they are timestamps and include fractions of a second) or there are isolated cases of invalid data.
Either the length, or the format, or the specific values don't compute. The error message you got suggests that at least some of the data is "too long", and the code in my comment should help you locate that.
I am writing a query with a where clause which needs to return the final day of the previous month in the previous year. So if I ran it today (18/06/2014), it would return 31/05/2013. If I ran it next month on 19/07/2014 it would return 30/06/2013 and so on.
I can use
select DATEADD(day,-1,DATEadd(MONTH,datediff(month,0,GETDATE()),0))
which returns the final date of last month but I can't work out how to adapt this to give me the date a year previously.
I can add - 365 to the end of the above code which does what I want but wouldn't account for leap years. Although I don't expect my reports to be in use for long enough for this to necessarily matter it would be good to find a solution that works nicely (and to think that they might be).
Any solutions greatly appreciated.
Just add another dateadd(YY, -1, {your code}) around your getdate() to roll the year back first, then look to previous month (per comments below).
select DATEADD(day,-1,DATEadd(MONTH,datediff(month,0,DATEADD(YY,-1,GETDATE())),0))
I am new to T-SQL and needed urgent assistance here.
I am trying to get the week number from a given date.
I understand that there is a build in function for it but the value return is not exactly what I wanted.
For e.g., by using select datepart(wk, '2013-01-07'), it would return me '2'.. but the actually fact is it should return '1' instead of '2'.
Any ideas how to correct this issue?
You can use dy datepart specifier to get dayOfYear number and divide it by 7:
select (datepart(dy, '2013-01-05') - 1) / 7 + 1;
Working DEMO.
Try this
SELECT DATEPART(WEEK,GETDATE())
This depends on hop you define the first week. Does it always start on the same weekday? or does it always start on the first of January? If you want it to always start on the same weekday, then use Set datefirst to tell T-SQL what weekdaty you want to define as the start of the week. If you want it to always start on Jan 1, then just use day of year instead of week, subtract 1, integer divide by 7 and add 1.
declare #dat DateTime = getdate()
Select Select (datepart(dy, #dat)-1) / 7 + 1
Although going from memory, I believe the ISO standard for the first week of the year is the week in the year that the first Thursday of the year is in. This would possibly explain why the built in function gives a result different to that you require.