I need to add the week day from previous year on the 3rd column.
Form example 06/02/2021 was on a Wednesday. I need the week day of 06/02/2021.
Thanks
try this:
SELECT DATENAME(weekday,DATEADD(YEAR, -1, '06-02-2021'))
Tableau is notoriously difficult to figure out work days. As some have suggested, it may be easier to use the underlying database to calculate this. However, if you need to do this in Tableau it can be accomplished like this:
DATEADD(
"day"
,CASE LEFT(LOWER(DATENAME("weekday",[DateFieldUsed])), 3)
WHEN 'fri' THEN 3
WHEN 'sat' THEN 2
ELSE 1
END
,[DateFieldUsed]
)
Basically, if it is Friday you need to add 3 days to the date, Saturday you need to add, and in any other situations just add 1 date. (please test, not doing this in Tableau so the numbers may be off).
I may have misunderstood the question, the example isn't exactly what you're asking. If your current value is 6/2/2021, and you are trying to find the week day of 6/2/2020, I would do this:
DATENAME("weekday", DATEADD(year, -1, [DateFieldUsed])
Related
I'm trying to figure out how to translate this line of PROC SQL code into Snowflake SQL but haven't found a way yet.
%LET last_post_dt = %SYSFUNC(INTNX(WEEK.3,%SYSFUNC(TODAY(),),-2,B),DATE9.);
Basically, its subtracting 2 weeks from the third day of the current week (monday = 1).
I've tried altering the session by using WEEK_START and other functions such as DATEADD, but haven't been able to solve this.
Thanks in advance!
You can use date_trun() to get whatever today is to a Monday. Then add 2 for 3rd day of the week, and then subtract 2 weeks from that.
SELECT DATE_TRUNC('week',current_date()) as week_start,
DATEADD('day',2,week_start) as week_3rd_day,
DATEADD('week',-2,week_3rd_day) as weeks_ago,
DATEADD('week',-2,DATEADD('day',2,DATE_TRUNC('week',current_date()))) as all_in_one_line_date
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'm currently studying SQL Server using the book Ben-Gan, Itzik. T-SQL Fundamentals. Below is a query used to select order placed at end of the month. (I know that function EOMONTH() can also be used)
SELECT orderid, orderdate, custid, empid
FROM Sales.Orders
WHERE orderdate = DATEADD( month, DATEDIFF( month, '18991231', orderdate), '18991231');
The author's explanation is:
This expression first calculates the difference in terms of whole
months between an anchor last day of some month (December 31, 1899, in
this case) and the specified date. Call this difference diff. By
adding diff months to the anchor date, you get the last day of the
target month.
However, I'm still a bit confused as to how it actually works. Would someone kindly explain it?
That seems like a rather arcane way to do this. What the code is doing is calculating the number of months since the last day of some month. Then, it adds this number of months to that date. Because of the rules of dateadd(), the month remains the last date.
However, I prefer a simpler method:
where day(dateadd(day, 1, orderdate)) = 1
I find this much clearer.
select DATEDIFF(MONTH, '20160131', '20160201')
give us 1 month and
SELECT DATEADD(month, 1, '20160131')
give us 2016-02-29 00:00:00.000
that's ok
I tried out the query myself and seem to have got the hang of it. here is what i wrote just in case anyone else is interested
SELECT DATEADD(month, DATEDIFF(MONTH, '20160131', '20160201'), '20160131');
result:
2016-02-29 00:00:00.000
so my interpretation is that adding one or more "month" to a particular date in which the last date of the month is 31 will always return a date in which the date is the last day of the month. if this sentence makes any sense...
I'm pretty new to SQL so I hope this isn't a dumb question, tried to google but couldn't find anything.
I'm summing sales of departments per week in SQL and am using TD_SYSFNLIB.WEEKNUMBER_OF_YEAR (trans_dt) to get the week number.
I think everything is working except I'd like to change the format of the weeks to the start date of the week, e.g. week 1 = 1/4/15
Also, i'm not sure how to handle the very first of the year week 0 since I think that should be grouped up with week 52 of last year.
The following date math trick should get you Beginning of Week as an actual date without having to join to the SYS_CALENDAR view or using a function:
SELECT CURRENT_DATE - ((CURRENT_DATE - DATE '0001-01-07) MOD 7) AS BOW;
Starting with TD14 there's NEXT_DAY which returns the following weekday, if you subtract 7 days you get the previous day:
next_day(trans_dt - 7, 'sunday')
I want to find first day month of month and also like 3rd day or 5th day ,15th day or any day of the month .So how to find through query.I know how to find first day and last day of month.Mainy I want find other days.
For those of you following along who may not know how to get the First Day of the month in SQL Server you can do so with something like this. This will also give you the 5th, 10th or whatever you need.
DECLARE #FirstDay DATETIME
SET #FirstDay = (DATEADD(MONTH, DATEDIFF(MONTH, -1, GETDATE()) - 1, -1) + 1)
SELECT GETDATE() AS CurrentDay
, #FirstDay AS FirstDay
, DATEADD(d, 10, #FirstDay-1) AS TenthDay
The -1 after the #FirstDay in the DateAdd is because the DateAdd will add the numbers of days onto the firstday, which will give you the 11th in that example. Of course you could just add one less day to make it work without the -1 but I prefer including it. Suit yourself.
If you know how to find the first day of a month, you can add the 2-day, the 4-day or the 14-day interval to the first day of the month to get, respectively, the 3rd, the 5th or the 15th day of the month.
Similarly you can get any day of the month by simply adding the proper number of days.
Different RDBMSs may offer different syntax to achieve the goal. Assuming #MonthBeginning to be a date or datetime value representing the first day of a month, here's how you can get, for example, the 5th day of the same month in Microsoft SQL Server:
SELECT DATEADD(day, 4, #MonthBeginning) AS FifthDay
Again, it may not be the way you should do that if your RDBMS is not MS SQL Server.