In SQL Server, trying to write a age-off report for inventory purposes. Each week, the inventory system marks thousands of rows for deletion. This takes place on Sundays # 06:00:00 as part of weekly SQL DB purge schedule.
Using (yyyy-mm-dd hh:mm:ss:ms) format for closed_time, how can I calculate the numbers of days between that date, until next Sunday of the current week? And to be more elaborate, is there a way to narrow it down to the exact DD:HH:MM? The problem is the each client's Sunday DB schedule for purge varies. So that might be difficult to compute. Might be easier to just calculate whole days until Sunday 00:00:00. I tried using the DATEDIFF function with no success.
SELECT
Yada
DATEDIFF(DAY, closed_time,DW) AS Days_Until_Purged
FROM DB1
WHERE closed_time DESC
Thx in advance
If you choose any Sunday in the past (Such as 06:00 Sunday 2nd January 2000), you can calculate time that has GONE BY since then.
Then, if you take that and do modulo 7-days you get the time that has gone by since the most recent Sunday.
Then, if you do 7 - time_gone_by_since_last_sunday you get the time until the next sunday.
I'm going to do this in minutes to cope with a client that has a setting of 06:30.
DECLARE
#batch_processing_time SMALLDATETIME
SET
#batch_processing_time = '2000-01-02 06:00'
SELECT
(60*24*7) - DATEDIFF(minute, #batch_processing_time, closed_time) % (60*24*7)
FROM
yourTable
That's the number of minutes from each record's closed_time until the next #batch_processing_time.
Divide by (24*60) to get it in days.
try this:
select 8-DATEpart(w, closed_time) AS Days_Until_Purged from DB1 ...
This should solve your problem
SET DATEFIRST 1
Select DATEDIFF(dd,GETDATE(),DATEADD(DAY , 7-DATEPART(WEEKDAY,GETDATE()),GETDATE()))
Related
I have an requirement in hive to calculate Saturday as week start date for a given date in hive sql.
Eg)
Date week_start
03-27-2021 03-27-2021
03-28-2021 03-27-2021
03-31-2021 03-27-2021
04-07-2021 O4-03-2021
04-09-2021. 04-03-2021
I tried using pmod and other date functions but not getting desired output. Any insight is much appreciated.
Hive offers next_day(), which can be adapted for this purpose. I think the logic you want is:
select date_add(next_day(date, 'SAT'), -7)
This is a little arcane. next_day() gets the next date after the argument date with a given day of the week. So, go to the next Saturday and then subtract 7 days for the start of the week.
I work for a company where everyday I modify a query by changing the date of the day before, because the report is always from the previous day.
I want to automate the date change. I have made a table with two columns, one with all dates from this year and another with bits where if 0 is a working day and 1 if is a holiday.
I have successfully automated a little bit by telling if the day before is a working day then subtract 1 from the date (This is what happens everyday). But the problem is, that if is Monday appears as Friday, because Saturday and Sunday are not billable. And let's also say, that if today is Thursday and Wednesday and Tuesday we're holidays, then the report will run on Monday. I will leave you a picture, that shows how the table is made with dates.
Remembering, that if there is no holidays in the middle of the week, always will be subtract one.
The way to do this sort of thing is close to what you have done, but just extend it further. Create a BusinessDate table that has every date, and then every rule you have implemented inside it. You can go so far as to include a column such as ReportDate which will return,for every date, what date the report should be run for.
Do this once, and it will work forever more. You may have to update for future holidays once a year, but better than once a day!
It will also allow you to update things specific for your business, like quarter dates, company holidays, etc.
If you want to know more on the subject, look up topics around creating a date dimension in a data warehouse. Its the same general issue you are facing.
Too complicated for a comment and it involves a lot of guessing.
So everyday, your process starts by first determining if "today" is a work day. So you would do something like:
if exists (select * from <calendar> where date = cast (getdate() as date) and IsWorkday = 1")
begin
<do stuff>
end;
The "do stuff" section would then run a report or your query (or something that isn't very clear) using the most recent work day prior to the current date. You find that date using something like:
declare #targetdate date;
set #targetdate = (select max(date) from <calendar>
where date < cast (getdate() as date)
and IsWorkday = 1);
if #targetdate is not null
<run your query using #targetdate>
That can be condensed into less code but it is easier to understand when the logic is written step-by-step.
SQL server DATEPART function has two options to retrieve week number;
ISO_WEEK and WEEK. I Know the difference between the two, I want to have week numbers based on Sunday start standard as followed in the US; i.e. WEEK. But it doesn't handles partial weeks the way I expected. e.g.
SELECT DATEPART(WEEK,'2015-12-31') --53
SELECT DATEPART(WEEK,'2016-01-01') --1
SELECT DATEPART(WEEK,'2016-01-03') --2
gives two different week numbers for a single week, divided in two years. I wanted to implement something like in the following link for week days.
Week numbers according to US standard
Basically I would like something like this;
SELECT DATEPART(WEEK,'2015-12-31') --1
SELECT DATEPART(WEEK,'2016-01-01') --1
SELECT DATEPART(WEEK,'2016-01-03') --2
EDIT:
Basically I am not good with the division of a single week into two, I have to perform some calculations based on week numbers and the fact that a single week to be divided isn't acceptable. So if above isn't possible.
Is it possible that the week number one would start from 2016-01-03. i.e. what I would in that case would be something like this:
SELECT DATEPART(WEEK,'2015-12-31') --53
SELECT DATEPART(WEEK,'2016-01-01') --53
SELECT DATEPART(WEEK,'2016-01-03') --1
If you want the US numbering, you can do this by taking the WEEK number of the end of the week rather than the date itself.
First ensure that the setting for first day of the week is in fact Sunday on your system. You can verify this by running SELECT ##DATEFIRST; this should return 7 for Sunday. If it doesn't, run SET DATEFIRST 7; first.
SELECT
end_of_week=DATEADD(DAY, 7-(DATEPART(WEEKDAY, '20151231')), '20151231'),
week_day=DATEPART(WEEK, DATEADD(DAY, 7-(DATEPART(WEEKDAY, '20151231')), '20151231'));
Which will return 2016/01/02 - 1.
If you wish generate week number of a date, it will return the week number of the year(input date)
Thus, I think sql server treat '2015-12-31' as the last week of 2015.
What does this do in a SQL Query? Can someone explain? What does the .5- represent?
WHERE ScheduleEntry.ScheduleDate >= getdate() and ScheduleEntry.ScheduleDate <= getDate() +.50
Think of date unit as 1 day. 0.50 of a day is 1/2 of a day. So this returns anything that has ScheduleDate within half a day from getdate() time forward.
It restricts the rows returned to rows where the ScheduleEntry.ScheduleDate is in the future and where it is not more than .50 units later than the current date. To find out how much time .50 units is equal to, run the following on your console:
SELECT getDate()
SELECT getDate() +.50
The difference between the dates should tell you the difference. Most likely, it is half a day.
GETDATE returns the current date and time in SQL.
You can use addition to "add days" to the current date.
Generally, it is better to use DATE_ADD instead of adding directly and is probably easier to read. Adding 0.5 is akin to adding half a day (or 12 hours).
To better illustrate, the .5 is half a day
SELECT DateDiff(HH,getDate(),getDate() +.50)
Returns 12 hours
I need to run a weekly report based on the arrival date. How can I set the arrival date in the where clause so that I can get the result only for each week. The hard part is I DO NOT want to modify the dates each week. I need the permanent where clause for the date. I have to provide a list of customers who arrived every week and I just want to run the same script without changing the week dates.
Please assist.
Thanks.
SELECT * FROM TABLE WHERE
(ARRIVAL_DATE>DATEFROMPARTS(YEAR(GETDATE()-7), MONTH(GETDATE()-7), DAY(GETDATE()-7)))//7 days before starting at midnight
AND
(ARRIVAL_DATE<DATEFROMPARTS(YEAR(GETDATE()), MONTH(GETDATE()), DAY(GETDATE()))) //NOW in the YYYY, MM,DD format
This will get everything that happened in the current calendar week (Mon-Sun)
SELECT * FROM Table1
WHERE ArrivalDate BETWEEN
CAST(DATEADD(dd,(((DATEPART(dw,getdate())+##DATEFIRST) % 7)+5) % 7 ,getdate()) as date) AND
CAST(DATEADD(dd,6+((((DATEPART(dw,getdate())+##DATEFIRST) % 7)+5) % 7 ),getdate()) as date)
Edit - Removed extra E in BETWEEN