My requirement is that I want to find business-week-ending (not the calender week) given a DATE column from the sales table in MSSQL.
Using different techniques I was able to find the [Calender] week-endings (and week-starting) dates corresponding to DATE in the table.
Since our business week ends on Wednesday [DOW 3 or 4 depending when the week started], I tried to deduct number of days from the week ending dates to pull it back to Wednesday. The idea did work pretty good with a flaw. Works fine as long as the Date in the table is greater than DOW 3 or 4. Any suggestion?
SELECT DateAdd(wk, DateDiff(wk, 0, Recons_Sales_Details.Recons_Date), 0) + 2
You need to look into SET DATEFIRST to do this:
SET DATEFIRST 4 --4 is Thursday week start
SQL Fiddle Demo
Related
I have one column in which I have to show the week of the another column [Sales Stage Date].
I used
DATEPART( wk, [Sales Stage Date])
But the issue is it takes week from Sunday to Saturday but I want it to take it as Monday to Sunday.
Example: 1st Aug 2021 should be in week 31 and not in week 32.
Sub a day off the date before you ask what week it's in
DATEPART( wk, DATEADD(day, -1, [Sales Stage Date]))
You'll need to handle the first day of the year directly with a CASE WHEN
Presumably the language your LOGIN is using (American) ENGLISH and thus the first day of the week is Sunday.
One method would be to change your language, but this feels excessive and would be specific to your LOGIN only.
Another is to explicitly set the DATEFIRST setting to 1 prior to the statement:
SET DATEFIRST 1;
SELECT DATEPART(WEEK, '20210802'),
DATEPART(WEEK, '20210101');
This, however, would be explicit to the connection where you changed the value of DATEFIRST so if you use this logic a lot you'd need to continuously force the value of DATEFIRST (which could break other code).
Depending on your business neeeds, however, you might be better off investing in creating a calendar table. Then you can JOIN to that table, and get the values you need agnostic of the LANGUAGE and date settings.
Snowflake Database:
I have a day level table , i am trying use the day column to create a week column with the value of first Monday's of the week as a week value using following function. even though the table have a data for all seven days ( Monday through Sunday ) of the week. The following function will create a week column with five working days ( Monday through Friday) only rolled up to a week leaving Saturday and Sunday. is there any function which i can use to grab all seven days of data under one week in SNOWFLAKE DATABASE (first monday of the week).
CURRENT FUNCTION USED:
select dateadd('day', (extract('dayofweek_iso', current_date()) * -1) +1 , current_date() );
Snowflake provides DATE_TRUNC() to roll up date to corresponding year/month/week/day.
For the mentioned use case following query will roll up to Monday of the week.
select date_trunc('WEEK',current_date()) ;
Snowflake last_day() function can be used to achieve that. You can easily get the last day of week (Sunday) and subtract 6 days to get the first day of that week (Monday) like that:
select last_day(current_date(), 'week')-6 as first_day_of_week;
Given a (for us) large datastream, of about 2GB/day. This data has to be incorporated/structured/cleansed into a destination database (this part is solved).
Until now we were dropping and creating a new datatable every day for the whole dataset of the current month, and then union into a materialized table of the previous months' dataset.
Besides, that this is a very un-elegant way, also unfortunately by the end of the month this query is becoming quite massive and draining the server for up to 1,5 hours daily.
I thought I do a parameterized query for date by grabbing all the data for DateMax
(
declare #ReportDateMax datetime
set #ReportDateMax = dateadd(dd,-1, cast(getdate() as date)
)
however it is not good enough-
the problem is the following:
1. on monday we need to download the data for friday
2. on tuesday for monday
3. wednesday for tuesday
4. thursday for wednesday
5. friday for thursday
which means that I cannot just download the information for yesterday, as than on monday I would not get anything.
Also ideally, if we just daily data I would like to run twice a day a select query for the previous reporting day.
desired solution:
design a query which selects the previous working day (public holidays are irrelevant for the topic).
I have found a solution eventually in another post.
In case somebody else is interested:
SELECT DATEADD(DAY, CASE DATENAME(WEEKDAY, GETDATE())
WHEN 'Sunday' THEN -2
WHEN 'Monday' THEN -3
ELSE -1 END, DATEDIFF(DAY, 0, GETDATE()))
Sorry if the Title is confusing but it's hard to explain what I'm after in one phrase.
I'm currently producing a report based on the production for the week. I start off my CTE construction with the following to get the days Monday to Friday of the current week:
WITH
cte_Date AS
(
SELECT
CAST(DateTime AS date) AS Date
FROM
( VALUES
(GETDATE()
)
, (DATEADD(day,-1,GETDATE()))
, (DATEADD(day,-2,GETDATE()))
, (DATEADD(day,-3,GETDATE()))
, (DATEADD(day,-4,GETDATE()))
, (DATEADD(day,-5,GETDATE()))
, (DATEADD(day,-6,GETDATE())) ) AS LastSevenDays(DateTime)
WHERE
DATENAME(weekday, DateTime) = 'Monday'
UNION ALL
SELECT
DATEADD(day,1,Date)
FROM
cte_Date
WHERE
DATENAME(weekday,Date) <> 'Friday'
)
This is working fine. I have made the report available to users so they can run it anytime however sometimes nobody is available to run it last thing Friday. This means they don't get to see the full production for Friday and then the following week the CTE days change.
I'm trying to keep this a one-click affair so rather than introduce date parameters I proposed to the users that we adjust the query such that if they run the report before midday on "Monday" then it will show them last week's figures and they were happy with this (me and my big mouth). I put Monday in quotes because what we really mean of course is the first production day of the week.
My primary data table (which we'll call MyData) has a datetime field named DateTime (really!) that I can reference to determine the first day of production for the week.
One final caveat: Due to the layout of the report the users insisted that they always want to see the five days Monday to Friday, even if there is no production on a given day. (Consequently I do a LEFT JOIN from cte_Date to all other tables required.) So to be clear, right now as I'm typing this it's 11:45am local time on Tuesday and yesterday happened to be a public holiday here so running the report now should return Monday to Friday last week, but running it in 20 minutes time should return Monday to Friday this week.
Please help, my poor brain is getting twisted trying to figure it out.
There are a few different ways you can tackle this, but they all boil down to the same thing: you need a way of figuring out whether it's before or after 12pm on the first working day of the current week, then you need to get the Monday of the current "production week".
Let's just say, for simplicity's sake, you have some sort of table that contains public holidays (or non-production days). To find out whether it's the first day of the current production week, you basically just have to add the number of days in a row since the start of the week that have been public holidays.
Then you need to figure out whether it's before or after 12pm of that day.
If it's before you want last week's Monday-Friday. If it's after, you want this week's Monday-Friday.
Here's one way you might do this:
DECLARE #NonProductionDays TABLE (NPD DATE UNIQUE NOT NULL); -- Public holiday table.
INSERT #NonProductionDays (NPD) VALUES ('2017-09-25');
DECLARE #i INT = -- You don't need a variable for this, but just to keep things simple...
(
SELECT COUNT(*) -- Extract number of public holidays in a row this week before current date.
FROM #NonProductionDays AS N
WHERE DATEDIFF(WEEK, 0, N.NPD) = DATEDIFF(WEEK, 0, GETDATE())
AND N.NPD <= GETDATE()
AND (DATENAME(WEEKDAY, N.NPD) = 'Monday' OR EXISTS (SELECT 1 FROM #NonProductionDays AS N2 WHERE N2.NPD = DATEADD(DAY, -1, N.NPD)))
);
SELECT D = CAST(DATEADD(DAY, T.N, DATEADD(WEEK, DATEDIFF(HOUR, DATEADD(DAY, #i, '1900-01-01 12:00:00'), GETDATE()) / 24 / 7, '1900-01-01')) AS DATE)
FROM (VALUES (0), (1), (2), (3), (4)) AS T(N);
/*
Breaking this down:
X = DATEADD(DAY, #i, '1900-01-01 12:00:00')
-- Adds the number of NPD days this week to '1900-01-01 12:00:00'
-- So, for example, X would be '1900-01-02 12:00:00' this week
Y = DATEDIFF(HOUR, X, GETDATE()) / 24 / 7
-- The number of weeks between X and now, by taking the number of hours and dividing by 24 then by 7
-- The division is necessary to compare the hour.
-- So, for example, as of 11am on the September 26 2017, you'd get 6142.
-- As of 12pm on September 26 2017, you'd get 6143.
Z = DATEADD(WEEK, Y, '1900-01-01')
-- Just adds Y weeks to 1900-01-01, which was a Monday. This tells you the Monday of the current "production week".
-- So, for example, as of 11am on September 26 2017, you'd get '2017-09-18 00:00:00.000'.
-- As of 12pm on September 26 2017, you'd get '2017-09-25 00:00:00.000'.
Then we cast this as a date and add 0/1/2/3/4 days to it to get Monday, Tuesday, Wednesday, Thursday and Friday of the current "production week".
*/
I'm not sure I came up with the most efficient approach, but after a week of tossing it about in my brain this is what I came up with. I approached the problem from the opposite direction of that suggested by #ZLK.
My existing logic was already giving me the Monday of this week so in a subquery I looked for the first production record after Monday, stripped off the time with a DATEDIFF and made it midday with a DATEADD. I was then able to compare the current Date/Time with midday of the first production day to determine whether to reduce the date by one week.
I replaced this SELECT clause:
SELECT
CAST(DateTime AS date) AS Date
with this one:
SELECT -- Monday this week if it's after midday on the first production day otherwise Monday last week
DATEADD(week,IIF(GETDATE()>=DATEADD(hour,12,(
SELECT DATEDIFF(day,0,MIN(DateTime))
FROM MyData
WHERE CAST(MyData.DateTime AS date) >= CAST(LastSevenDays.DateTime AS date)
)),0,-1),CAST(LastSevenDays.DateTime AS date)) AS Date
To cater for the case where a new week has commenced but the operator runs the report before production starts I carefully arranged the boolean condition inside my IIF clause so that the empty result set from the subquery would mean the test returned FALSE and the operator would still see last week's figures.
(#ZLK, Thanks for your input - you did help my thinking a bit but I don't think your answer should be marked as correct. What I've come up with here is what I was originally requesting and didn't require the use of a static table.)
Why is this returning 4 instead of 6?
SET DATEFIRST 1
SELECT DATEDIFF(WEEK, CAST('2017-01-01' AS DATE), CAST('2017-01-31' AS DATE))
Is week as datepart calculating only weeks from monday - sunday (whole weeks)? How to get all weeks - including those which doesn't have seven days ? In the case above answer should be 6.
DATEDIFF counts transitions, not periods (e.g. look at DATEDIFF(year,'20161231','20170101')). It also treats Sunday as the first day of the week. So, how do we compensate for these features? First, we shift our dates so that Mondays are the new Sundays, and second we add 1 to compensate for the Fence-Post error:
declare #Samples table (
StartAt date not null,
EndAt date not null,
SampleName varchar(93) not null
)
insert into #Samples (StartAt,EndAt,SampleName) values
('20170101','20170131','Question - 6'),
('20170102','20170129','Exactly 4'),
('20170102','20170125','3 and a bit, round to 4'),
('20170101','20170129','4 and 1 day, round to 5')
--DATEDIFF counts *transitions*, and always considers Sunday the first day of the week
--We subtract a day from each date so that we're effectively treating Monday as the first day of the week
--We also add one because DATEDIFF counts transitions but we want periods (FencePost/FencePanel)
select *,
DATEDIFF(WEEK, DATEADD(day,-1,StartAt), DATEADD(day,-1,EndAt)) +1
as NumWeeks
from #Samples
Results:
StartAt EndAt SampleName NumWeeks
---------- ---------- -------------------------- -----------
2017-01-01 2017-01-31 Question - 6 6
2017-01-02 2017-01-29 Exactly 4 4
2017-01-02 2017-01-25 3 and a bit, round to 4 4
2017-01-01 2017-01-29 4 and 1 day, round to 5 5
If this doesn't match what you want, perhaps you can adopt and adapt my #Samples table to show the results you do expect.
What you ask though, is how many weeks are covered by a range, not how many weeks are between two dates.
DATEDIFF always uses Sunday when calculating week transitions.
This isn't a bug, it's done to ensure the function is deterministic and returns the same value, for every query, no matter the DATEFIRST setting. From the documentation
Specifying SET DATEFIRST has no effect on DATEDIFF. DATEDIFF always uses Sunday as the first day of the week to ensure the function is deterministic.
One solution would be to calculate the difference between the week numbers of the start and end dates, when the first day is Monday. 1 is added to the difference to take account of the first week as well:
SET DATEFIRST 1;
select 1 +datepart(WEEK,'20170131') - datepart(WEEK,'20170101')
That's a fragile calculation though that breaks if DATEFIRST changes or if one of the dates is on a different year.
You could use ISO Weeks to get rid of SET DATEFIRST:
select 1 +datepart(ISO_WEEK,'20170131') - datepart(ISO_WEEK,'20170101')
but that would fail for 2017-01-01 because Sunday is counted as Week 52 of the previous year.
A far better solution though would be to count the distinct week numbers using a Calendar table that contains dates and different week numbers to cover multiple business requirements, eg both normal and ISO Week numbers, or business calendars based on a 4-4-5 calendar.
In this case, you could just count distinct week numbers:
SELECT COUNT(DISTINCT Calendar.IsoWeek )
from Calendar
where date between '20170101' and '20170131'
If the table doesn't have an ISO Week column, you can use DATEPART:
select count (distinct datepart(ISO_WEEK,date) )
from Calendar
where date between '20170101' and '20170131'