Calculate working hours between two dates based on business hours - sql-server

I have a table in SQL Server representing service requests with their created date and LastUpdateAt which represents the end date if exist, if null it means the request is open and not yet solved.
I want to calculate the working hours for each request to be closed or if it's not yet closed, then the hours from the created date until current time, with weekends excluded and only calculating working hours each working day, from 8am to 4pm "8 hours per working day".
What I have tried so far:
WITH CTE AS
(
SELECT
Id,
CreatedAt AS DYN_DATE,
CreatedAt AS START_DATE,
(CASE
WHEN LastUpdateAt IS NULL
THEN GETDATE()
ELSE LastUpdateAt
END) AS END_DATE
FROM
Request
WHERE
Id IN ('14578')
)
SELECT
(DATEDIFF(HOUR, START_DATE, END_DATE) - (DATEDIFF(DAY, START_DATE, END_DATE) * 16 )) -
((SELECT COUNT(*) FROM CTE
WHERE DYN_DATE BETWEEN START_DATE AND END_DATE
AND DATEPART(WEEKDAY, DYN_DATE) NOT IN (6, 7)) * 24) AS B_DURATION
FROM
CTE
In this query I am trying to calculate the working hours for the request ID : 14578.
In the last part of the query, I was trying to calculate the total hours then subtract the number of non-working hours then also subtract the number of (weekends * 24).
Probably my method is wrong from the beginning, or I am thinking inside the box, so I am open to any suggestion or solution.

There are multiple phases to this calculation. It may be helpful to look at a calendar for the explanation.
Weeks between start and end
DATEDIFF(WEEK, cte2.START_DATE, cte2.END_DATE) As NumWeeks
This is calculating the difference between the rows of a calendar, a week. (a 7 day period generally represented as a single row on a calendar). Saturday and the following Sunday may only be one day apart, but they are in separate weeks, and therefore are in separate rows on the calendar.
Week Days between start and end.
DATEPART(WEEKDAY, cte2.END_DATE) - DATEPART(WEEKDAY, cte2.START_DATE) as NumDays
Step 1 calculated the difference in rows of the calendar, step 2 we're calculating the difference in columns. This will account for partial week differences.
Tuesday to the following Monday is six days, one day less than a week. Step one returned 1 week. Since we're shifting one column to the left, we adjusting the week by -1 days. If the end date had been Wednesday, it would be 1 week plus 1 day, but since it is Monday, it is 1 week minus 1 day.
Normalizing the Start/End time outside of working hours
CAST(CASE WHEN CAST(CTE.START_DATE AS TIME) < '08:00' THEN '08:00'
WHEN CAST(CTE.START_DATE AS TIME) > '16:00' THEN '16:00'
ELSE CAST(CTE.START_DATE AS TIME)
END AS DATETIME) as StartTimeOnly,
CAST(CASE WHEN CAST(CTE.END_DATE AS TIME) < '08:00' THEN '08:00'
WHEN CAST(CTE.END_DATE AS TIME) > '16:00' THEN '16:00'
ELSE CAST(CTE.END_DATE AS TIME)
END AS DATETIME) as EndTimeOnly
This calculation is only interested in the time, so we cast to Time, then back to DateTime. This sets the Date component for both to 1900-01-01.
Similar to the week & day relationship, an End Time that occurs the next day, but before the Start Time will subtract hours credited from the number of days. For example 1/2 at 12:00 to 1/3 at 10:00 would be 1 day(1/3 - 1/2), or 8 hours, - 2 hours (10-12) = 6 hours of business time.
Calculating the difference in minutes, to ensure partial hours are considered correctly, then converting to hours. This ensures times only a couple minutes apart across an hour boundary don't get counted as a full hour. Of course, the trade off is 59 minutes rounds down to 0 hours.
DATEDIFF(MINUTE, StartTimeOnly, EndTimeOnly)/60 as NumHours
If rounding at the half hour...
CAST(ROUND(DATEDIFF(MINUTE, StartTimeOnly, EndTimeOnly) / 60.0, 0) AS INT) AS RoundedHours
Wrap it all together
,
cte2 AS
(
SELECT CTE.START_DATE,
CTE.END_DATE,
StartTimeOnly = CAST(CASE WHEN CAST(CTE.START_DATE AS TIME) < '08:00' THEN '08:00'
WHEN CAST(CTE.START_DATE AS TIME) > '16:00' THEN '16:00'
ELSE CAST(CTE.START_DATE AS TIME)
END AS DATETIME),
EndTimeOnly = CAST(CASE WHEN CAST(CTE.END_DATE AS TIME) < '08:00' THEN '08:00'
WHEN CAST(CTE.END_DATE AS TIME) > '16:00' THEN '16:00'
ELSE CAST(CTE.END_DATE AS TIME)
END AS DATETIME)
FROM CTE
),
CTE3 AS
(
SELECT START_DATE = CAST(cte2.START_DATE AS DATETIME2(0)),
END_DATE = CAST(cte2.END_DATE AS DATETIME2(0)),
NumWeeks = DATEDIFF(WEEK, cte2.START_DATE, cte2.END_DATE),
NumDays = DATEPART(WEEKDAY, cte2.END_DATE) - DATEPART(WEEKDAY, cte2.START_DATE),
NumHours = DATEDIFF(MINUTE, cte2.StartTimeOnly, cte2.EndTimeOnly)/60
FROM cte2
)
SELECT CTE3.START_DATE,
CTE3.END_DATE,
CTE3.NumWeeks,
CTE3.NumDays,
CTE3.NumHours,
TotalBusinessHours = (CTE3.NumWeeks * 5 * 8) + (CTE3.NumDays * 8) + (CTE3.NumHours )
FROM CTE3
;
For more accurate results, you'll also want to add a table containing your holidays. You'll then subtract the number of holidays found between your start and end date from your total number of days, before converting it to hours.
A question you may still need to answer... what happens if the start and end dates occur during non-working hours? e.g. Start at 19:00 and finish at 20:00. Is that 0 business hours to resolve?

Related

How to find the past 4 weeks of the same weekday value starting today

I am trying to select records from today and the same day of each week for the last 4 weeks.
Today (Tuesday)
Last Tuesday
The Tuesday before that
The Tuesday before that
I need this to be tied to current date because I am going to run this query every day so I don't want to use a between or something where I manually specify the date range.
Everything I have found or tried so far has pulled the last month of data but not the last 4 weeks of the same weekday.
select *
from table
where thedatecolumn >= DATEADD(mm, -1, GETDATE())
This works but pulls everything from the last month.
If today's date is 7/10/2019 I need
Data from 7/10/2019
Data from 7/3/2019
Data from 6/26/2019
Data from 6/19/2019
Every day I will run this query, so I need it to be dynamic based on the current date.
I believe you want to look back 21 days and then filter those dates that have the same day of week:
select * from table
where thedatecolumn >= DATEADD(DAY, -21, CAST(GETDATE() AS DATE))
and DATEPART(WEEKDAY, thedatecolumn) = DATEPART(WEEKDAY, GETDATE())
You Can try using a recursive cte which starts today and repeatedly substracts 7 days - so you ensure you always land on the same weekday. Following an example:
WITH cteFromToday AS(
SELECT 0 AS WeeksBack, GETDATE() AS MyDate
UNION ALL
SELECT WeeksBack + 1 AS WeeksBack, DATEADD(d, -7, MyDate) AS MyDate
FROM cteFromToday
)
SELECT TOP 5 *
FROM cteFromToday
OPTION ( MaxRecursion 0 );
This is quite simple. Substitute CURRENT_TIMESTAMP here for any given date.
SELECT CONVERT(DATE,CURRENT_TIMESTAMP) AS Today,
DATEADD(DAY,-7,CONVERT(DATE,CURRENT_TIMESTAMP)) AS LastWeek ,
DATEADD(DAY,-14,CONVERT(DATE,CURRENT_TIMESTAMP)) AS TwoWeeksAgo,
DATEADD(DAY,-21,CONVERT(DATE,CURRENT_TIMESTAMP)) AS ThreeWeeksAgo
SO, if you want to get data for a set of ranges for one entire day with those dates:
SELECT something
WHERE
datetimecolumn >= CONVERT(DATE,CURRENT_TIMESTAMP) AND datetimecolumn < DATEADD(DAY,1, CONVERT(DATE,CURRENT_TIMESTAMP)) -- Todays range,
OR datetimecolumn >= DATEADD(DAY,-7,CONVERT(DATE,CURRENT_TIMESTAMP)) AND datetimecolumn < DATEADD(DAY,1,DATEADD(DAY,-7,CONVERT(DATE,CURRENT_TIMESTAMP)))-- LastWeek ,
OR datetimecolumn >= DATEADD(DAY,-14,CONVERT(DATE,CURRENT_TIMESTAMP)) AND datetimecolumn < DATEADD(DAY,1,DATEADD(DAY,-14,CONVERT(DATE,CURRENT_TIMESTAMP)))-- TwoWeeksAgo,
OR datetimecolumn >= DATEADD(DAY,-21,CONVERT(DATE,CURRENT_TIMESTAMP)) AND datetimecolumn < DATEADD(DAY,1, DATEADD(DAY,-21,CONVERT(DATE,CURRENT_TIMESTAMP))) -- ThreeWeeksAgo

SQL Server Parameter Date Range Issues

I am using the following query to get the difference between two dates. The date ranges are tolling 12 months interval.
CY stand for Current year while PY stands for Previous Year. The dates in Current year are used to calculate the previous year dates
When I execute my query I have the following output, where the month is 11 and day 364. But I want my months to be twelve and the day 365 or (366 for leap year).
DECLARE #CY_StartDate date =CAST(DATEADD(MONTH, DATEDIFF(MONTH, -1, GETDATE())-13, 0) AS DATE),
#CY_EndDate date =CAST(DATEADD(MONTH, DATEDIFF(MONTH, -1, GETDATE())-1, -1) AS DATE); --- Rolling 12 months
DECLARE #PY_startDate date =DATEADD(YEAR,-1,#CY_StartDate),
#PY_EndDate date =DATEADD(YEAR,-1,#CY_EndDate)
SELECT
#CY_StartDate AS CY_Start,
#CY_EndDate AS CY_End,
#PY_StartDate AS PY_Start,
#PY_EndDate AS PY_End,
DATEDIFF(year, #CY_StartDate, #CY_EndDate) AS yr,
DATEDIFF(month, #CY_StartDate, #CY_EndDate) AS month,
DATEDIFF(day, #CY_StartDate, #CY_EndDate) AS day
Current Output
CY_Start CY_End PY_Start PY_End yr month day
2017-10-01 2018-09-30 2016-10-01 2017-09-30 1 11 364
Expected output
CY_Start CY_End PY_Start PY_End yr month day
2017-10-01 2018-09-30 2016-10-01 2017-09-30 1 12 365
The values you are getting make sense. DATEDIFF counts the ticks between 2 dates, where a tick is the value of the first parameter. So, for example: DATEDIFF(MONTH, '20180101','20180228') will return 1, as only 1 tick has occured (2 - 1 = 1). Seems, here, you simply need to add 1:
DECLARE #CY_StartDate date =CAST(DATEADD(MONTH, DATEDIFF(MONTH, -1, GETDATE())-13, 0) AS DATE),
#CY_EndDate date =CAST(DATEADD(MONTH, DATEDIFF(MONTH, -1, GETDATE())-1, -1) AS DATE); --- Rolling 12 months
DECLARE #PY_startDate date =DATEADD(YEAR,-1,#CY_StartDate),
#PY_EndDate date =DATEADD(YEAR,-1,#CY_EndDate)
select
#CY_StartDate as CY_Start,
#CY_EndDate AS CY_End,
#PY_StartDate AS PY_Start,
#PY_EndDate AS PY_End,
DATEDIFF(year,#CY_StartDate,DATEADD(DAY,1,#CY_EndDate)) as yr,
DATEDIFF(month,#CY_StartDate,DATEADD(DAY,1,#CY_EndDate)) as month,
DATEDIFF(day,#CY_StartDate,DATEADD(DAY,1,#CY_EndDate)) as day
The reason I used a further DATEADD is because this makes it consistent with every expression. The value of yr was correct, however, for dates like 20170101 and 20171231, the value of yr would be 0. Hence adding 1 day the the value of #CY_EndDate makes this far more reliable, should the dates move.
Common sense. How many numbers are there between 1 and 10 including both? You might say that there are 10 - 1 = 9 which is incorrect. The correct answer is (10 - 1) + 1 = 10.
Likewise, if you have two inclusive dates e.g. 2017-10-01 and 2018-09-30 you add one to DATEDIFF(DAY, '2017-10-01', '2018-09-30') to get 365 instead of 364.
However, as suggested in the other answer, it is much better to the end date exclusive (not counted) which makes date calculations straight forward. In your example, you should add 1 day to the last date so that you have [2017-10-01, 2018-10-01) and DATEDIFF will produce desired results.

MSSQL order by previous 7 days

I run this query in MSSQL to get the items, grouping by the last 7 days of the week:
SELECT COUNT(Date_Entered), DATENAME(WEEKDAY, Date_Entered)
FROM my_table
WHERE Board_Name = 'Board'
AND DATEDIFF(DAY,Date_Entered,GETDATE()) <= 7
GROUP BY DATENAME(WEEKDAY, Date_Entered)
In the result, days of the week are sorted in alphabetical order: Friday > Monday > Saturday > Sunday > Thursday > Tuesday > Wednesday
How do I sort by the normal/correct/common sense order, starting with the weekday of 7 days ago and ending with yesterday?
Ordering by MAX(Date_Entered) should work too:
SELECT
COUNT(Date_Entered),
DATENAME(WEEKDAY, Date_Entered)
FROM my_table
WHERE Board_Name = 'Board' AND DATEDIFF(DAY,Date_Entered,GETDATE()) <= 7
GROUP BY DATENAME(WEEKDAY, Date_Entered)
ORDER BY MAX(Date_Entered);
Normally you would want to order by the date ascending, but since you use an aggregate function you would need to group by the date which would ruin it, but since the max(date) in every group is the date you can do max(date) to order.
DATEPART is your friend, try it like this:
SELECT COUNT(Date_Entered), DATENAME(WEEKDAY, Date_Entered),DATEPART(WEEKDAY,Date_Entered)
FROM my_table
WHERE Board_Name = 'Board'
AND DATEDIFF(DAY,Date_Entered,GETDATE()) <= 7
GROUP BY DATEPART(WEEKDAY,Date_Entered),DATENAME(WEEKDAY, Date_Entered)
ORDER BY DATEPART(WEEKDAY,Date_Entered)
If you can't count on data being available for every week then you'd need to do something more based on date calculations. Off the top of my head I think this will be more reliable:
ORDER BY (DATEDIFF(dd, MAX(Date_Entered), CURRENT_TIMESTAMP) + 77777) % 7
EDIT: I wrote that not realizing that the data was already limited to a single week. I thought the intention was to group in buckets by day of week for a longer range of dates.
I'll also comment that to me it is more natural to do the grouping on cast(Date_Entered as date) rather than on a string value and I wouldn't be surprised if it's a more efficient query.

SQL Server : calculating days elapsed

I need to get the number of elapsed days between any two dates with respect to the current date. IE:
mm/dd/yyyy
Current day = 07/10/2015
07/08/2013 ... 07/11/2013 - 4 days elapsed
Current day = 07/10/2015
07/08/2015 ... 07/11/2015 - 2 days have elapsed
I've tried several combinations using DATEDIFF with day as the date part, however, I can't seem to get a clean way to get the days elapsed when the date could be past or present.
EDIT
I know the start date and the end date of a certain business process. They could be this year, last year, two years ago and so on. I need a way via SQL Server functions to figure out the days total elapsed. If it's not the current year, obviously the entire span/range would have elapsed. If it's the current year, perhaps the entire span/range hasn't elapsed and it needs to say how many days are "into the process" based on the respected start time, end time and current time.
Hopefully this makes more sense?
Please help.
I used #Sean Lange, with a small tweak:
DATEDIFF(DAY, #StartDate, case when #EndDate < GETDATE() then #EndDate + 1 else GETDATE() end)
Thanks all.
This is pretty similar to the answer provided by Stan but here is my take on this.
with Something as
(
select CAST('2013-07-08' as datetime) as StartDate
, CAST('2013-07-11' as datetime) as EndDate
union all
select '2015-07-08', '2015-07-11'
)
select *
, DATEDIFF(DAY, StartDate, case when EndDate < GETDATE() then EndDate else GETDATE() end)
from Something
How about this:
Given:
CREATE TABLE dbo.test ( ChildID INT Identity,
Start DateTime
, Finish DateTime
)
and your test data:
insert into dbo.test (start,finish) values('07/08/2013','07/11/2013')
insert into dbo.test (start,finish) values('07/08/2015','07/11/2015')
then
select start,finish
, DATEDIFF(DAY, start, CASE WHEN GETDATE() BETWEEN start and finish
THEN GETDATE() - 1 ELSE finish END) + 1 as elapsed
from dbo.test
gives the result from your example.
You might have to tweak if there are other adjustments for how the current date fits between the range.

get first and last date of 4-week period

I am trying to create a calender table that contains indivudual dates.
For each date a column is present that tells you in which week, isowk, month, quarter and so on thast date belongs.
That is not a big problem. But now we are working with 4-week periods, based on isowk.
So
date 2014-12-30 belongs to period 1 which starts on 2014-12-30 and ends on 2014-01-26
date 2014-02-01 belongs to period 2 which starts on 2014-01-27 and ends on 2014-03-02
date 2014-08-05 belongs to period 8 which starts on 2014-07-14 and ends on 2014-08-11
It is easy to calculate which period a date belongs to:
period_number = ceiling(cast(datepart(isowk,#date) as float) /4 )
Finding the start and end of that particular week is not that hard either:
start_week = cast(DATEADD(wk,DATEDIFF(wk,0,#date),0) as date)
end_week = cast(DATEADD(wk,DATEDIFF(wk,0,#date),6) as date)
but how do I calculate the first date of that 4-week period?
thanks for thinking with me
If the start week/end week logic works, then this might work:
start_week = cast(DATEADD(wk, 4 * (DATEDIFF(wk, 0, #date) / 4), 0) as date)
end_week = cast(DATEADD(wk, 4 * (DATEDIFF(wk, 0, #date) / 4), 27) as date)
I'm not sure why you are using wk instead of isowk if you want ISO weeks.

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