Can someone suggest me how do we consider week to start on Sunday and end on Saturday, while numbering them backwards in a 52 week rolling report like week1, week2.. week52
I want to count my current week as Week1 starting on Sunday, so even if its partial week its still week1 and last week Sunday-Saturday is week2 and so on until 52nd week last year (that would roughly be in September counting backwards). I need this as I am working on a daily report that will look for sales for current week and past 51 (full) weeks. My report should also return any week without sales '0' without skipping it.
Here is a way. Note I created the recursive CTE to populate some dates. You won't have to do this step, and real only need the YourWeekOrder = ... part.
declare #startDate date = dateadd(year,-1,getdate())
declare #endDate date = getdate()
;with cte as(
select #startDate as TheDate
union all
select dateadd(day,1,TheDate)
from cte
where TheDate < #endDate)
select
TheDate
,TheWeekOfYear = datepart(week,TheDate)
,YourWeekOrder = dense_rank() over (order by cast(datepart(year,TheDate) as char(4)) + case when len(datepart(week,TheDate)) = 1 then '0' + cast(datepart(week,TheDate) as char(2)) else cast(datepart(week,TheDate) as char(2)) end desc)
from cte
order by
TheDate
option(maxrecursion 0)
SEE IT IN ACTION HERE
Related
I would like to calculate a due date.
If the start date is 30/12/2020, I need to count 20 working days from start date and display the date for the 20th working date from start date.
For example if the start date is 30/12/2020 must give me 28/01/2021 (excludes saturdays and sundays and finds the 20th working day from 30/12/2020).
But I am unable to exclude the weekends.
SELECT
DATEADD(DAY,20,CAST(CAST('2020-12-30' AS DATE) AS DATETIME))
-(CASE WHEN DATENAME(dw,'2020-12-30') = 'Sunday' THEN 1 ELSE 0 END)
-(CASE WHEN DATENAME(dw,'2020-12-30') = 'Saturday' THEN 1 ELSE 0 END) AS DueDate
thanks
Your best pick would be to build a calendar table, with a boolean flag that indicates whether each day is a working day or not. Then you would just do something like:
select dt
from calendar
where dt >= '20201231' and is_working_day = 1
order by dt
offset 19 rows fetch next 1 row only
The way your question is asked, however, one option enumerates the days in a subquery until the count of working days is reached:
declare #dt date = '20201231';
declare #no_days int = 20;
with cte as (
select #dt as dt, #no_days as no_days
union all
select dateadd(day, 1, dt),
case when datename(weekday, dt) in ('Saturday', 'Sunday')
then no_days
else no_days - 1
end
from cte
where no_days > 1
)
select max(dt) as res from cte
If you don't care about holidays, then the 20th working day is exactly 28 days later. So you can use:
dateadd(day, 28, '2020-12-30')
Note: This assumes that the starting date is a working day.
Hi please let me know how to extract the last day of Financial year in sql server.my financial year start from 2016-04-01 to 2017-03-31
Closest you can use is End Of Month for that you need to provide one date to that month as below:
select eomonth('2017-03-01')
To get the last day of the financial year for any date, you need to find the last of march if before march, or the last of march next year if after march:
declare #yourdate datetime = getdate();
select case when month(#yourdate) < 4 then CONVERT(datetime,cast(YEAR(#yourdate) as char(4)) + '-03-31' ,120)
else CONVERT(datetime,cast(YEAR(#yourdate) + 1 as char(4)) + '-03-31' ,120)
end as financial_year_end
Edit:
If you want last date derived based on from_date, then use something like this
Rextester Demo
select
case when datepart(mm,from_date) <=3 then
cast(concat(year(from_date),'-03-31') as datetime)
else
dateadd(year,1,cast(concat(year(from_date),'-03-31') as datetime))
end as last_date_fin
from
(select '2017-04-30' as from_date union all
select '2017-01-13') t;
This way from_date between Jan - Mar will give same year's 31st march. Else it will give next year's 31st March.
Previous answer:
http://rextester.com/AXVM26769
If you want to get last day of march for same year as passed, then use
select cast(concat(given_year,'-03-31') as datetime)
from
(select '2017' as given_year) t
If you want to pass 2016 and then get 2017-03-31 then use. You can change the year in derived table and change the output based on that.
select dateadd(year,1,cast(concat(given_year,'-03-31') as datetime))
from
(select '2016' as given_year) t;
This Code will work to find the last date of Financial Year.
For Previous Year case matches and 'THEN' part will Execute and for current year 'ELSE'
part will execute.
select CASE WHEN (MONTH(GETDATE())) <= 3
THEN convert(varchar(4), YEAR(GETDATE())-1) + '-' + '03-31'
ELSE convert(varchar(4),YEAR(GETDATE()))+ '-' + '03-31'
end
> LastDayOfYearFY] =
> eomonth( dateadd(month, 5,
> dateadd(year, datepart(year, (dateadd(month, 6, [date])) ) -1900, 0)))
Idea extension taken from return-first-day-of-financial-year
You can select all the dates order them descendant and take the first one.
SELECT date
FROM table
ORDER BY date desc
LIMIT 1;
I'm not really good when it comes to database...
I'm wondering if it is possible to get the weeks of a certain month and year..
For example: 1 (January) = month and 2016 = year
Desired result will be:
week 1
week 2
week 3
week 4
week 5
This is what I have tried so far...
declare #date datetime = '01/01/2016'
select datepart(day, datediff(day, 0, #date) / 7 * 7) / 7 + 1
This only returns the total of the weeks which is 5.
declare #MonthStart datetime
-- Find first day of current month
set #MonthStart = dateadd(mm,datediff(mm,0,getdate()),0)
select
Week,
WeekStart = dateadd(dd,(Week-1)*7,#MonthStart)
from
( -- Week numbers
select Week = 1 union all select 2 union all
select 3 union all select 4 union all select 5
) a
where
-- Necessary to limit to 4 weeks for Feb in non-leap year
datepart(mm,dateadd(dd,(Week-1)*7,#MonthStart)) =
datepart(mm,#MonthStart)
Got the answer in the link: http://www.sqlservercentral.com/Forums/Topic1328013-391-1.aspx
Here is one way to approach this:
A month has a minimum of 29 or more days, and a max of 31 or less. Meaning there are almost always 5 weeks a month, with the exception of a non-leap year's feburary, and in those cases, 4 weeks a month.
You can refer to this to find out which years are "leap".
Check for leap year
Hope this helps!
The following code will allow you to select a start and end date, and output one row per week, with numbered weeks, between those dates:
declare #start date = '1/1/2016'
declare #end date = '5/1/2016'
;with cte as (select #start date
, datename(month, #start) as month
union all
select dateadd(dd, 7, date)
, datename(month, dateadd(dd, 7, date))
from CTE
where date <= #end
)
select *, 'week '
+ cast(row_number() over (partition by month order by date) as varchar(1))
from CTE
order by date
I want to calculate the number of days per-quarter if start date and finish dates are given.
for example, one table has two columns, start date and finish date.
start date = 1st september and finish is 14th november.
I want to calculate the number of days present in between these two days that are present in each quarter
Q3 - 30 days
Q4 - 45 days (for this scenario)
Regards.
declare #StartDate date='2012-09-01';
declare #EndDate date='2012-11-14';
select CEILING(month(dateadd(q,datediff(q,0,dateadd(dd,number ,#StartDate)),0))/3.0) as QuarterNo,
COUNT(*) as 'number of days'
from master..spt_values
where type='p'
and dateadd(dd,number ,#StartDate)<=#EndDate
group by dateadd(q,datediff(q,0,dateadd(dd,number ,#StartDate)),0)
SQL fiddle demo
You can use a recursive query to get this. This generates the list of dates between your start and end date and then gets the count of days per quarter:
;with cte (start, enddate) as
(
select startdate, enddate
from yourtable
union all
select dateadd(dd, 1, start), enddate
from cte
where dateadd(dd, 1, start) <= enddate
)
select datepart(q, start) Quarter, count(datepart(q, start)) NoDays
from cte
group by datepart(q, start)
See SQL Fiddle with Demo
For example,
the start date = '20100530' and
the end date = '20100602'
How can I write a SQL to display the below result?
month: may, 2 days
month: June, 2 days
Use a recursive CTE to generate all of the dates between the start and end, and then do a simple group and count (caution, not tested, but should be close if not exactly right):
with dates (the_date) as (
select #start_date
UNION ALL
select dateadd(dd, 1, the_date) from dates where the_date <= #end_date
)
select
datepart(mm, the_date) month,
count(*) num_days
from
dates
group by
datepart(mm, the_date)
TBH, you really need to provide more schema and information about the source data. However, given what little we know, you should be able to write:
Select DateName('month', [Date]) As Month
, Cast(DateDiff(d, #StartDate, #EndDate) As varchar(10) - 1) + ' days'
From Table
Where [Date] Between #StartDate And #EndDate
What we'd need to know to refine our solutions is exactly how 2 days is supposed to be calculated. Is it the days between the start and end date? Is it the day of the second parameter?