Related
I have a table, with types int, datetime, datetime:
id start date end date
-- ---------- ----------
1 2019-04-02 2020-09-17
2 2019-08-10 2020-08-10
Here is create/insert:
CREATE TABLE dbo.something
(
id int,
[start date] datetime,
[end date] datetime
);
INSERT dbo.something(id,[start date],[end date])
VALUES(1,'20190402','20200917'),(2,'20190810','20200810');
What is a SQL query that can produce these results:
id Year Quarter
-- ---- ----------
1 2019 2
1 2019 3
1 2019 4
1 2020 1
1 2020 2
1 2020 3
2 2019 3
2 2019 4
2 2020 1
2 2020 2
2 2020 3
Just use a recursive CTE. This version switches to counting quarters from year 0:
with cte as (
select id,
year(start_date) * 4 + datepart(quarter, start_date) - 1 as yyyyq,
year(end_date) * 4 + datepart(quarter, end_date) - 1 as end_yyyyq
from t
union all
select id, yyyyq + 1, end_yyyyq
from cte
where yyyyq < end_yyyyq
)
select id, yyyyq / 4 as year, (yyyyq % 4) + 1 as quarter
from cte;
Here is a db<>fiddle.
If you cannot make another reference table/etc, you can use DATEDIFF (and DATEPART) using quarters, and then some simple date arithmetic.
The logic below is simply to find, for each startdate, the first quarter and then the number of additional quarters to get to the maximum. Then do a SELECT where the additional quarters are added to the startdate, to get each quarter.
The hardest part of the query to understand imo is the WITH numberlist section - all this does is generate a series of integers between 0 and the maximum number of quarters difference. If you already have a numbers table, you can use that instead.
Key code part is below, and here's a full DB_Fiddle with some additional test data.
CREATE TABLE #yourtable (id int, startdate date, enddate date)
INSERT INTO #yourtable (id, startdate, enddate) VALUES
(1, '2019-04-02', '2020-09-17'),
(2, '2019-08-10', '2020-08-20')
; WITH number_list AS
-- list of ints from 0 to maximum number of quarters
(SELECT n
FROM (SELECT ones.n + 10*tens.n AS n
FROM (VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) ones(n),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) tens(n)
) AS a
WHERE n <= (SELECT MAX(DATEDIFF(quarter,startdate,enddate)) FROM #yourtable)
)
SELECT id,
YEAR(DATEADD(quarter, number_list.n, startdate)) AS [Year],
DATEPART(quarter, DATEADD(quarter, number_list.n, startdate)) AS [Quarter]
FROM (SELECT id, startdate, DATEDIFF(quarter,startdate,enddate) AS num_additional_quarters FROM #yourtable) yt
CROSS JOIN number_list
WHERE number_list.n <= yt.num_additional_quarters
DROP TABLE #yourtable
First create a date dimension table which contains date, corresponding quarter and year. Then use below query to get the result. Tweak column and table name according to your schema.
with q_date as
(
select 1 as id, '2019-04-02' :: date as start_date, '2020-09-17' :: date as end_date
UNION ALL
select 2 as id, '2019-08-10' :: date as start_date, '2020-08-10' :: date as end_date
)
select qd.id, dd.calendar_year, dd.calendar_quarter_number
from dim_date dd, q_date qd
where dd.date_dmk between qd.start_date and qd.end_date
group by qd.id, dd.calendar_year, dd.calendar_quarter_number
order by qd.id, dd.calendar_year, dd.calendar_quarter_number;
I have a list of dates in a table. For this examples the 1st day of each month. Let's call it table timeperiod with column endTime
endTime
1-1-2019
2-1-2019
3-1-2019
4-1-2019
I want to find all dates x number of days after each date in a list. Lets say x = 4. Then the list should be:
1-1-2019
1-2-2019
1-3-2019
1-4-2019
2-1-2019
2-2-2019
2-3-2019
2-4-2019
3-1-2019
3-2-2019
3-3-2019
3-4-2019
4-1-2019
4-2-2019
4-3-2019
4-4-2019
I have found solutions to find all dates between dates but I keep getting "Subquery returned more than 1 value" error when I try to use it with a list of dates.
Here is an example of something I tried but doesn't work
declare #days DECIMAL = 4
declare #StartDate date = (select convert(varchar, DATEADD(Day, +0, endTime),101) from timeperiod
declare #EndDate date = (select convert(varchar, DATEADD(Day, +#days, endTime),101) from timeperiod;
;WITH cte AS (
SELECT #StartDate AS myDate
UNION ALL
SELECT DATEADD(day,1,myDate) as myDate
FROM cte
WHERE DATEADD(day,1,myDate) <= #EndDate
)
SELECT myDate
FROM cte
OPTION (MAXRECURSION 0)
Here is a row generator that generates 5 rows, 0 to 4:
WITH rg AS (
SELECT 0 AS rn
UNION ALL
SELECT rg.rn + 1
FROM rg
WHERE rn < 4
)
Here we join it with your existing table that has firsts of the month and use DATEADD to add rn numbers of days (between 0 and 4) to the endPeriod. CROSS JOINing it caused the rows in timePeriod to repeat 5 times each:
SELECT
DATEADD(DAY, rg.rn, timePeriod.endTime) as fakeEndTime
FROM
rg CROSS JOIN timePeriod
I wasn't really clear when you say "days X days after the date, say x = 4" - to me if there is a day that is 1-Jan-2000, then the date 4 days after this is 5-Jan-2000
If you only want the 1,2,3 and 4 of Jan make the row generator < 3 instead of < 4
Already +1'd on Caius Jard's recursive cte.
Here is yet another option using an ad-hoc tally table in concert with a CROSS JOIN
Example
Declare #YourTable Table ([endTime] date)
Insert Into #YourTable Values
('1-1-2019')
,('2-1-2019')
,('3-1-2019')
,('4-1-2019')
Select NewDate = dateadd(DAY,N-1,EndTime)
From #YourTable A
Cross Join (
Select Top (4) N=row_number() over (order by (select null))
From master..spt_values N1
) B
Returns
NewDate
2019-01-01
2019-01-02
2019-01-03
2019-01-04
2019-02-01
2019-02-02
2019-02-03
2019-02-04
2019-03-01
2019-03-02
2019-03-03
2019-03-04
2019-04-01
2019-04-02
2019-04-03
2019-04-04
I have a table
DATE Val
01-01-2020 1
01-02-2020 3
01-05-2020 2
01-07-2020 8
01-13-2020 3
...
I want to summarize these values by the following Sunday. For example, in the above example:
1-05-2020, 1-12-2020, and 1-19-2020 are Sundays, so I want to summarize these by those dates.
The final result should be something like
DATE SUM
1-05-2020 6 //(01-01-2020 + 01-02-2020 + 01-05-2020)
1-12-2020 8
1-19-2020 3
I wasn't certain if the best place to start would be to create a temp calendar table, and then try to join backwards based on that? Or if there was an easier way involving DATEDIFF. Any help would be appreciated! Thanks!
Here's a solution that uses DATEADD & DATEPART to calculate the closest Sunday.
With a correction for a different setting of ##datefirst.
(Since the datepart weekday values are different depending on the DATEFIRST setting)
Sample data:
create table #TestTable
(
Id int identity(1,1) primary key,
[Date] date,
Val int
);
insert into #TestTable
([Date], Val)
VALUES
('2020-01-01', 1)
, ('2020-01-02', 3)
, ('2020-01-05', 2)
, ('2020-01-07', 8)
, ('2020-01-13', 3)
;
Query:
WITH CTE_DATA AS
(
SELECT [Date], Val
, DATEADD(day,
((7-(##datefirst+datepart(weekday, [Date])-1)%7)%7),
[Date]) AS Sunday
FROM #TestTable
)
SELECT
Sunday AS [Date],
SUM(Val) AS [Sum]
FROM CTE_DATA
GROUP BY Sunday
ORDER BY Sunday;
Date | Sum
:--------- | --:
2020-01-05 | 6
2020-01-12 | 8
2020-01-19 | 3
db<>fiddle here
Extra:
Apparently the trick of adding the difference of weeks from day 0 to day 6 also works independently from the DATEFIRST setting.
So this query will return the same result for the sample data.
WITH CTE_DATA AS
(
SELECT [Date], Val
, CAST(DATEADD(week, DATEDIFF(week, 0, DATEADD(day, -1, [Date])), 6) AS DATE) AS Sunday
FROM #TestTable
)
SELECT
Sunday AS [Date],
SUM(Val) AS [Sum]
FROM CTE_DATA
GROUP BY Sunday
ORDER BY Sunday;
The subtraction of 1 day makes sure that if the date is already a Sunday that it isn't calculated to the next Sunday.
Here is a way to do it:
nb:1-13-2020 wont show cuz its not a sunday
with cte as
(
select cast('01-01-2020'as Date) as Date, 1 as Val
union select '01-02-2020' , 3
union select '01-05-2020' , 2
union select '01-07-2020' , 8
)
select Date, max(dateadd(dd,number,Date)), sum(distinct Val) as SUM
from master..spt_values a inner join cte on Date <= dateadd(dd,number,Date)
where type = 'p'
and year(dateadd(dd,number,Date))=year(Date)
and DATEPART(dw,dateadd(dd,number,Date)) = 7
group by Date
Output:
Date (No column name) SUM
2020-01-01 2020-12-26 1
2020-01-02 2020-12-26 3
2020-01-05 2020-12-26 2
2020-01-07 2020-12-26 8
Here is a simple solution. Putting your values into a temporary table and viewing the results on that table:
DECLARE #dates TABLE
(
mDATE DATE,
Val INT,
Sunday DATE
)
INSERT INTO #dates (mDATE,Val) VALUES
('01-01-2020',1),('01-02-2020',3),('01-05-2020',2),('01-07-2020',8),('01-13-2020',3)
UPDATE #dates
SET Sunday = dateadd(week, datediff(week, 0, mDATE), 6)
SELECT Sunday,SUM(Val) AS Val FROM #dates
GROUP BY Sunday
OUTPUT:
Sunday Val
2020-01-05 4
2020-01-12 10
2020-01-19 3
I havea table as table_A . table_A includes these columns
-CountryName
-Min_Date
-Max_Date
-Number
I want to duplicate data with seperating by months. For example
Argentina | 2015-01-04 | 2015-04-07 | 100
England | 2015-02-08 | 2015-03-11 | 90
I want to see a table as this (Monthly seperated)
Argentina | 01-2015 | 27 //(days to end of the min_date's month)
Argentina | 02-2015 | 29 //(days full month)
Argentina | 03-2015 | 31 //(days full month)
Argentina | 04-2015 | 7 //(days from start of the max_date's month)
England | 02-2015 | 21 //(days)
England | 03-2015 | 11 //(days)
I tried too much thing to made this for each records. But now my brain is so confusing and my project is delaying.
Does anybody know how can i solve this. I tried to duplicate each rows with datediff count but it is not working
WITH cte AS (
SELECT CountryName, ISNULL(DATEDIFF(M,Min_Date ,Max_Date )+1,1) as count FROM table_A
UNION ALL
SELECT CountryName, count-1 FROM cte WHERE count>1
)
SELECT CountryName,count FROM cte
-Generate all the dates between min and max dates for each country.
-Then get the month start and month end dates for each country,year,month.
-Finally get the date differences of the month start and month end.
WITH cte AS (
SELECT Country, min_date dt,min_date,max_date FROM t
UNION ALL
SELECT Country, dateadd(dd,1,dt),min_date,max_date FROM cte WHERE dt < max_date
)
,monthends as (
SELECT country,year(dt) yr,month(dt) mth,max(dt) monthend,min(dt) monthstart
FROM cte
GROUP BY country,year(dt),month(dt))
select country
,cast(mth as varchar(2))+'-'+cast(yr as varchar(4)) yr_month
,datediff(dd,monthstart,monthend)+1 days_diff
from monthends
Sample Demo
EDIT: Another option would be to generate all the dates once (the example shown here generates 51 years of dates from 2000 to 2050) and then joining it to the table to get the days by month.
WITH cte AS (
SELECT cast('2000-01-01' as date) dt,cast('2050-12-31' as date) maxdt
UNION ALL
SELECT dateadd(dd,1,dt),maxdt FROM cte WHERE dt < maxdt
)
SELECT country,year(dt) yr,month(dt) mth, datediff(dd,min(dt),max(dt))+1 days_diff
FROM cte c
JOIN t on c.dt BETWEEN t.min_date and t.max_date
GROUP BY country,year(dt),month(dt)
OPTION (MAXRECURSION 0)
I think you have the right idea. But you need to construct the months:
WITH cte AS (
SELECT CountryName, Min_Date as dte, Min_Date, Max_Date
FROM table_A
UNION ALL
SELECT CountryName, DATEADD(month, 1, dte), Min_Date, Max_Date
FROM cte
WHERE dte < Max_date
)
SELECT CountryName, dte
FROM cte;
Getting the number of days in the month is a bit more complicated. That requires some thought.
Oh, I forgot about EOMONTH():
select countryName, dte,
(case when dte = min_date
then datediff(day, min_date, eomonth(dte)) + 1
when dte = max_date
then day(dte)
else day(eomonth(dte))
end) as days
from cte;
Using a Calendar Table makes this stuff pretty easy. RexTester: http://rextester.com/EBTIMG23993
begin
create table #enderaric (
CountryName varchar(16)
, Min_Date date
, Max_Date date
, Number int
)
insert into #enderaric values
('Argentina' ,'2015-01-04' ,'2015-04-07' ,'100')
, ('England' ,'2015-02-08' ,'2015-03-11' ,'90')
end;
-- select * from #enderaric
--*/"
declare #FromDate date;
declare #ThruDate date;
set #FromDate = '2015-01-01';
set #ThruDate = '2015-12-31';
with x as (
select top (cast(sqrt(datediff(day, #FromDate, #ThruDate)) as int) + 1)
[number]
from [master]..spt_values v
)
/* Date Range CTE */
,cal as (
select top (1+datediff(day, #FromDate, #ThruDate))
DateValue = convert(date,dateadd(day,
row_number() over (order by x.number)-1,#FromDate)
)
from x cross join x as y
order by DateValue
)
select
e.CountryName
, YearMonth = convert(char(7),left(convert(varchar(10),DateValue),7))
, [Days]=count(c.DateValue)
from #enderaric as e
inner join cal c on c.DateValue >= e.min_date
and c.DateValue <= e.max_date
group by
e.CountryName
, e.Min_Date
, e.Max_Date
, e.Number
, convert(char(7),left(convert(varchar(10),DateValue),7))
results in:
CountryName YearMonth Days
---------------- --------- -----------
Argentina 2015-01 28
Argentina 2015-02 28
Argentina 2015-03 31
Argentina 2015-04 7
England 2015-02 21
England 2015-03 11
More about calendar tables:
Aaron Bertrand - Generate a set or sequence without loops
generate-a-set-1
generate-a-set-2
generate-a-set-3
David Stein - Creating a Date Table/Dimension on SQL 2008
Michael Valentine Jones - F_TABLE_DATE
I have the following customer table:
ID | StartDate | WeekCount
1 | 01.12.2015 | 2
2 | 03.12.2015 | 4
3 | 06.06.2014 | 8
The Startdate represents the date the customer gets the first visit, WeekCount is for the next visit (every X Weeks)
I want to query the next visit dates for a timespawn.
Lets say the first visit is 03.12.2015 then I query for March 2016 so the expected date should be 03.03.2015.
So basically StartDate+WeekCount and then the Date between filter.
I think recursive CTE will help you to solve your problem.
DECLARE #to_date DATETIME
SET #to_date = N'2016.03.01'
;WITH test_data AS(
SELECT 1 AS id, CAST(N'2015.12.01' AS DATETIME) AS startDate, 2 AS weekCount
UNION ALL
SELECT 2 AS id, CAST(N'2015.12.03' AS DATETIME) AS startDate, 4 AS weekCount
UNION ALL
SELECT 3 AS id, CAST(N'2014.06.06' AS DATETIME) AS startDate, 8 AS weekCount
),
result_tbl AS(
SELECT id, startDate, weekCount FROM test_data
UNION ALL
SELECT id, DATEADD(ww, R.weekCount, R.startDate), weekCount FROM result_tbl AS R
WHERE R.startDate < #to_date
)
SELECT * FROM result_tbl
ORDER BY id
Provided the datatype is date/datetime
Select columns from your_table
where StartDate>='20160301' and StartDate<'20160401'