How do I get the following query give me the same date for the last N years?
declare @step int = 0;
declare @datobserve date = '2021-11-03';
with dates as
(
select dateadd(year, @step, @datobserve) datobserve, @step step
union all
select dateadd(year, step, datobserve) as datobserve, step - 1
from dates
where 1=1
--and step = step 1
and step > -4
)
select * from dates
The result I am getting is:
instead of:
CodePudding user response:
You need to make two changes in the recursive member of your statement:
- Use
@datobserve
instead ofdatobserve
- Use
step - 1
instead ofstep
declare @step int = 0;
declare @datobserve date = '2021-11-03';
with dates as
(
select dateadd(year, @step, @datobserve) datobserve, @step step
union all
select dateadd(year, step - 1, @datobserve) as datobserve, step - 1
from dates
where step > -4
)
select *
from dates
Result:
datobserve step
2021-11-03 0
2020-11-03 -1
2019-11-03 -2
2018-11-03 -3
2017-11-03 -4
CodePudding user response:
Just another option using an ad-hoc tally table
Example
Declare @Years int = 4
Declare @Date date ='2021-11-03'
Select dateobserve = dateadd(year,N,@Date)
,Step = N
From ( Select Top (@Years 1) N=1-Row_Number() Over (Order By (Select NULL))
From master..spt_values n1
) NT
Results
dateobserve Step
2021-11-03 0
2020-11-03 -1
2019-11-03 -2
2018-11-03 -3
2017-11-03 -4
CodePudding user response:
When you know it's only a handful of rows, like 5, I find it simpler to not bother with recursion:
DECLARE @datobserve date = '20211103';
;WITH n(n) AS
(
SELECT n FROM (VALUES(1),(2),(3),(4),(5)) AS n(n)
)
SELECT datobserve = DATEADD(YEAR, 1-n, @datobserve), step = 1-n
FROM n
ORDER BY datobserve DESC;
If the list is larger or variable, I still like to get the numbers part recursively, and work the dates around that output:
DECLARE @steps int = 5,
@datobserve date = '20211103';
;WITH n(n) AS
(
SELECT 1 UNION ALL SELECT n 1 FROM n WHERE n < @steps
)
SELECT datobserve = DATEADD(YEAR, 1-n, @datobserve), step = 1-n
FROM n
ORDER BY datobserve DESC;
- Examples @ db<>fiddle