SQL Server Select the most recent past date if no future date available - sql

Question

I have a table structure as below,
CREATE TABLE #CustOrder ( CustId INT, OrderDate DATE )
INSERT #CustOrder ( CustId, OrderDate )
VALUES ( 1, '2016-11-01' ),
( 1, '2019-09-01' ),
( 2, '2019-07-01' ),
( 2, '2019-11-01' ),
( 3, '2017-01-01' ),
( 4, '2016-12-01' ),
( 4, '2017-01-01' )
I want to list the customer with their future order dates, if they do not have a future order I want to list their last or most recent order. I have the following query.
; WITH LastOrder AS
(
SELECT
CO.CustId,
CO.OrderDate,
ROW_NUMBER() OVER(PARTITION BY CO.CustId ORDER BY ABS(DATEDIFF(DAY, CO.OrderDate, GETUTCDATE()))) AS RowNum
FROM #CustOrder AS CO
)
SELECT LO.CustId, LO.OrderDate
FROM LastOrder AS LO
WHERE LO.RowNum = 1
This query gives me the result as,
CustId | OrderDate
--------+-------------
1 | 2016-11-01
2 | 2019-07-01
3 | 2017-01-01
4 | 2017-01-01
However, I need the result as,
CustId | OrderDate
--------+-------------
1 | 2019-09-01
2 | 2019-07-01
3 | 2017-01-01
4 | 2017-01-01
As
Customer 1 has a future order on 2019-09-01
Customer 2 has two future order but the first one is on 2019-07-01
Customer 3 has no more than 1 order, it should just return 2017-01-01
Customer 4 has two past orders but the most recent is 2017-01-01

Answer 1

rextester: http://rextester.com/PBKNA95127
CREATE TABLE #CustOrder ( CustId INT, OrderDate DATE )
INSERT #CustOrder ( CustId, OrderDate )
VALUES ( 1, '2016-11-01' ),
( 1, '2019-09-01' ),
( 2, '2019-07-01' ),
( 2, '2019-11-01' ),
( 3, '2017-01-01' ),
( 4, '2016-12-01' ),
( 4, '2017-01-01' )
; WITH LastOrder AS
(
SELECT
CO.CustId,
CO.OrderDate,
ROW_NUMBER() OVER(PARTITION BY CO.CustId
ORDER BY case when co.OrderDate > getdate() then 0 else 1 end
, abs(DATEDIFF(DAY, getdate(),CO.OrderDate)) asc
) AS RowNum
FROM #CustOrder AS CO
)
SELECT LO.CustId, LO.OrderDate
FROM LastOrder AS LO
WHERE LO.RowNum = 1
results:
+--------+------------+
| CustId | OrderDate |
+--------+------------+
| 1 | 2019-09-01 |
| 2 | 2019-07-01 |
| 3 | 2017-01-01 |
| 4 | 2017-01-01 |
+--------+------------+

Answer 2

You can use the MAX function to check if the latest date is in the future. If so, get the MIN date after today using MIN. Else get the latest date.
SELECT CUSTID,OrderDate
FROM (SELECT CustId,
OrderDate,
CASE WHEN MAX(orderdate) OVER(PARTITION BY CustId) > GETUTCDATE()
THEN MIN(case when orderdate >getutcdate() then orderdate end) OVER(PARTITION BY CustId)
ELSE MAX(orderdate) OVER(PARTITION BY CustId) end as latest_date
FROM #CustOrder) T
WHERE latest_date=orderDate

Answer 3

Min, Max, UNION approach
select custID, MIN(OrderDate)
from #CustOrder
where OrderDate > '2017-02-17'
group by custID
union all
select co1.custID, max(co1.OrderDate)
from #CustOrder co1
where not exists ( select 1
from #CustOrder co2
where co2.CustId = co1.CustId
and co2.OrderDate > '2017-02-17'
)
group by co1.custID

Answer 4

Start your ORDER BY with a CASE expression that prefers future over past, and then use the ABS DATEDIFF (like you have now) as the second condition in the ORDER BY.

Answer 5

Maybe create another column and use the LAG() window function to grab the last date function and then put a conditional/case statement within the select portion? https://msdn.microsoft.com/en-us/library/hh231256.aspx