How can we select distinct *, From and get only Distinct IDs? - sql

Question

I have a small dataset that looks like this.
SELECT *
INTO Order_Table
FROM (VALUES
(1, 456, 'repair', 'House'),
(2, 456, 'paint', 'House'),
(3, 678, 'repair', 'Fence'),
(4, 789, 'repair', 'House'),
(5, 789, 'paint', 'House'),
(6, 789, 'repair', 'Fence'),
(7, 789, 'paint', 'Fence')
)
v (OrderNum, CustomerNum, OrderDesc, Structure)
SELECT *
INTO Veg_Table
FROM (VALUES
(1, '12/01/2020'),
(2, '12/02/2020'),
(3, '12/03/2020'),
(4, '12/04/2020'),
(5, '12/05/2020'),
(6, '12/06/2020'),
(7, '12/07/2020'),
(1, '12/10/2020'),
(2, '12/11/2020'),
(3, '12/12/2020')
)
v (ID, MyDate)
I have a query that looks something like this...
Select Distinct CTE.ID, *
From (
Select *
From Order_Table as Hist
Inner Join Veg_Table As Veg
On Hist.OrderNum = Veg.ID) as CTE
How can this query be modified to give only unique IDs? I always get duplicate IDs.
I also tried: Where In (Select Distinct ID From Event_View)
That didn't work either.
I want to end up with something like this.
OrderNum CustomerNum OrderDesc Structure ID MyDate
1 456 repair House 1 12/1/2020
2 456 paint House 2 12/2/2020
3 678 repair Fence 3 12/3/2020
4 789 repair House 4 12/4/2020
5 789 paint House 5 12/5/2020
6 789 repair Fence 6 12/6/2020
7 789 paint Fence 7 12/7/2020
I suppose Row_Number() Over (Partition By ID) would do it, but I was hoping for a simpler solution using 'Distinct'.

Answer 1

Using a regular GROUP BY and MIN appears to give you what you want.
SELECT Hist.OrderNum, Hist.CustomerNum, Hist.OrderDesc, Hist.Structure, MIN(Veg.MyDate)
FROM #Order_Table AS Hist
INNER JOIN #Veg_Table AS Veg ON Hist.OrderNum = Veg.ID
GROUP BY Hist.OrderNum, Hist.CustomerNum, Hist.OrderDesc, Hist.Structure;