SQL SELECT ID WHERE rows with the same ID have different Values - sql

Question

I need some help creating a SQL statement across rows.
SELECT SZ.Stammindex AS ID, S.sEbene1, S.sEbene2, S.sEbene3
FROM SuchbaumZuordnung SZ
LEFT JOIN Suchbaum S
ON SZ.gSuchbaumID = S.gID
WHERE (S.sEbene1 IN ('Test1')
AND (S.sEbene2 IN ('Test2') OR S.sEbene2 IS NULL)
AND S.sEbene3 IS NULL)
As you can see in the screenshot, I selected ID=10004 and ID=10005. But actually I only want ID=10005 to show up. I am trying to filter across Rows as already mentioned.
My goal is to get all the IDs, where all the conditions are connected with "AND", something like this:
WHERE (sEbene1 IN ('Test1')
AND (sEbene2 IN ('Test2') *AND* sEbene2 IS NULL)
AND sEbene3 IS NULL)
But this will return nothing.
Edit
I hope you guys can help me.

Answer 1

I suspect that you want:
SELECT SZ.Stammindex AS ID
FROM SuchbaumZuordnung SZ
WHERE EXISTS (SELECT 1
FROM Suchbaum S
WHERE SZ.gSuchbaumID = S.gID AND
S.sEbene1 IN ('Test1') AND
sEbene2 IN ('Test2')
) AND
EXISTS (SELECT 1
FROM Suchbaum S
WHERE SZ.gSuchbaumID = S.gID AND
S.sEbene2 IS NULL AND
S.sEbene3 IS NULL
);
This is looking for two different rows in Suchbaum, each one matching one of the conditions.

Answer 2

Considering you only have 3 columns you want to check different rows, it seems like this would be easily serviced with a CTE and a Windowed COUNT:
WITH CTE AS(
SELECT SZ.Stammindex AS ID,
S.sEbene1, --Guessed the table alias
S.sEbene2, --Guessed the table alias
S.sEbene3, --Guessed the table alias
COUNT(DISTINCT CONCAT(ISNULL(S.S.sEbene1,'-'),ISNULL(S.sEbene2,'-'),ISNULL(S.sEbene3,'-'))) OVER (PARTITION BY SZ.Stammindex) AS DistinctRows
FROM SuchbaumZuordnung SZ
LEFT JOIN Suchbaum S ON SZ.gSuchbaumID = S.gID) --This was missing the ON in your sample
SELECT C.Stammindex,
C.sEbene1,
C.sEbene2,
C.sEbene3
FROM CTE C
WHERE C.DistinctRows > 1;
If it's purely where an ID has more than 1 rows (which could be identical) then you can just use COUNT:
WITH CTE AS(
SELECT SZ.Stammindex AS ID,
S.sEbene1, --Guessed the table alias
S.sEbene2, --Guessed the table alias
S.sEbene3, --Guessed the table alias
COUNT(*) OVER (PARTITION BY SZ.Stammindex) AS [Rows]
FROM SuchbaumZuordnung SZ
LEFT JOIN Suchbaum S ON SZ.gSuchbaumID = S.gID)
SELECT C.Stammindex,
C.sEbene1,
C.sEbene2,
C.sEbene3
FROM CTE C
WHERE C.[Rows] > 1;