Comparing SQL Table to itself (Self-join)

Question

I'm trying to find duplicate rows based on mixed columns. This is an example of what I have:

CREATE TABLE Test
(
   id INT PRIMARY KEY,
   test1 varchar(124),
   test2 varchar(124)
)

INSERT INTO TEST ( id, test1, test2 ) VALUES ( 1, 'A', 'B' )
INSERT INTO TEST ( id, test1, test2 ) VALUES ( 2, 'B', 'C' )

Now if I run this query:

SELECT [LEFT].[ID] 
FROM [TEST] AS [LEFT] 
   INNER JOIN [TEST] AS [RIGHT] 
   ON [LEFT].[ID] != [RIGHT].[ID] 
WHERE [LEFT].[TEST1] = [RIGHT].[TEST2]

I would expect to get back both id's. (1 and 2), however I only ever get back the one row.

My thoughts would be that it should compare each row, but I guess this is not correct? To fix this I had changed my query to be:

SELECT [LEFT].[ID] 
FROM [TEST] AS [LEFT] 
   INNER JOIN [TEST] AS [RIGHT] 
   ON [LEFT].[ID] != [RIGHT].[ID] 
WHERE [LEFT].[TEST1] = [RIGHT].[TEST2] 
OR [LEFT].[TEST2] = [RIGHT].[TEST1]

Which gives me both rows, but the performance degrades extremely quickly based on the number of rows.

The final solution I came up for for performance and results was to use a union:

SELECT [LEFT].[ID] 
FROM [TEST] AS [LEFT] 
   INNER JOIN [TEST] AS [RIGHT] 
   ON [LEFT].[ID] != [RIGHT].[ID] 
WHERE [LEFT].[TEST1] = [RIGHT].[TEST2] 
UNION
SELECT [LEFT].[ID] 
FROM [TEST] AS [LEFT] 
   INNER JOIN [TEST] AS [RIGHT] 
   ON [LEFT].[ID] != [RIGHT].[ID] 
WHERE [LEFT].[TEST2] = [RIGHT].[TEST1]

But overall, I'm obviously missing an understanding of why this is not working which means that I'm probably doing something wrong. Could someone point me in the proper direction?

Answer 1

Do not JOIN on an inequality; it seems that the JOIN and WHERE conditions are inverted.

SELECT t1.id
FROM Test t1
INNER JOIN Test t2
ON ((t1.test1 = t2.test2) OR (t1.test2 = t2.test1))
WHERE t1.id <> t2.id

Should work fine.