SQL How to filter table with values having more than one unique value of another column

Question

I have data table Customers that looks like this:

ID | Sequence No |
1  |  1          |
1  |  2          |
1  |  3          |
2  |  1          |
2  |  1          |
2  |  1          |
3  |  1          |
3  |  2          |

I would like to filter the table so that only IDs with more than 1 distinct count of Sequence No remain.

Expected output:

ID | Sequence No |
1  |  1          |
1  |  2          |
1  |  3          |
3  |  1          |
3  |  2          |

I tried

select ID, Sequence No 
from Customers 
where count(distinct Sequence No) > 1
order by ID

but I'm getting error. How to solve this?

Answer 1

Using analytic functions, we can try:

WITH cte AS (
    SELECT *, MIN([Sequence No]) OVER (PARTITION BY ID) min_seq,
              MAX([Sequence No]) OVER (PARTITION BY ID) max_seq
    FROM Customers
)

SELECT ID, [Sequence No]
FROM cte
WHERE min_seq <> max_seq
ORDER BY ID, [Sequence No];

Demo

We are checking for a distinct count of sequence number by asserting that the minimum and maximum sequence numbers are not the same for a given ID . The above query could benefit from the following index:

CREATE INDEX idx ON Customers (ID, [Sequence No]);

This would let the min and max values be looked up faster.

Answer 2

You can get the desired result by using the below query. This is similar to what you were trying -

Sample Table & Data

Declare @Data table
(Id int,  [Sequence No] int)

Insert into @Data
values
(1  ,  1   ),
(1  ,  2   ),
(1  ,  3   ),
(2  ,  1   ),
(2  ,  1   ),
(2  ,  1   ),
(3  ,  1   ),
(3  ,  2   )

Query

Select * from @Data 
where ID in(
            select ID
            from @Data 
            Group by ID
            Having count(distinct [Sequence No]) > 1
            )