SQL Server-如果列包含值，则按ID分组

Question

I have following table:

ID   | NR | Status
1000 | 1  | A
1000 | 2  | A
1001 | 3  | A
1002 | 4  | A
1002 | 5  | N
1003 | 6  | N

I need to an output which groups these by ID's . The NR column can be ignored. If one of the records with those ID's contains Status A, That status will be given as result.

So my output would be:

ID   | Status
1000 | A 
1001 | A 
1002 | A 
1003 | N 

Any suggestions/ideas?

Answer 1

Do a GROUP BY , use MIN() to pick minimum status value for each id, and A < N!

select id, min(status)
from tablename
group by id

Answer 2

Although min() is the simplest method, it is not easily generalizable. Another method is:

select id
       (case when sum(case when status = 'A' then 1 else 0 end) > 0
             then 'A'
             else 'N' -- or whatever
        end) as status
from t
group by id;

Or, if you have a table with one row per id , then I would use exists :

select ids.id,
       (case when exists (select 1 from t where t.id = ids.id and t.status = 'A')
             then 'A' else 'N'
        end) as status
from ids;

This saves on the group by aggregation and can use an index on (id, status) for optimal performance.

Answer 3

You want exactly the records that match the predicate "If one of the records with those ID's contains Status A, that status will be given as result." ? The query can be written simply as:

Select distinct ID, STATUS from [your working TABLE] where STATUS = 'A'.

Hope this can help.