SQL Select distinct rule out a row based on value

Question

I have a bit of a challenge right now. I have the following table:

 | ID | STATUS   | 
   12   COMPLETE 
   12   FAIL  
   11   COMPLETE

What I need to select is the distinct ID where the status is complete. If the status is "Fail", I don't want to see that ID returned to me.

I have tried this:

 select distinct id where status = 'COMPLETE'

This returns to me however all rows since the status for id 12 is complete on one row. I want it not to return that row since 12 also has a fail (I only want to return the row where all statuses for a given ID is "COMPLETE").

Does anybody have any pointers on this? Any help is greatly appreciated!

Best regards, Prince of Sweden

Answer 1

Just add a validation to make sure all rows have status "COMPLETE"

 SELECT id 
 FROM status      
 GROUP BY id
 HAVING count(id) = count( CASE WHEN status = 'COMPLETE' THEN 1 END )

Answer 2

i would try:

SELECT id 
FROM status 
WHERE status = 'COMPLETE' 
GROUP BY id

Answer 3

This would work. The idea is to select the rows where status = complete as first phase. Then second phase is to validate if there is another status for the same ID in the table, where it is <> than complete. That means that it will only return the iteration where there is only one status of complete.

declare @table table (id integer, status varchar(max))
insert into @table select 12,'complete'
insert into @table select 12,'fail'
insert into @table select 11,'complete'

select ID, status
from @table t1
where not exists (select 1 from @table t2 where t2.id =t1.id and t1.status <> t2.status)
and t1.status = 'complete'