SQL Server 多连接返回重复值

Question

I have 4 tables under my database TEST:

1. USER
2. USER_GROUP_PERMISSION
3. PERMISSION_TYPE
4. GROUPS

when I join the above tables using inner join and cross join instead of returning 520 rows it returns 2600 rows, most of the values duplicated

I need a final output without duplicates

Query:

select user.*
from TEST.USER user 
    inner join TEST.USER_GROUP_PERMISSION user_grp1
        on user.APPLICATION_ID = user_grp1.APPLICATION_ID 
    inner join TEST.PERMISSION_TYPE permission2_
        on user_grp1.PERMISSION_TYPE_ID = permission2_.PERMISSION_TYPE_ID
    inner join TEST.GROUPS groups
        on user_grp1.GROUP_ID = groups.GROUP_ID
    cross join TEST.PERMISSION_TYPE permission4_ 
where user_grp1.PERMISSION_TYPE_ID = permission4_.PERMISSION_TYPE_ID
    and groups.GROUP_ID in (101)
    and permission4_.PERMISSION_TYPE in (0 , 1 , 2 , 3 , 4 , 5 , 6)
    and user.NAME = 'ROBIN'
    and user.ACTIVE = '1'
order by upper(user.DISPLAY_VERSION) asc

Answer 1

If you want to show users without duplicates, then select from users only. Don't join! This is what the query should look like:

select * from users where ...

Your query is hard to understand, especially with the permission type joined twice, once even with an inner join disguised as a cross join. It looks like you want the user ROBIN provided he is active and his application has a group permission for group 101 and type 1 to 6. This would be:

select *
from test.user
where name = 'ROBIN'
and active = '1'
and application_id in
(
  select application_id
  from test.user_group_permission
  where group_id = 101
  and permission_type_id in 
  (
    select permission_type_id
    from test.permission_type
    where permission_type in (0, 1, 2, 3, 4, 5, 6)
  )
)
order by upper(display_version);

(You can also join permission_type to user_group_permission , if you like that better than IN .)

Answer 2

I would phrase the query using exists :

select user.*
from TEST.USER user 
where user.NAME = 'ROBIN'
      user.ACTIVE = '1' and
      exists (select 1
              from TEST.USER_GROUP_PERMISSION user_grp1 join
                   TEST.PERMISSION_TYPE permission2_
                   on user_grp1.PERMISSION_TYPE_ID = permission2_.PERMISSION_TYPE_ID join
                   TEST.GROUPS groups
                   on user_grp1.GROUP_ID = groups.GROUP_ID join 
                   TEST.PERMISSION_TYPE permission4_ 
                   on user_grp1.PERMISSION_TYPE_ID = permission4_.PERMISSION_TYPE_ID
              where user.APPLICATION_ID = user_grp1.APPLICATION_ID 
                    groups.GROUP_ID in (101) and
                    permission4_.PERMISSION_TYPE in (0 , 1 , 2 , 3 , 4 , 5 , 6)
       )
order by upper(user.DISPLAY_VERSION) asc

Then, the most important index for this is on users(name, active, upper(DISPLAY_VERSION)) .