假设我有以下表RIGHTS和数据：
ID NAME OWNER_ID ACL_ID ACL_NAME
--------------------------------------------------
100 Entity_1 1 1 g1
100 Entity_1 2 2 g2
100 Entity_1 3 3 g3
200 Entity_2 1 1 g1
200 Entity_2 2 2 g2
300 Entity_3 1 1 g1
300 Entity_3 2 2 g2
300 Entity_3 4 NULL NULL
400 Entity_4 1 1 g1
400 Entity_4 2 2 g2
400 Entity_4 3 3 g3
400 Entity_4 4 NULL NULL
500 Entity_5 4 NULL NULL
500 Entity_5 5 NULL NULL
500 Entity_5 6 NULL NULL
600 Entity_6 NULL NULL NULL
如何选择除了具有OWNER_ID = NULL的那些行之外甚至没有单个ACL_ID = NULL行的所有（ID，NAME）记录。在这个特定的例子中，我想选择3行：
（100，Entity_1） - 因为所有3行都有ACL_ID！= NULL（1,2,3）
（200，Entity_2） - 因为所有2行都有ACL_ID！= NULL（1,2）
（600，Entity_6） - 因为OWNER_ID = NULL
现在我使用的是SQL Server，但我希望它能在Oracle上运行，如果可能的话。
更新 我道歉我不得不提到这个表数据只是一个带连接的查询的结果，所以必须考虑到它：
SELECT DISTINCT
EMPLOYEE.ID
,EMPLOYEE.NAME
, OWNERS.OWNER_ID as OWNER_ID
, GROUPS.GROUP_ID as ACL_ID
, GROUPS.NAME as ACL_NAME
from EMPLOYEE
inner join ENTITIES on ENTITIES.ENTITY_ID = ID
left outer join OWNERS on (OWNERS.ENTITY_ID = ID and OWNERS.OWNER_ID != 123)
left outer join GROUPS on OWNERS.OWNER_ID = GROUPS.GROUP_ID
where
ENTITIES.STATUS != 'D'
答案 0 :(得分：1)
试试这个：
select s.id, s.name
from
(select id,name,max(coalesce(owner_id,-1)) owner_id, min(coalesce(acl_id,-1)) acl_id
from yourtable
group by id,name) as s
where s.owner_id = -1
or (s.owner_id > -1 and s.acl_id > -1)
我们使用COALESCE将空值默认设置为-1（假设列是整数），然后根据唯一owner_id获取acl_id和id-name的最小值组合。如果owner_id的最大值为-1，则owner列为null。同样，如果acl_id的最小值为-1，则至少存在一个空值行。基于这两个条件，我们过滤列表以获得所需的id-name对。请注意，在这种情况下，我只选择-1作为默认值，因为我假设您不使用负数作为ID。如果这样做，您可以选择合适的“不可能”值作为COALESCE函数的默认值。
这应该适用于SQL Server和Oracle。
答案 1 :(得分：0)
这是我在Oracle上的解决方案。
SELECT DISTINCT
EMPLOYEE.ID
,EMPLOYEE.NAME
, OWNERS.OWNER_ID as OWNER_ID
, GROUPS.GROUP_ID as ACL_ID
, GROUPS.NAME as ACL_NAME
from EMPLOYEE
inner join ENTITIES on ENTITIES.ENTITY_ID = ID
left outer join OWNERS on (OWNERS.ENTITY_ID = ID and OWNERS.OWNER_ID != 123)
left outer join GROUPS on OWNERS.OWNER_ID = GROUPS.GROUP_ID
where ENTITIES.STATUS != 'D'
and EMPLOYEE.ID not in (select id from EMPLOYEE
where GROUPS.GROUP_ID is null
and OWNERS.OWNER_ID is not null);
您只需要在我之前的答案中附加内部子查询，您就会得到解决方案。
答案 2 :(得分：0)
在where子句中使用简单过滤器：
with tab(ID,NAME,OWNER_ID,ACL_ID,ACL_NAME) as (
select 100, 'Entity_1', 1,1, 'g1' from dual union all
select 100, 'Entity_1', 2,2, 'g2' from dual union all
select 100, 'Entity_1', 3,3, 'g3' from dual union all
select 200, 'Entity_2', 1,1, 'g1' from dual union all
select 200, 'Entity_2', 2,2, 'g2' from dual union all
select 300, 'Entity_3', 1,1, 'g1' from dual union all
select 300, 'Entity_3', 2,2, 'g2' from dual union all
select 300, 'Entity_3', 4,NULL, NULL from dual union all
select 400, 'Entity_4', 1,1, 'g1' from dual union all
select 400, 'Entity_4', 2,2, 'g2' from dual union all
select 400, 'Entity_4', 3,3, 'g3' from dual union all
select 400, 'Entity_4', 4,NULL,NULL from dual union all
select 500, 'Entity_5', 4,NULL,NULL from dual union all
select 500, 'Entity_5', 5,NULL,NULL from dual union all
select 500, 'Entity_5', 6,NULL,NULL from dual union all
select 600, 'Entity_6', NULL,NULL,NULL from dual)
--------------------------------
---End of data preparation here
--------------------------------
select a.id, a.name
from tab a
where ((a.ACL_ID is not null and a.ACL_NAME is not NULL) or a.OWNER_ID is null)
and not exists (select 'x'
from tab b
where b.id = a.id
and (b.ACL_ID is null or b.ACL_NAME is null)
and b.owner_id is not null)
group by a.id, a.name;
输出：
ID NAME
------------
200 Entity_2
100 Entity_1
600 Entity_6
但我仍然想知道，在有数据的情况下，你的逻辑是什么：
ID NAME OWNER_ID ACL_ID ACL_NAME
--------------------------------------------------
600 Entity_1 null null null
600 Entity_1 2 null null
____