选择基于一个字段关联与另一个字段中的多个值

Question

我需要开发一个脚本，将捕获所有领域时，一辆车被绑定到多种颜色。

如果一辆车不止一次绑定一种颜色，只有当该汽车被绑上其他颜色时才需要捕获。

如果一辆车多次绑定一种颜色，并且不需要捕获其他颜色。

{CREATE TABLE test2 
(
    ID  NUMBER(9), 
    CAR NUMBER(9), 
    COLOR NUMBER(9) 
); 


Insert into test2 (ID, CAR, COLOR) Values (1, 5, 10); 
Insert into test2 (ID, CAR, COLOR) Values (2, 5, 11); 
Insert into test2 (ID, CAR, COLOR) Values (3, 5, 10); 
Insert into test2 (ID, CAR, COLOR) Values (4, 9, 6); 
Insert into test2 (ID, CAR, COLOR) Values (5, 9, 6); 
Insert into test2 (ID, CAR, COLOR) Values (6, 8, 4); 
Insert into test2 (ID, CAR, COLOR) Values (7, 8, 9); 
Insert into test2 (ID, CAR, COLOR) Values (8, 12, 9); 
COMMIT;} 



--expected results 
    ID   CAR   COLOR 
    1   5    10 
    2   5    11 
    3   5    10 
    6   8    4 
    7   8    4 

所有见解和建议深表赞赏。

Answer 1

我会用任何一个in条款或相关exists条款。后者应履行好于前者：

select id, car, color from test2 
where car in (
    select car 
    from test2 
    group by car 
    having count(distinct color) > 1 
) 

select id, car, color from test2 t 
where exists (
    select car 
    from test2 
    where car = t.car 
    group by car 
    having count(distinct color) > 1 
) 

Sample SQL Fiddle

Answer 2

您需要执行count两次：

with cte as 
( select CAR,COLOR,count(*) cn 
    from test2 
    group by CAR,COLOR 
) 
select t.id,t.car,t.color 
from test2 t 
join( 
    select car,count(*) 
    from cte 
    group by CAR 
    having count(*)>1)q 
on t.car=q.car 
order by 1 

OUTPUT：

ID CAR COLOR 
1 5 10 
2 5 11 
3 5 10 
6 8 4 
7 8 9 

Answer 3

我认为你需要，除了那些所有的车都只有一个不同的颜色：
（替代其他的答案，但很简单）

select * 
from test2 
where not car in (select car from test2 group by car having count(distinct color = 1))