SQL 查询仅查找具有 2 个源值的客户 ID

Question

我有 2 个表，一个存储客户 ID，另一个存储客户 ID 以及有关使用该客户信息的不同来源的信息。例子： 表A

Customer Id
1
2
3
..

表 B

Customer Id Source
1            'AA'
2            'AA'
1            'AB'
2            'AB'
2            'AC'
3            'AA'
3            'AB'
3            'AE'
4            'AA'
4            'AB'

我想编写一个 SQL 查询，它返回只有 AA 和 AB 作为来源（没有其他来源）的记录

我已经编写了以下查询，但它不能正常工作：

select a.customer_id
  from A a, B b
 where a.customer_id = b.customer_id
   and b.source IN ('AA','AB')
 group by a.customer_id
having count(*) = 2;

Answer 1

一个相当有效的解决方案是几个exists 子查询：

select a.*
from a
where
    exists(select 1 from b where b.customer_id = a.customer_id and b.source = 'AA')
    and exists(select 1 from b where b.customer_id = a.customer_id and b.source = 'AB')
    and not exists(select 1 from b where b.customer_id = a.customer_id and b.source not in ('AA', 'AB'))

在b(customer_id, source) 上有一个索引，这应该会运行得很快。

另一种选择是聚合：

select customer_id
from b
group by customer_id
having
    max(case when source = 'AA' then 1 else 0 end) = 1
    and max(case when source = 'AB' then 1 else 0 end) = 1
    and max(case when source not in ('AA', 'AB') then 1 else 0 end) = 0

Answer 2

这假设 customer_id/source 组合没有重复项

select a.customer_id
  from A a join B b
    on a.customer_id = b.customer_id
 group by a.customer_id
-- both 'AA' and 'AB', but no other
having sum(case when b.source IN ('AA','AB') then 1 else -1 end) = 2

在加入之前聚合可能更有效：

select a.customer_id
from A a join 
  ( select customer_id
    from B b
    group by customer_id
    -- both 'AA' and 'AB', but no other
    having sum(case when source IN ('AA','AB') then 1 else -1 end) = 2
  ) b
on a.customer_id = b.customer_id

Answer 3

你可以使用聚合：

select b.customer_id
from b
where b.source in ('AA', 'AB')
group by b.customer_id
having count(distinct b.source) = 2;

也就是说，您的版本应该可以工作。但是，您应该学会使用正确、明确、标准、可读的JOIN 语法。但是，在这种情况下不需要连接。

如果你想要只这两个来源，你需要调整逻辑：

select b.customer_id
from b
group by b.customer_id
having sum(case when b.source = 'AA' then 1 else 0 end) > 0 and  -- has AA
       sum(case when b.source = 'AB' then 1 else 0 end) > 0 and  -- has AB
       count(distinct b.source) = 2;