SQL查询，仅查找具有2个源值的客户ID

我有两个表，一个存储客户id，另一个存储客户id以及使用该客户信息的不同来源的信息。例子： 表A

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，返回只有AA和AB作为源的记录（没有其他源）

我已编写了以下查询，但无法正常工作：

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;

---

您可以使用聚合：

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联接

如果只需要这两个源，则需要调整逻辑：

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;

---

您可以使用聚合：

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联接

如果只需要这两个源，则需要调整逻辑：

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;

---

一个相当有效的解决方案是两个

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'))

如果索引位于

b（客户id，源代码）

，则应该可以快速运行

另一种选择是侵害：

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

---

一个相当有效的解决方案是两个

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'))

如果索引位于

b（客户id，源代码）

，则应该可以快速运行

另一种选择是侵害：

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

---

这假设客户id/源组合没有重复项

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

在联接之前进行聚合可能更有效：

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

---

这假设客户id/源组合没有重复项

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

在联接之前进行聚合可能更有效：

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

---

此解决方案是否确保除“AA”和“AB”之外不存在其他源？第一个源为AA，第二个源为AC的记录也不会返回任何其他源。@AlokMathur。我错过了要求的这一部分，但更新了答案。此解决方案是否确保除“AA”和“AB”之外不存在其他来源？第一个来源为AA，第二个来源为AC的记录，也不会返回任何其他来源。@AlokMathur。我错过了这部分的要求，但更新了答案。谢谢，这应该有效，存在和不存在的组合。我将学习更多关于这方面的知识。谢谢，这应该是有效的，存在和不存在的组合。我将在这方面了解更多。