SQL 仅返回具有特定状态的特定行

Question

下面这张表有两列

Order_No     Order_Status
A            Receiving
A            Active
A            Retired
A            Ordering
B            Receiving
B            Ordering 
C            Active
C            Retired
D            Receiving
E            Ordering 

我想获取 Order_no B、D 和 E 记录，因为它的订单状态是（接收/订购）。它应该过滤掉 A 和 C，因为它们都有 Active 和 Retired 状态。

我尝试了以下查询，但没有显示结果。

select ORDER_NUMBER
from table ror
where ror.use_Status 
Order_Status not in ('Active', 'Retired')
and Order_Status in ('Receiving', 'Ordering').

谁能告诉我我做错了什么或者我缺少任何连接？

Answer 1

您可以使用group by 和having：

select order_no
from mytable
group by order_no
having max(case when status = 'Receiving'            then 1 else 0 end) = 1
   and max(case when status = 'Ordering'             then 1 else 0 end) = 1
   and max(case when status not in ('Receiving', 'Ordering') then 1 else 0 end) = 0

这句话是：获取所有同时具有“接收”和“订购”状态的订单，并且没有其他状态。

如果一个给定的订单不能有两次相同的状态，那么having可以简化一点：

having sum(case when status     in ('Receiving', 'Ordering') then 1 else 0 end) = 2
   and sum(case when status not in ('Receiving', 'Ordering') then 1 else 0 end) = 0

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编辑 - 如果您希望订单具有任一“接收”和“订购”状态（不一定两者），那么条件充分：

having max(case when status not in ('Receiving', 'Ordering') then 1 else 0 end) = 0