Question

我有一些SQL语句的问题。我需要在以下设置中为每个DeliveryDate找到下一个Customer。

表格

Customer (id)
DeliveryOrder (id, deliveryDate)
DeliveryOrderCustomer (customerId, deliveryOrderId)

每个Customer可能在同一DeliveryOrders上有多个deliveryDate。我只是弄清楚如何只为每个客户获得一个deliveryDate。该日期应该是今天之后的下一个DeliveryDate。我觉得我需要某种“为每个人”，但我不知道如何在SQL中解决它。

Answer 1

这将使用subselect给出预期结果。考虑到 current_date 可能是特定于rdbms的，它适用于Oracle。

select c.id, o.date 
from customer c 
    inner join deliveryordercustomer co o on co.customerId = c.id 
    inner join deliveryorder o on co.deliveryOrderId = o.id 
where o.date = 
     (select min(o2.date) 
      from deliveryorder o2 
      where o2.id = co.deliveryOrderId and o2.date > current_date)

Answer 2

你需要使用一个分组。有很多方法可以做到这一点，这是我的解决方案，在同一天为客户考虑多个订单，并允许您查询不同的交付时段，第一，第二等。这假设Sql Server 2005及更高版本。

;with CustomerDeliveries as
(
  Select c.id, do.deliveryDate, Rank() 
               over (Partition BY c.id order by do.deliveryDate) as DeliverySlot
  From   Customer c
         inner join DeliveryOrderCustomer doc on c.id = doc.customerId
         inner join DeliveryOrder do on do.id = doc.deliveryOrderId
  Where  do.deliveryDate>GETDATE()
  Group By c.id, do.deliveryDate
)
Select id, deliveryDate
From   CustomerDeliveries
Where  DeliverySlot = 1

Answer 3

另一个更简单的版本

select c.id, min(o.date)
from customer c 
    inner join deliveryordercustomer co o on co.customerId = c.id 
    inner join deliveryorder o on co.deliveryOrderId = o.id and o.date>getdate()
group by c.id

SQL查找每个客户的下一个日期，每个SQL？

3 个答案: