Question

我想查询列出花费比普通猫要长的猫？我有五张桌子：

Animal，Sale，AnimalOrderItem，AnimalOrder和SaleAnimal

动物表：AnimalID，名称，类别   （猫，狗，鱼）

SaleAnimal表：SaleID，AnimalID，   SalePrice

销售表：SaleID，日期，employeeID，   客户id

AnimalOrderItem表：OrderID，   AnimalID，费用

AnimalOrder：OrderID，OrderDate，   ReceivingDate，        SupplierID，ShippingCost，EmployeeID

还有其他表我认为它们对查询没有影响。

我想到了以下内容......查询计算所有前期销售天数：

[SaleDate]-[ReceiveDate] AS DaysToSell
Have the INNER JOIN built:
Sale INNER JOIN ((AnimalOrder INNER JOIN (Animal INNER JOIN AnimalOrderItem
ON Animal.AnimalID = AnimalOrderItem.AnimalID) ON AnimalOrder.
OrderID = AnimalOrderItem.OrderID) INNER JOIN SaleAnimal ON Animal.
AnimalID = SaleAnimal.AnimalID) ON Sale.SaleID = SaleAnimal.SaleID

根据上述查询

创建另一个查询

SELECT AnimalID, Name, Category, DaysToSell
WHERE Category="Cat" AND DaysToSell>
(SELECT Avg(DaysToSell)
FROM the earlier query
WHERE Category="Cat"
ORDER BY DaysToSell DESC;

运行查询后，我收到错误

ORA-00921：SQL意外结束命令

任何建议！请

Answer 1

查询可以与子查询结合使用。例如，

select *
from (
    select * 
    from mytable
) subquery

将此模式应用于您的问题似乎相当简单。

Answer 2

我没有看到与select avg匹配的闭括号

Answer 3

好的，我想出了这个：

    SELECT AnimalID, Name, Category,
           [SaleDate]-[ReceiveDate] AS DaysToSell
    FROM   Sale INNER JOIN ((AnimalOrder INNER JOIN (Animal INNER JOIN AnimalOrderItem ON Animal.AnimalID = AnimalOrderItem.AnimalID) ON AnimalOrder.OrderID = AnimalOrderItem.OrderID)
           INNER JOIN SaleAnimal ON Animal.AnimalID = SaleAnimal.AnimalID) ON Sale.SaleID = SaleAnimal.SaleID
    WHERE  Category = "Cat"
    AND    ([SaleDate]-[ReceiveDate]) > (SELECT AVG([SaleDate]-[ReceiveDate])
                                         FROM   Sale INNER JOIN ((AnimalOrder INNER JOIN (Animal INNER JOIN AnimalOrderItem ON Animal.AnimalID = AnimalOrderItem.AnimalID) ON AnimalOrder.OrderID = AnimalOrderItem.OrderID)
                                         INNER JOIN SaleAnimal ON Animal.AnimalID =SaleAnimal.AnimalID) ON Sale.SaleID = SaleAnimal.SaleID
                                         WHERE Category = "Cat")
    ORDER BY ([SaleDate]-[ReceiveDate]) DESC;

组合多个SQL查询

3 个答案: