我想创建一个 SQL 查询来列出所有订单数量高于平均水平的客户答案

【问题标题】：I wanted to create a SQL query to list of all customers who have placed an above average number of orders我想创建一个 SQL 查询来列出所有订单数量高于平均水平的客户
【发布时间】：2016-12-04 05:44:11
【问题描述】：

SQL 查询列出所有下单数量高于平均水平的客户。

订单详情显示在 NW_orders 表中，客户信息显示在 NW_Customers 表中。

首先，我计算了所有客户的平均订单数。然后我只想拉下订单大于平均订单数的客户。

我的查询：

SELECT 
    C.customerid, C.companyname, COUNT(O.orderid) AS cnt
FROM 
    NW_customers C
LEFT JOIN 
    NW_orders O ON O.customerID = C.Customerid 
GROUP BY 
    C.customerid 
HAVING 
   cnt > (SELECT COUNT(O.OrderID) / COUNT(DISTINCT(c.customerid)) AS Avg
          FROM NW_orders O
          LEFT JOIN NW_customers C ON O.customerID = C.Customerid)

我收到一个错误

ORA-00904：“CNT”：无效标识符

任何人都可以帮助纠正错误吗？

【问题讨论】：

标签： sql oracle join

【解决方案1】：

使用公用表表达式：

WITH cte AS (
    SELECT o.customerid, COUNT(o.orderid) AS cnt
    FROM NW_orders o
    GROUP BY o.customerid
)

SELECT t.customerid
FROM cte t
WHERE t.cnt > (SELECT AVG(cnt) FROM cte)

如果要引入实际的客户信息，可以在上面的查询中加入join：

SELECT t1.*, t2.*
FROM cte t1
INNER JOIN NW_customers t2
    ON t1.customerid = t2.customerid
WHERE t1.cnt > (SELECT AVG(cnt) FROM cte)

【讨论】：

Tim，撇开不使用分析函数不谈，如果您只想检索customerid，则根本不需要NW_customers，因为只有有订单的客户可能会下达高于平均水平的订单订单数量。

【解决方案2】：

这是解析函数的经典之作。

客户标识

select      customerID

from       (select      customerID
                       ,count(*)                as customer_orders
                       ,avg  (count(*)) over () as avg_customer_orders

            from        NW_orders

            group by    customerID
            )

where       customer_orders > avg_customer_orders
;

完整的客户信息

select      *

from        NW_customers

where       customerID in
            (
                select      customerID

                from       (select      customerID
                                       ,count(*)                as customer_orders
                                       ,avg  (count(*)) over () as avg_customer_orders

                            from        NW_orders

                            group by    customerID
                            )

                where       customer_orders > avg_customer_orders
            )
;

完整的客户信息+订单信息

select      o.customer_orders
           ,o.avg_customer_orders
           ,c.*

from                    NW_customers    c

            join        (select     customerID
                                   ,count(*)                as customer_orders
                                   ,avg  (count(*)) over () as avg_customer_orders

                        from        NW_orders

                        group by    customerID
                        ) o

            on          o.customerID    =
                        c.customerID


where       o.customer_orders > o.avg_customer_orders
;

【讨论】：