Question

我在sql开发人员中使用了两个表，以确定哪个客户已经下了最多的订单，直到找出每个客户下达的订单总数，但无法弄清楚如何只展示一个最多订单数量......

例如 - 这将为我提供所有下订单的客户和每个订单的订单数量列表

 SELECT 
  customer.cust_num, customer.cust_bizname,
  COUNT(invoice.inv_num) AS "TOTAL ORDERS"
FROM customer INNER JOIN invoice ON customer.cust_num = invoice.cust_num
GROUP BY customer.cust_num, customer.cust_bizname;

如果我尝试使用max和count ...

 SELECT 
      customer.cust_num, customer.cust_bizname,
      MAX(COUNT(invoice.inv_num)) AS "TOTAL ORDERS"
    FROM customer INNER JOIN invoice ON customer.cust_num = invoice.cust_num
    GROUP BY customer.cust_num, customer.cust_bizname;

我收到一条错误，指出“不是单组群组功能”。如何有效地计算订单数量并仅显示订单数量最多的客户？

Answer 1

将Order BY与ROWNUM

一起使用

select * 
from
(SELECT customer.cust_num, 
        customer.cust_bizname,
        COUNT(invoice.inv_num) AS "TOTAL ORDERS"
FROM customer 
INNER JOIN invoice 
ON customer.cust_num = invoice.cust_num
GROUP BY customer.cust_num, customer.cust_bizname
Order by "TOTAL ORDERS" DESC
)
Where ROWNUM =1

或使用Row_Number()分析函数

select customer.cust_num, 
       customer.cust_bizname,
       "TOTAL ORDERS"
FROM
(
SELECT Row_number() over(order by COUNT(invoice.inv_num) DESC) As RN
        customer.cust_num, 
        customer.cust_bizname,
        COUNT(invoice.inv_num) AS "TOTAL ORDERS"
FROM customer 
INNER JOIN invoice 
ON customer.cust_num = invoice.cust_num
GROUP BY customer.cust_num, customer.cust_bizname
)
Where RN=1

Answer 2

你不能那样使用max。相反，按降序排序并获得第一条记录，如下所示：

SELECT * FROM
(select customer.cust_num, customer.cust_bizname,
  COUNT(invoice.inv_num) AS "TOTAL ORDERS"
FROM customer INNER JOIN invoice ON customer.cust_num = invoice.cust_num
GROUP BY customer.cust_num, customer.cust_bizname
ORDER BY "TOTAL ORDERS" desc)
WHERE rownum = 1

Answer 3

在Oracle 12中，您可以使用ANSI标准fetch first 1 row only：

SELECT c.cust_num, c.cust_bizname, COUNT(i.inv_num) AS "TOTAL ORDERS"
FROM customer c INNER JOIN
     invoice i
     ON c.cust_num = i.cust_num
GROUP BY c.cust_num, c.cust_bizname
ORDER BY COUNT(i.inv_num) DESC
FETCH FIRST 1 ROW ONLY;

确定大多数订单的客户

3 个答案: