Mysql 如何计算每个城市最大订单的顶级客户

我创建了表名

customers

orders

和

orderdetails

，如下所示

CREATE TABLE `customers` (
  `customerNumber` int(11) NOT NULL,
  `customerName` varchar(50) NOT NULL,
  `contactLastName` varchar(50) NOT NULL,
  `contactFirstName` varchar(50) NOT NULL,
  `phone` varchar(50) NOT NULL,
  `addressLine1` varchar(50) NOT NULL,
  `addressLine2` varchar(50) default NULL,
  `city` varchar(50) NOT NULL,
  `state` varchar(50) default NULL,
  `postalCode` varchar(15) default NULL,
  `country` varchar(50) NOT NULL,
  `salesRepEmployeeNumber` int(11) default NULL,
  `creditLimit` double default NULL,
  PRIMARY KEY  (`customerNumber`)
)

CREATE TABLE `orders` (
  `orderNumber` int(11) NOT NULL,
  `orderDate` datetime NOT NULL,
  `requiredDate` datetime NOT NULL,
  `shippedDate` datetime default NULL,
  `status` varchar(15) NOT NULL,
  `comments` text,
  `customerNumber` int(11) NOT NULL,
  PRIMARY KEY  (`orderNumber`)
) 

CREATE TABLE `orderdetails` (
  `orderNumber` int(11) NOT NULL,
  `productCode` varchar(15) NOT NULL,
  `quantityOrdered` int(11) NOT NULL,
  `priceEach` double NOT NULL,
  `orderLineNumber` smallint(6) NOT NULL,
  PRIMARY KEY  (`orderNumber`,`productCode`)
)

之后，我必须找出每个城市中订单最多的客户。我已经计算了每个客户的总订单，如下所示，但我不知道如何从每个订单最多的城市中选择订单最多的客户。 所以问题是我怎样才能做到这一点

SELECT o.customerNumber, ROUND( SUM( od.priceEach * od.quantityOrdered ) , 2 ) AS totalOrder
FROM orders AS o
INNER JOIN orderDetails AS od ON o.orderNumber = od.orderNumber
GROUP BY o.customerNumber
ORDER BY totalOrder DESC 

我正在使用mysql。

试试这个：

Select tempTable.custNum, MAX(tempTable.totalOrder), customers.city
from
(
SELECT 
o.customerNumber as custNum, 
ROUND( SUM( od.priceEach * od.quantityOrdered ) , 2 ) AS totalOrder,
city
FROM orders AS o
INNER JOIN orderDetails AS od ON o.orderNumber = od.orderNumber
GROUP BY o.customerNumber
ORDER BY totalOrder DESC) as tempTable
 inner join customers on (tempTable.custNum = customers.customerNumber)
group by tempTable.city

您可以按照建议执行此操作，但需要对该答案进行一些编辑

SELECT x.customerName, MAX( p.totalOrder ) , x.city
FROM (

SELECT o.customerNumber, ROUND( SUM( od.priceEach * od.quantityOrdered ) , 2 ) AS totalOrder
FROM orders AS o
INNER JOIN orderDetails AS od ON o.orderNumber = od.orderNumber
GROUP BY o.customerNumber
ORDER BY totalOrder DESC
)p
INNER JOIN customers x ON p.customerNumber = x.customerNumber
GROUP BY city

---

phpmyadmin说，

orders

和

orderDetails

表上没有

city

字段我想你可以与客户的表进行连接，请不要指望我们会提供端到端的解决方案，顺便说一句，已经更新了查询，现在检查我还听说了

分区

。。。我们是否可以使用

分区
来解决这个问题。如果我们可以解决，那么对于大数据表，哪种方法在时间方面是有效的。。。
SELECT x.customerName, MAX( p.totalOrder ) , x.city
FROM (

SELECT o.customerNumber, ROUND( SUM( od.priceEach * od.quantityOrdered ) , 2 ) AS totalOrder
FROM orders AS o
INNER JOIN orderDetails AS od ON o.orderNumber = od.orderNumber
GROUP BY o.customerNumber
ORDER BY totalOrder DESC
)p
INNER JOIN customers x ON p.customerNumber = x.customerNumber
GROUP BY city