MySQL的 - 选择所有客户和每个客户的订单总数和总价值

Question

对每一个客户，我想回到：ID，姓名，total_orders，total_value

客户：

╔════╦═════════════╗ 
║ ID ║ NAME  ║ 
╠════╬═════════════╣ 
║ 1 ║ John Smith ║ 
║ 2 ║ Jim Jimmers ║ 
╚════╩═════════════╝ 

订单：

╔═══════╦══════════╦═══════╗ 
║ ID ║ CUSTOMER ║ VALUE ║ 
╠═══════╬══════════╬═══════╣ 
║ 34656 ║  1 ║ 20 ║ 
║ 37345 ║  2 ║ 25 ║ 
║ 38220 ║  1 ║ 15 ║ 
║ 39496 ║  1 ║ 38 ║ 
║ 41752 ║  1 ║  9 ║ 
║ 43734 ║  2 ║ 20 ║ 
╚═══════╩══════════╩═══════╝ 

如何选择喜欢的结果：

╔════╦═════════════╦═════════════╦═════════════╗ 
║ ID ║ NAME  ║ TOTALORDERS ║ TOTAL_VALUE ║ 
╠════╬═════════════╬═════════════╬═════════════╣ 
║ 1 ║ John Smith ║   4 ║   82 ║ 
║ 2 ║ Jim Jimmers ║   2 ║   45 ║ 
╚════╩═════════════╩═════════════╩═════════════╝ 

Answer 1

SELECT a.ID, 
     a.Name, 
     COUNT(b.Customer) totalOrders, 
     SUM(b.value) total_value 
FROM Customers a 
     LEFT JOIN Orders b 
      ON a.ID = b.Customer 
GROUP BY a.ID, 
     a.Name 

• SQLFiddle Demo

OR

SELECT a.ID, 
     a.Name, 
     COUNT(b.Customer) totalOrders, 
     COALESCE(SUM(b.value), 0) total_value 
FROM Customers a 
     LEFT JOIN Orders b 
      ON a.ID = b.Customer 
GROUP BY a.ID, 
     a.Name 

为了进一步获得更多的知识有关加入，请访问以下链接：

• Visual Representation of SQL Joins

结果，

╔════╦═════════════╦═════════════╦═════════════╗ 
║ ID ║ NAME  ║ TOTALORDERS ║ TOTAL_VALUE ║ 
╠════╬═════════════╬═════════════╬═════════════╣ 
║ 1 ║ John Smith ║   4 ║   82 ║ 
║ 2 ║ Jim Jimmers ║   2 ║   45 ║ 
╚════╩═════════════╩═════════════╩═════════════╝ 

Answer 2

这个查询应该给你deriderated输出：

SELECT c.id, c.name, count(*) AS total_orders, sum(o.value) AS total_value 
FROM Customers AS c 
    LEFT JOIN Orders AS o ON c.id = o.customer 
GROUP BY c.id 

Answer 3

使用内通过分组用户ID加入再总和。

SQL连接用于通过平等点合并表，你应该给看看这个http://dev.mysql.com/doc/refman/5.0/en/join.html

对不起，我没有一个MySQL服务器现在BU是这样的：

选择（选择customers.id作为id，customers.name作为名称，orders.value作为客户的内部连接订单的值，从customers.id = orders.customer获得）的id，name，sum（value）

几乎肯定不是很正确，但在某个地方接近。