MySQL的 - 選擇所有客戶和每個客戶的訂單總數和總價值

Question

對每一個客戶，我想回到：ID，姓名，total_orders，total_value

客戶：

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

訂單：

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

如何選擇喜歡的結果：

╔════╦═════════════╦═════════════╦═════════════╗ 
║ 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 

爲了進一步獲得更多的知識有關加入，請訪問以下鏈接：

• Visual Representation of SQL Joins

結果，

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

Answer 2

這個查詢應該給你deriderated輸出：

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

使用內通過分組用戶ID加入再總和。

SQL連接用於通過平等點合併表，你應該給看看這個http://dev.mysql.com/doc/refman/5.0/en/join.html

對不起，我沒有一個MySQL服務器現在BU是這樣的：

選擇（選擇customers.id作爲id，customers.name作爲名稱，orders.value作爲客戶的內部連接訂單的值，從customers.id = orders.customer獲得）的id，name，sum（value）

幾乎肯定不是很正確，但在某個地方接近。