mysql - mysql + JOIN 查询的问题

Question

好的，我很抱歉问这个问题，因为我看到了几个 mysql JOIN 示例，但我似乎无法让它工作。

“销售量”

----------------------
idcustomer | datecode 
----------------------
 1         | 20120503 
 1         | 20120503 
 1         | 20120503 
 2         | 20120503 
 3         | 20120503 

我想知道谁是最大的买家......就客户在特定日期从我这里购买东西的次数而言（是的，我使用一些奇怪的格式来表示我知道的日期，请不要介意）......所以我做：

SELECT idcustomer, COUNT(idcustomer) FROM sales WHERE datecode = 20120503 GROUP BY idcustomer ORDER BY COUNT(idcustomer) DESC

我得到：

-----------------------------
idcustomer | Count(idcustomer)
-----------------------------
 1         | 3
 2         | 1
 3         | 1

问题是......因为我也有桌子：

“顾客”

----------------------
| name | id_customer |
----------------------
 Jess  | 1
 Matt  | 2
 Perry | 3 

以下是我想要实现的目标......如何做到这一点？

---------------------------------------------
customer.name | idcustomer | Count(idcustomer)
---------------------------------------------
 Jess         | 1          | 3
 Matt         | 2          | 1
 Perry        | 3          | 1

Answer 1

SELECT customer.name, idcustomer, COUNT(idcustomer)
FROM sales
JOIN customer
ON sales.idcustomer = customer.id_customer
WHERE datecode = 20120503
GROUP BY idcustomer
ORDER BY COUNT(idcustomer) DESC

在线查看它：sqlfiddle

Answer 2

SELECT 
  t1.name,t2.count_buy,t2.idcustomer 
   FROM customer as t1,
 (SELECT 
    idcustomer,COUNT(idcustomer) as count_buy 
       FROM `sales`  
        WHERE datecode = 20120503
        GROUP BY idcustomer 
         ORDER BY COUNT(idcustomer) DESC) t2
  WHERE 
   t1.idcustomer =t2.idcustomer

Answer 3

希望这会有所帮助::

Select cust_ref.idcustomer, cust_ref.COUNT(idcustomer), customer.name
from 
(SELECT idcustomer, COUNT(idcustomer) FROM sales WHERE datecode = 20120503 
 GROUP BY idcustomer) cust_ref
 inner join customer on (sales.idcustomer = customer.id_customer)
ORDER BY COUNT(idcustomer) DESC

Answer 4

你需要这样做 -

SELECT idcustomer, c.name, COUNT(idcustomer) 
FROM sales s inner join customer c on (s.idcoustomer=c.id_customer)  
WHERE datecode = 20120503 
GROUP BY idcustomer, c.name  
ORDER BY COUNT(idcustomer) DESC