SQL：MySQL中的GROUP BY問題

Question

這是數據庫表

╔════╦═════════════╦══════════════════╦═══════╗
║ id ║ customer_id ║    last_seen     ║ param ║
╠════╬═════════════╬══════════════════╬═══════╣
║  1 ║       12345 ║ 2012-08-01 12:00 ║     1 ║
║  2 ║       22345 ║ 2012-08-01 12:00 ║     1 ║
║  3 ║       32345 ║ 2012-08-01 12:00 ║     1 ║
║  4 ║       42345 ║ 2012-08-01 12:00 ║     1 ║
║  5 ║       52345 ║ 2012-08-01 12:00 ║     1 ║
║  6 ║       12345 ║ 2012-09-01 12:00 ║     2 ║
║  7 ║       12345 ║ 2012-10-01 12:00 ║     3 ║
╚════╩═════════════╩══════════════════╩═══════╝

其中id是AUTO INCREMENT主鍵。

我想要實現的是獲取每個customer_id的最后一條記錄。 預期結果 ：

╔════╦═════════════╦══════════════════╦═══════╗
║ id ║ customer_id ║    last_seen     ║ param ║
╠════╬═════════════╬══════════════════╬═══════╣
║  2 ║       22345 ║ 2012-08-01 12:00 ║     1 ║
║  3 ║       32345 ║ 2012-08-01 12:00 ║     1 ║
║  4 ║       42345 ║ 2012-08-01 12:00 ║     1 ║
║  5 ║       52345 ║ 2012-08-01 12:00 ║     1 ║
║  7 ║       12345 ║ 2012-10-01 12:00 ║     3 ║
╚════╩═════════════╩══════════════════╩═══════╝

我試過這個SQL，但它返回錯誤的結果：

SELECT customer_id, param, last_seen 
FROM `my_table` 
GROUP BY customer_id 
ORDER BY last_seen DESC 

我在這里錯過了什么？

更新 ：表結構（ DESC my_table輸出）

+--------------+--------------+------+-----+---------+----------------+
| Field        | Type         | Null | Key | Default | Extra          |
+--------------+--------------+------+-----+---------+----------------+
| id           | int(11)      | NO   | PRI | NULL    | auto_increment |
| customer_id  | varchar(8)   | NO   |     | NULL    |                |
| last_seen    | datetime     | NO   |     | NULL    |                |
| param        | int(11)      | NO   |     | NULL    |                |
+--------------+--------------+------+-----+---------+----------------+

Answer 1

子查詢背后的想法是它分別獲取每個Customer_ID的最新last_seen值。 然后，子查詢的結果將與原始表連接，前提是必須滿足條件： CustomerID和日期必須相互匹配。

SELECT  a.*
FROM    Customer a
        INNER JOIN
        (
            SELECT Customer_ID, MAX(last_seen) maxDate
            FROM Customer
            GROUP BY Customer_ID
        ) b ON a.Customer_ID = b.Customer_ID AND
                a.last_seen = b.maxDate
ORDER BY a.ID

• SQLFiddle演示

后續問題：請您檢查記錄ID 6的last_seen值是否正確？

Answer 2

這是一個SQLFiddle演示

在MySQL中，您可以這樣做：

select ID,CUSTOMER_ID,LAST_SEEN,PARAM from
(
   select t.*,
       if(@i=customer_id,0,1) isLast,
       @i:=customer_id 
       from `my_table` t,(select @i:=0) t1 
   order by customer_id,last_seen desc
) t2 where isLAst=1