如何选择MySQL中的最新行？

Question

我想选择MySQL中有5个项目的特定表的最新行.该表看起来像:

• id(自动增加)
• 至
• 从
• 时间戳
• 文本

数据类似于:

|id | to     | from   | time stamp | text
| 1 | user01 | user02 | 2011-09-01 | text1
| 2 | user01 | user02 | 2011-09-02 | text2
| 3 | user02 | user01 | 2011-09-02 | text3
| 4 | user01 | user03 | 2011-09-03 | text4
| 5 | user01 | user04 | 2011-09-03 | text5
| 6 | user01 | user03 | 2011-09-04 | text6
| 7 | user03 | user01 | 2011-09-05 | text7

我想select * WHERE to = 'user01'和最新数据(可能是"id"或"时间戳")."from"可以很多,但每个相同的"from"数据只能出现一次.

---

无论如何,所选数据将是:

| 2 | user01 | user02 | 2011-09-02 | text2
| 5 | user01 | user04 | 2011-09-03 | text5
| 6 | user01 | user03 | 2011-09-04 | text6

可以吗？感谢您花时间阅读我的问题:)

Answer 1

SELECT t.* 
FROM
      TableX AS t
  JOIN
      ( SELECT DISTINCT `from` AS f
        FROM TableX
        WHERE `to` = 'user01'
      ) AS df
    ON 
      t.id = ( SELECT tt.id
               FROM TableX AS tt
               WHERE tt.`to` = 'user01'
                 AND tt.`from` = df.f
               ORDER BY tt.`timestamp` DESC
               LIMIT 1
             )

最好避免使用诸如to,from和 之类的关键字来命名表和字段timestamp。