返回 MySQL 中每个“分组依据"的“最后"行

Returning the #39;last#39; row of each #39;group by#39; in MySQL(返回 MySQL 中每个“分组依据的“最后行)

问题描述

是否有更有效的方法来执行以下操作?

Is there a more efficient way of doing the following?

select * 
    from foo as a
    where a.id = (select max(id) from foo where uid = a.uid group by uid)
    group by uid;
)

这个答案看起来很相似，但这个答案是最好的方法吗 - 如何在 MySQL 中为每个组选择第一行?

This answer looks similar, but is this answer the best way of doing this - How to select the first row for each group in MySQL?

谢谢，

克里斯.

附言表格看起来像:

CREATE TABLE foo (
    id INT(10) NOT NULL AUTO_INCREMENT,
    uid INT(10) NOT NULL,
    value VARCHAR(50) NOT NULL,
    PRIMARY KEY (`id`),
    INDEX `uid` (`uid`)
)

数据:

id, uid, value
 1,   1, hello
 2,   2, cheese
 3,   2, pickle
 4,   1, world

结果:

id, uid, value
 3,   2, pickle
 4,   1, world

参见 http://www.barricane.com/2012/02/08/mysql-select-last-matching-row.html 了解更多详情.

See http://www.barricane.com/2012/02/08/mysql-select-last-matching-row.html for more details.

推荐答案

试试这个查询 -

SELECT t1.* FROM foo t1
  JOIN (SELECT uid, MAX(id) id FROM foo GROUP BY uid) t2
    ON t1.id = t2.id AND t1.uid = t2.uid;

然后使用EXPLAIN来分析查询.

SELECT t1.* FROM foo t1
  LEFT JOIN foo t2
    ON t1.id < t2.id AND t1.uid = t2.uid
WHERE t2.id is NULL;

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