本文介绍了返回 MySQL 中每个“分组依据"的“最后"行的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着跟版网的小编来一起学习吧！
问题描述
限时送ChatGPT账号..
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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