SQL MAX（）问题

Question

uid     timestamp
1   1242420497
1   1243534661
1   1243534858
1   1243611312
1   1243611511
3   1244817764
3   1244819093
1   1244749446

我有这张桌子，我想抓住时间戳最高的那一行。 我尝试使用

SELECT uid,max(timestamp) FROM `node_revisions` WHERE nid=51

但是返回了

uid timestamp
1   1244819093

你可以看到

有错误的uid。如何让它从正确的行中抓取uid？感谢

Answer 1

你错过了GROUP BY子句。

SELECT
    uid,
    max(timestamp) as max_time
FROM
    node_revisions
WHERE
    nid = 51
GROUP BY
    uid
ORDER BY 
    max_time DESC 
LIMIT 1

Answer 2

SELECT * FROM node_revisions WHERE nid = 51 ORDER BY timestamp DESC LIMIT 1

Answer 3

首先，nid未出现在您的表格中。

第二，我怀疑你想要group by uid

Answer 4

您的示例使用nid = 51而不是uid = 51。此代码是否直接从您正在运行的代码中复制？如果有一个nid字段，这可能是您的问题。你需要一个group by子句。

SELECT uid, max(timestamp) as max_time
FROM 'node_revisions'
WHERE uid = 51
GROUP BY uid

Answer 5

这样就可以了。确定

select uid FROM `node_revisions`
WHERE uid=51 and timestamp = (select max(timestamp) where uid = 51)
group by uid

Answer 6

SELECT uid, timestamp 
FROM node_revisions 
WHERE timestamp = (SELECT MAX(timestamp) FROM node_revisions);

根据Ryan Oberoi的评论

更新了;因为我们只得到一条记录，所以外部查询中的MAX（）是不必要的。感谢。