GROUP BY with MAX(DATE)

我正在尝试在表格中列出每列火车的最新目的地(最长出发时间),例如:

I'm trying to list the latest destination (MAX departure time) for each train in a table, for example:

Train    Dest      Time
1        HK        10:00
1        SH        12:00
1        SZ        14:00
2        HK        13:00
2        SH        09:00
2        SZ        07:00

期望的结果应该是:

Train    Dest      Time
1        SZ        14:00
2        HK        13:00

我尝试过使用

SELECT Train, Dest, MAX(Time)
FROM TrainTable
GROUP BY Train

我收到ora-00979 不是 GROUP BY 表达式"错误,提示我必须在 group by 语句中包含Dest".但这肯定不是我想要的……

by I got a "ora-00979 not a GROUP BY expression" error saying that I must include 'Dest' in my group by statement. But surely that's not what I want...

是否可以在一行 SQL 中完成?

Is it possible to do it in one line of SQL?

推荐答案

您不能在结果集中包含未分组的非聚合列.如果一列火车只有一个目的地,那么只需将目的地列添加到您的 group by 子句中,否则您需要重新考虑您的查询.

You cannot include non-aggregated columns in your result set which are not grouped. If a train has only one destination, then just add the destination column to your group by clause, otherwise you need to rethink your query.

试试:

SELECT t.Train, t.Dest, r.MaxTime
FROM (
      SELECT Train, MAX(Time) as MaxTime
      FROM TrainTable
      GROUP BY Train
) r
INNER JOIN TrainTable t
ON t.Train = r.Train AND t.Time = r.MaxTime

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