只返回每组中的最高值oracle

Question

给定一个表格，

NAME        DATE_OCCURRED
a           2020-05-14 23:48.07
a           2020-05-14 23:48.07
a           2020-05-14 23:48.08
b           2020-05-14 23:48.08
b           2020-05-14 23:48.08
b           2020-05-14 23:48.08
b           2020-05-14 23:48.09

我想为每个人返回 NAME，最多记录存在的时间；以及该时间存在多少记录。

NAME       MAXCOUNT_PER_SECOND   DATE_OCCURRED
a          2                     2020-05-14 23:48.07
b          3                     2020-05-14 23:48.08

我已经找到了用DATE_OCCURRED和NAME来分组的SQL语句。

SELECT COUNT(*) AS COUNT_PER_SECOND, NAME, DATE_OCCURRED FROM TABLE GROUP BY NAME, DATE_OCCURRED ORDER BY NAME ASC, COUNT_PER_SECOND DESC

但我现在想再选择一次，只选每个名字的最大值。我已经试过了。

SELECT MAX(COUNT_PER_SECOND) AS MAXCOUNT_PER_SECOND, NAME FROM (the above query) GROUP BY NAME; 这样就能得到 MAXCOUNT_PER_SECOND 和 NAME但只要我一尝试，也会得到哪种。DATE_OCCURRED 导致 MAXCOUNT_PER_SECOND 值，我最终在运行SQL时出现分组错误；或者得不到我期望的结果。

即

SELECT MAX(COUNT_PER_SECOND) AS MAXCOUNT_PER_SECOND, NAME, DATE_OCCURRED FROM (the above query) GROUP BY NAME; -> Not a GROUP BY expression

SELECT MAX(COUNT_PER_SECOND) AS MAXCOUNT_PER_SECOND, NAME, DATE_OCCURRED FROM (the above query) GROUP BY NAME, DATE_OCCURRED ; -> 运行，但给我所有日期的结果，而不是只给最大日期。

Answer 1

你可以使用aggreagation和window函数。

select name, date_occured, no_records
from (
    select 
        name, 
        date_occured, 
        count(*) no_records, 
        rank() over(partition by name order by count(*) desc) rn
    from mytable
    group by name, date_occured
) t
where rn = 1

子查询的aggreagates是 name 和 date_occured，对每组记录进行统计，并对具有相同记录的组进行排名。name 按降序计数。然后，外部查询会对最上面的记录进行过滤，每条记录的数量为 name. 由于我们使用 rank()，可能的顶部连接将被包含在结果集中（如果您不希望这样，请使用 row_number() 而不是）。)