只返回每组oracle中的最大值

Posted 2023-03-31

【中文标题】只返回每组oracle中的最大值

【英文标题】：Only return the highest value in each group oracle

【发布时间】：2020-06-02 12:24:19

【问题描述】：

给定一张桌子：

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NAME 返回记录最多的时间；以及当时存在多少条记录：

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

我找到了按 DATE_OCCURRED 和 NAME 对计数进行分组的 SQL：

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

但我现在想再次选择，只选择每个名称的最大值。我试过了：

SELECT MAX(COUNT_PER_SECOND) AS MAXCOUNT_PER_SECOND, NAME FROM (the above query) GROUP BY NAME;，这可以给我MAXCOUNT_PER_SECOND和NAME；但是一旦我尝试并得到哪个DATE_OCCURRED 导致了MAXCOUNT_PER_SECOND 值，我在运行SQL 时最终会出现分组错误；或者我没有得到我期望的结果。

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，而不仅仅是最大值。

【参考方案1】：

您可以使用聚合和窗口函数：

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

子查询按name 和date_occured 聚合，计算每组的记录数，并按降序排列具有相同name 的记录组。然后外部查询过滤每个name 的顶部记录。由于我们使用rank()，因此可能会在结果集中包含可能的排名靠前（如果您不希望这样，请改用row_number()）。