mysql - 如何使用sql select获取一组记录中一列的最大值结果？

Question

我在表格中保存了这些数据。

date       | name        | code
2020-01-01 | category1   | 0
2020-01-02 | category1   | 1
2020-01-03 | category1   | 2
2020-01-04 | category1   | 3
2020-01-05 | category1   | 0
2020-01-06 | category1   | 1
2020-01-07 | category1   | 2
2020-01-08 | category1   | 0
2020-01-01 | category2   | 0
2020-01-02 | category2   | 0
2020-01-03 | category2   | 1
2020-01-04 | category2   | 0
2020-01-05 | category2   | 1
2020-01-06 | category2   | 2
2020-01-07 | category2   | 0
2020-01-08 | category2   | 1
......

集合中的列代码增加1。

需要获取一组记录中某一列具有最大值的记录。我的示例的结果必须是:

date       | name        | code
2020-01-04 | category1   | 3
2020-01-07 | category1   | 2
2020-01-03 | category2   | 1
2020-01-06 | category2   | 2
2020-01-08 | category2   | 1
......

Answer 1

这是你想要的吗？

select date, name, code
from (
    select 
        t.*,
        lead(code) over(partition by name order by date) lead_code
    from mytable t
) t 
where not code <= lead_code

对于您的示例数据，this produces :

date       | name      | code
:--------- | :-------- | ---:
2020-01-04 | category1 |    3
2020-01-07 | category1 |    2
2020-01-03 | category2 |    1
2020-01-06 | category2 |    2