如何使用sql select通过一组记录中的一列获取最大值的结果？

Question

我在表中有这些数据。

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
......

集合中的列码加1。

需要获取一组记录中一列具有最大值的记录。 我的例子的结果必须是：

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， 这会产生：

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

Answer 2

WITH cte AS (
SELECT date, 
       name, 
       code, 
       LEAD(code) OVER (PARTITION BY name ORDER BY date) next_code,
       LEAD(name) OVER (PARTITION BY name ORDER BY date) next_name
FROM source_table
)
SELECT date, name, code
FROM cte
WHERE (    name = next_name 
        OR next_name IS NULL )
  AND (    code > next_code
        OR next_code IS NULL )  
  AND code > 0;

小提琴。

附注。 感谢GMB - 从他的回答中窃取了小提琴的基础。