从表中获取记录？
[英]Fetch record from table?
问题描述
I have table in which data is like this
我有一个表格，其中的数据是这样的
Mapping Relationship映射关系
1 task_id is mapped to multiple task_detail_id ( 1 to Many ) 1 task_id 映射到多个 task_detail_id ( 1 to Many )
Task_detail_id Task_id Creation_date
1 20 2020-05-02 20:28:23.354
2 21 2020-05-02 20:28:23.354
3 22 2020-05-02 19:28:23.354
4 22 2020-05-02 18:28:23.354
5 22 2020-05-02 17:28:23.354
6 22 2020-05-02 16:28:23.354
7 22 2020-05-02 15:28:23.354
8 23 2020-05-02 10:28:23.354
9 24 2020-05-02 09:28:23.354
10 24 2020-05-02 08:28:23.354
11 24 2020-05-02 07:28:23.354
What I want is to traverse the table and fetch record as if same task_id exist more than 2 times then fetch top 2 (latest) records for that task_id我想要的是遍历表并获取记录，就好像相同的 task_id 存在超过 2 次然后获取该 task_id 的前 2 个（最新）记录
Sample Output样品 Output
Task_detail_id Task_id Creation_date
1 20 2020-05-02 20:28:23.354
2 21 2020-05-02 20:28:23.354
3 22 2020-05-02 19:28:23.354
4 22 2020-05-02 18:28:23.354
8 23 2020-05-02 10:28:23.354
9 24 2020-05-02 09:28:23.354
10 24 2020-05-02 08:28:23.354
1 个解决方案
解决方案1
0 已采纳 2020-05-07 10:23:11
This is typically done with a window function:这通常使用 window function 完成：
select task_detail_id, task_id, creation_date
from (
select task_detail_id, task_id, creation_date,
row_number() over (partition by task_id order by creation_date desc) as rn
from data
) t
where rn <= 2
order by task_detail_id, task_id;
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