SQL根据行显示第一个唯一值，然后显示第二个唯一值

Question

我有一个类似的表

表名称：任务

+---------+-------------+-------------
| Name    | taskName    | issueDate  |
+---------+-------------+-------------
| A       | AA          | 4/11/2022  |
| A       | AB          | 4/11/2022  |
| A       | AC          | 4/11/2022  |
| B       | BA          | 3/11/2022  |
| B       | BB          | 3/11/2022  |
| B       | BC          | 3/11/2022  |
| C       | CA          | 2/11/2022  |
| C       | CB          | 2/11/2022  |
+---------+-------------+-------------

结果表将类似于此

+---------+-------------+------------
| Name    | taskName    | issueTime |
+---------+-------------+------------
| A       | AA          | 4/11/2022 |
| B       | BA          | 3/11/2022 |
| C       | CA          | 2/11/2022 |
| A       | AB          | 4/11/2022 |
| B       | BB          | 3/11/2022 |
| C       | CB          | 2/11/2022 |
| A       | AC          | 4/11/2022 |
| C       | CB          | 2/11/2022 |
+---------+-------------+-----------|

按照issueTime的降序，首先显示唯一名称。

Answer 1

首先按照发布日期和任务名称的升序为每个"name"组提供一个行号，然后根据行号对结果集进行排序。

查询

select t.name, t.taskname, t.issuedate from(
select row_number() over(
partition by name
order by issuedate, taskname
) as rn, *
from tasks
) as t
order by t.rn, t.name;

Answer 2

此查询对于上述场景非常有用：

select name, taskname, issuedate from(
select name, taskname, issuedate, row_number() over(partition by name order by issuedate desc) as row_number 
from task)
order by rownum;