1. 题目描述
表: Views
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| article_id | int |
| author_id | int |
| viewer_id | int |
| view_date | date |
+---------------+---------+
此表可能会存在重复行
此表的每一行都表示某人在某天浏览了某位作者的某篇文章
请注意, 同一人的 author_id 和 viewer_id 是相同的
编写解决方案来找出在同一天阅读至少两篇文章的人
结果按照 id 升序排序
2. 测试用例
输入:
Views 表:
+------------+-----------+-----------+------------+
| article_id | author_id | viewer_id | view_date |
+------------+-----------+-----------+------------+
| 1 | 3 | 5 | 2019-08-01 |
| 3 | 4 | 5 | 2019-08-01 |
| 1 | 3 | 6 | 2019-08-02 |
| 2 | 7 | 7 | 2019-08-01 |
| 2 | 7 | 6 | 2019-08-02 |
| 4 | 7 | 1 | 2019-07-22 |
| 3 | 4 | 4 | 2019-07-21 |
| 3 | 4 | 4 | 2019-07-21 |
+------------+-----------+-----------+------------+
输出:
+------+
| id |
+------+
| 5 |
| 6 |
+------+
3. 解题思路
- 将
view_date, viewer_id进行分组统计group by view_date, viewer_id, 每人在同一天阅读至少两篇文章(需要考虑重复阅读文章场景,重复阅读的文章需要去重)having count(distinct article_id) >= 2
select viewer_id
from Views
group by view_date, viewer_id
having count(distinct article_id) >= 2
查询结果
+---------+
|viewer_id|
+---------+
|5 |
|6 |
+---------+
- 需要考虑
viewer_id去重, 并按照题目要求对viewer_id进行升序排序
select distinct viewer_id as id
from Views
group by view_date, viewer_id
having count(distinct article_id) >= 2
order by viewer_id
查询结果
+--+
|id|
+--+
|5 |
|6 |
+--+
