1148. 文章浏览 I

发表于 更新于

难度系数： 简单

Views 表：

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+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| article_id    | int     |
| author_id     | int     |
| viewer_id     | int     |
| view_date     | date    |
+---------------+---------+
此表无主键，因此可能会存在重复行。
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。

请编写一条 SQL 查询以找出所有浏览过自己文章的作者，结果按照 id 升序排列。

查询结果的格式如下所示：

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Views 表：
+------------+-----------+-----------+------------+
| article_id | author_id | viewer_id | view_date  |
+------------+-----------+-----------+------------+
| 1          | 3         | 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 |
+------------+-----------+-----------+------------+

结果表：
+------+
| id   |
+------+
| 4    |
| 7    |
+------+

SQL结构

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Create table If Not Exists Views (article_id int, author_id int, viewer_id int, view_date date);
Truncate table Views;
insert into Views (article_id, author_id, viewer_id, view_date) values ('1', '3', '5', '2019-08-01');
insert into Views (article_id, author_id, viewer_id, view_date) values ('1', '3', '6', '2019-08-02');
insert into Views (article_id, author_id, viewer_id, view_date) values ('2', '7', '7', '2019-08-01');
insert into Views (article_id, author_id, viewer_id, view_date) values ('2', '7', '6', '2019-08-02');
insert into Views (article_id, author_id, viewer_id, view_date) values ('4', '7', '1', '2019-07-22');
insert into Views (article_id, author_id, viewer_id, view_date) values ('3', '4', '4', '2019-07-21');
insert into Views (article_id, author_id, viewer_id, view_date) values ('3', '4', '4', '2019-07-21');

解法：

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SELECT DISTINCT author_id AS id
FROM Views 
WHERE author_id = viewer_id
ORDER BY author_id;

原题链接：https://leetcode.cn/problems/article-views-i/