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算法题（国外版）/user-activity-for-the-past-30-days-i.html Normal file
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<p>Table: <code>Activity</code></p>
<pre>
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| user_id | int |
| session_id | int |
| activity_date | date |
| activity_type | enum |
+---------------+---------+
There is no primary key for this table, it may have duplicate rows.
The activity_type column is an ENUM of type (&#39;open_session&#39;, &#39;end_session&#39;, &#39;scroll_down&#39;, &#39;send_message&#39;).
The table shows the user activities for a social media website.
Note that each session belongs to exactly one user.
</pre>
<p>&nbsp;</p>
<p>Write an SQL query to find the daily active user count for a period of <code>30</code> days ending <code>2019-07-27</code> inclusively. A user was active on someday if they made at least one activity on that day.</p>
<p>Return the result table in <strong>any order</strong>.</p>
<p>The query result format is in the following example.</p>
<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<pre>
<strong>Input:</strong>
Activity table:
+---------+------------+---------------+---------------+
| user_id | session_id | activity_date | activity_type |
+---------+------------+---------------+---------------+
| 1 | 1 | 2019-07-20 | open_session |
| 1 | 1 | 2019-07-20 | scroll_down |
| 1 | 1 | 2019-07-20 | end_session |
| 2 | 4 | 2019-07-20 | open_session |
| 2 | 4 | 2019-07-21 | send_message |
| 2 | 4 | 2019-07-21 | end_session |
| 3 | 2 | 2019-07-21 | open_session |
| 3 | 2 | 2019-07-21 | send_message |
| 3 | 2 | 2019-07-21 | end_session |
| 4 | 3 | 2019-06-25 | open_session |
| 4 | 3 | 2019-06-25 | end_session |
+---------+------------+---------------+---------------+
<strong>Output:</strong>
+------------+--------------+
| day | active_users |
+------------+--------------+
| 2019-07-20 | 2 |
| 2019-07-21 | 2 |
+------------+--------------+
<strong>Explanation:</strong> Note that we do not care about days with zero active users.
</pre>