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48 lines
1.7 KiB
HTML
48 lines
1.7 KiB
HTML
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<p>Table: <code>Activity</code></p>
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<pre>
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+--------------+---------+
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| Column Name | Type |
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+--------------+---------+
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| player_id | int |
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| device_id | int |
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| event_date | date |
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| games_played | int |
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+--------------+---------+
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（player_id，event_date）是此表的主键（具有唯一值的列的组合）。
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这张表显示了某些游戏的玩家的活动情况。
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每一行是一个玩家的记录，他在某一天使用某个设备注销之前登录并玩了很多游戏（可能是 0）。
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</pre>
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<p> </p>
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<p>编写解决方案，报告在首次登录的第二天再次登录的玩家的 <strong>比率</strong>，<strong>四舍五入到小数点后两位</strong>。换句话说，你需要计算从首次登录日期开始至少连续两天登录的玩家的数量，然后除以玩家总数。</p>
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<p>结果格式如下所示：</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<strong>输入：</strong>
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Activity table:
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+-----------+-----------+------------+--------------+
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| player_id | device_id | event_date | games_played |
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+-----------+-----------+------------+--------------+
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| 1 | 2 | 2016-03-01 | 5 |
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| 1 | 2 | 2016-03-02 | 6 |
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| 2 | 3 | 2017-06-25 | 1 |
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| 3 | 1 | 2016-03-02 | 0 |
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| 3 | 4 | 2018-07-03 | 5 |
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+-----------+-----------+------------+--------------+
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<strong>输出：</strong>
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+-----------+
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| fraction |
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+-----------+
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| 0.33 |
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+-----------+
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<strong>解释：</strong>
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只有 ID 为 1 的玩家在第一天登录后才重新登录，所以答案是 1/3 = 0.33
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</pre>
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