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https://gitee.com/coder-xiaomo/leetcode-problemset
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批量更新数据
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@@ -12,12 +12,12 @@
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"translatedContent": "<p>Table: <code>Activity</code></p>\n\n<pre>\n+--------------+---------+\n| Column Name | Type |\n+--------------+---------+\n| player_id | int |\n| device_id | int |\n| event_date | date |\n| games_played | int |\n+--------------+---------+\n(player_id,event_date)是此表的主键(具有唯一值的列的组合)。\n这张表显示了某些游戏的玩家的活动情况。\n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)。\n</pre>\n\n<p> </p>\n\n<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>\n\n<p>结果格式如下所示:</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入:</strong>\nActivity table:\n+-----------+-----------+------------+--------------+\n| player_id | device_id | event_date | games_played |\n+-----------+-----------+------------+--------------+\n| 1 | 2 | 2016-03-01 | 5 |\n| 1 | 2 | 2016-03-02 | 6 |\n| 2 | 3 | 2017-06-25 | 1 |\n| 3 | 1 | 2016-03-02 | 0 |\n| 3 | 4 | 2018-07-03 | 5 |\n+-----------+-----------+------------+--------------+\n<strong>输出:</strong>\n+-----------+\n| fraction |\n+-----------+\n| 0.33 |\n+-----------+\n<strong>解释:</strong>\n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\n</pre>\n",
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@@ -77,7 +77,7 @@
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"insert into Activity (player_id, device_id, event_date, games_played) values ('3', '4', '2018-07-03', '5')"
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