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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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@@ -7,17 +7,17 @@
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"boundTopicId": 90536,
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"title": "Movie Rating",
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"titleSlug": "movie-rating",
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"content": "<p>Table: <code>Movies</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| movie_id | int |\n| title | varchar |\n+---------------+---------+\nmovie_id is the primary key (column with unique values) for this table.\ntitle is the name of the movie.\n</pre>\n\n<p> </p>\n\n<p>Table: <code>Users</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| user_id | int |\n| name | varchar |\n+---------------+---------+\nuser_id is the primary key (column with unique values) for this table.\n</pre>\n\n<p> </p>\n\n<p>Table: <code>MovieRating</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| movie_id | int |\n| user_id | int |\n| rating | int |\n| created_at | date |\n+---------------+---------+\n(movie_id, user_id) is the primary key (column with unique values) for this table.\nThis table contains the rating of a movie by a user in their review.\ncreated_at is the user's review date. \n</pre>\n\n<p> </p>\n\n<p>Write a solution to:</p>\n\n<ul>\n\t<li>Find the name of the user who has rated the greatest number of movies. In case of a tie, return the lexicographically smaller user name.</li>\n\t<li>Find the movie name with the <strong>highest average</strong> rating in <code>February 2020</code>. In case of a tie, return the lexicographically smaller movie name.</li>\n</ul>\n\n<p>The result format is in the following example.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nMovies table:\n+-------------+--------------+\n| movie_id | title |\n+-------------+--------------+\n| 1 | Avengers |\n| 2 | Frozen 2 |\n| 3 | Joker |\n+-------------+--------------+\nUsers table:\n+-------------+--------------+\n| user_id | name |\n+-------------+--------------+\n| 1 | Daniel |\n| 2 | Monica |\n| 3 | Maria |\n| 4 | James |\n+-------------+--------------+\nMovieRating table:\n+-------------+--------------+--------------+-------------+\n| movie_id | user_id | rating | created_at |\n+-------------+--------------+--------------+-------------+\n| 1 | 1 | 3 | 2020-01-12 |\n| 1 | 2 | 4 | 2020-02-11 |\n| 1 | 3 | 2 | 2020-02-12 |\n| 1 | 4 | 1 | 2020-01-01 |\n| 2 | 1 | 5 | 2020-02-17 | \n| 2 | 2 | 2 | 2020-02-01 | \n| 2 | 3 | 2 | 2020-03-01 |\n| 3 | 1 | 3 | 2020-02-22 | \n| 3 | 2 | 4 | 2020-02-25 | \n+-------------+--------------+--------------+-------------+\n<strong>Output:</strong> \n+--------------+\n| results |\n+--------------+\n| Daniel |\n| Frozen 2 |\n+--------------+\n<strong>Explanation:</strong> \nDaniel and Monica have rated 3 movies ("Avengers", "Frozen 2" and "Joker") but Daniel is smaller lexicographically.\nFrozen 2 and Joker have a rating average of 3.5 in February but Frozen 2 is smaller lexicographically.\n</pre>\n",
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"content": "<p>Table: <code>Movies</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| movie_id | int |\n| title | varchar |\n+---------------+---------+\nmovie_id is the primary key (column with unique values) for this table.\ntitle is the name of the movie.\n</pre>\n\n<p> </p>\n\n<p>Table: <code>Users</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| user_id | int |\n| name | varchar |\n+---------------+---------+\nuser_id is the primary key (column with unique values) for this table.\nThe column 'name' has unique values.\n</pre>\n\n<p>Table: <code>MovieRating</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| movie_id | int |\n| user_id | int |\n| rating | int |\n| created_at | date |\n+---------------+---------+\n(movie_id, user_id) is the primary key (column with unique values) for this table.\nThis table contains the rating of a movie by a user in their review.\ncreated_at is the user's review date. \n</pre>\n\n<p> </p>\n\n<p>Write a solution to:</p>\n\n<ul>\n\t<li>Find the name of the user who has rated the greatest number of movies. In case of a tie, return the lexicographically smaller user name.</li>\n\t<li>Find the movie name with the <strong>highest average</strong> rating in <code>February 2020</code>. In case of a tie, return the lexicographically smaller movie name.</li>\n</ul>\n\n<p>The result format is in the following example.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nMovies table:\n+-------------+--------------+\n| movie_id | title |\n+-------------+--------------+\n| 1 | Avengers |\n| 2 | Frozen 2 |\n| 3 | Joker |\n+-------------+--------------+\nUsers table:\n+-------------+--------------+\n| user_id | name |\n+-------------+--------------+\n| 1 | Daniel |\n| 2 | Monica |\n| 3 | Maria |\n| 4 | James |\n+-------------+--------------+\nMovieRating table:\n+-------------+--------------+--------------+-------------+\n| movie_id | user_id | rating | created_at |\n+-------------+--------------+--------------+-------------+\n| 1 | 1 | 3 | 2020-01-12 |\n| 1 | 2 | 4 | 2020-02-11 |\n| 1 | 3 | 2 | 2020-02-12 |\n| 1 | 4 | 1 | 2020-01-01 |\n| 2 | 1 | 5 | 2020-02-17 | \n| 2 | 2 | 2 | 2020-02-01 | \n| 2 | 3 | 2 | 2020-03-01 |\n| 3 | 1 | 3 | 2020-02-22 | \n| 3 | 2 | 4 | 2020-02-25 | \n+-------------+--------------+--------------+-------------+\n<strong>Output:</strong> \n+--------------+\n| results |\n+--------------+\n| Daniel |\n| Frozen 2 |\n+--------------+\n<strong>Explanation:</strong> \nDaniel and Monica have rated 3 movies ("Avengers", "Frozen 2" and "Joker") but Daniel is smaller lexicographically.\nFrozen 2 and Joker have a rating average of 3.5 in February but Frozen 2 is smaller lexicographically.\n</pre>\n",
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"translatedTitle": "电影评分",
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"translatedContent": "<p>表：<code>Movies</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| movie_id | int |\n| title | varchar |\n+---------------+---------+\nmovie_id 是这个表的主键(具有唯一值的列)。\ntitle 是电影的名字。\n</pre>\n\n<p>表：<code>Users</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| user_id | int |\n| name | varchar |\n+---------------+---------+\nuser_id 是表的主键(具有唯一值的列)。\n</pre>\n\n<p>表：<code>MovieRating</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| movie_id | int |\n| user_id | int |\n| rating | int |\n| created_at | date |\n+---------------+---------+\n(movie_id, user_id) 是这个表的主键(具有唯一值的列的组合)。\n这个表包含用户在其评论中对电影的评分 rating 。\ncreated_at 是用户的点评日期。 \n</pre>\n\n<p> </p>\n\n<p>请你编写一个解决方案：</p>\n\n<ul>\n\t<li>查找评论电影数量最多的用户名。如果出现平局，返回字典序较小的用户名。</li>\n\t<li>查找在 <code>February 2020</code><strong> 平均评分最高</strong> 的电影名称。如果出现平局，返回字典序较小的电影名称。</li>\n</ul>\n\n<p><strong>字典序</strong> ，即按字母在字典中出现顺序对字符串排序，字典序较小则意味着排序靠前。</p>\n\n<p>返回结果格式如下例所示。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>\nMovies 表：\n+-------------+--------------+\n| movie_id | title |\n+-------------+--------------+\n| 1 | Avengers |\n| 2 | Frozen 2 |\n| 3 | Joker |\n+-------------+--------------+\nUsers 表：\n+-------------+--------------+\n| user_id | name |\n+-------------+--------------+\n| 1 | Daniel |\n| 2 | Monica |\n| 3 | Maria |\n| 4 | James |\n+-------------+--------------+\nMovieRating 表：\n+-------------+--------------+--------------+-------------+\n| movie_id | user_id | rating | created_at |\n+-------------+--------------+--------------+-------------+\n| 1 | 1 | 3 | 2020-01-12 |\n| 1 | 2 | 4 | 2020-02-11 |\n| 1 | 3 | 2 | 2020-02-12 |\n| 1 | 4 | 1 | 2020-01-01 |\n| 2 | 1 | 5 | 2020-02-17 | \n| 2 | 2 | 2 | 2020-02-01 | \n| 2 | 3 | 2 | 2020-03-01 |\n| 3 | 1 | 3 | 2020-02-22 | \n| 3 | 2 | 4 | 2020-02-25 | \n+-------------+--------------+--------------+-------------+\n<strong>输出：</strong>\nResult 表：\n+--------------+\n| results |\n+--------------+\n| Daniel |\n| Frozen 2 |\n+--------------+\n<strong>解释：</strong>\nDaniel 和 Monica 都点评了 3 部电影（\"Avengers\", \"Frozen 2\" 和 \"Joker\"） 但是 Daniel 字典序比较小。\nFrozen 2 和 Joker 在 2 月的评分都是 3.5，但是 Frozen 2 的字典序比较小。\n</pre>\n",
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"translatedContent": "<p>表：<code>Movies</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| movie_id | int |\n| title | varchar |\n+---------------+---------+\nmovie_id 是这个表的主键(具有唯一值的列)。\ntitle 是电影的名字。\n</pre>\n\n<p>表：<code>Users</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| user_id | int |\n| name | varchar |\n+---------------+---------+\nuser_id 是表的主键(具有唯一值的列)。\n'name' 列具有唯一值。</pre>\n\n<p>表：<code>MovieRating</code></p>\n\n<pre>\n+---------------+---------+\n| Column Name | Type |\n+---------------+---------+\n| movie_id | int |\n| user_id | int |\n| rating | int |\n| created_at | date |\n+---------------+---------+\n(movie_id, user_id) 是这个表的主键(具有唯一值的列的组合)。\n这个表包含用户在其评论中对电影的评分 rating 。\ncreated_at 是用户的点评日期。 \n</pre>\n\n<p> </p>\n\n<p>请你编写一个解决方案：</p>\n\n<ul>\n\t<li>查找评论电影数量最多的用户名。如果出现平局，返回字典序较小的用户名。</li>\n\t<li>查找在 <code>February 2020</code><strong> 平均评分最高</strong> 的电影名称。如果出现平局，返回字典序较小的电影名称。</li>\n</ul>\n\n<p><strong>字典序</strong> ，即按字母在字典中出现顺序对字符串排序，字典序较小则意味着排序靠前。</p>\n\n<p>返回结果格式如下例所示。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>\nMovies 表：\n+-------------+--------------+\n| movie_id | title |\n+-------------+--------------+\n| 1 | Avengers |\n| 2 | Frozen 2 |\n| 3 | Joker |\n+-------------+--------------+\nUsers 表：\n+-------------+--------------+\n| user_id | name |\n+-------------+--------------+\n| 1 | Daniel |\n| 2 | Monica |\n| 3 | Maria |\n| 4 | James |\n+-------------+--------------+\nMovieRating 表：\n+-------------+--------------+--------------+-------------+\n| movie_id | user_id | rating | created_at |\n+-------------+--------------+--------------+-------------+\n| 1 | 1 | 3 | 2020-01-12 |\n| 1 | 2 | 4 | 2020-02-11 |\n| 1 | 3 | 2 | 2020-02-12 |\n| 1 | 4 | 1 | 2020-01-01 |\n| 2 | 1 | 5 | 2020-02-17 | \n| 2 | 2 | 2 | 2020-02-01 | \n| 2 | 3 | 2 | 2020-03-01 |\n| 3 | 1 | 3 | 2020-02-22 | \n| 3 | 2 | 4 | 2020-02-25 | \n+-------------+--------------+--------------+-------------+\n<strong>输出：</strong>\nResult 表：\n+--------------+\n| results |\n+--------------+\n| Daniel |\n| Frozen 2 |\n+--------------+\n<strong>解释：</strong>\nDaniel 和 Monica 都点评了 3 部电影（\"Avengers\", \"Frozen 2\" 和 \"Joker\"） 但是 Daniel 字典序比较小。\nFrozen 2 和 Joker 在 2 月的评分都是 3.5，但是 Frozen 2 的字典序比较小。\n</pre>\n",
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"difficulty": "Medium",
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"likes": 55,
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"likes": 93,
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"dislikes": 0,
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"isLiked": null,
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@@ -59,7 +59,7 @@
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@@ -92,7 +92,7 @@
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"insert into MovieRating (movie_id, user_id, rating, created_at) values ('3', '2', '4', '2020-02-25')"
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