★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10470234.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

SQL架构

1 Create table If Not Exists cinema (id int, movie varchar(255), description varchar(255), rating float(2, 1))
2 Truncate table cinema
3 insert into cinema (id, movie, description, rating) values ('1', 'War', 'great 3D', '8.9')
4 insert into cinema (id, movie, description, rating) values ('2', 'Science', 'fiction', '8.5')
5 insert into cinema (id, movie, description, rating) values ('3', 'irish', 'boring', '6.2')
6 insert into cinema (id, movie, description, rating) values ('4', 'Ice song', 'Fantacy', '8.6')
7 insert into cinema (id, movie, description, rating) values ('5', 'House card', 'Interesting', '9.1')

X city opened a new cinema, many people would like to go to this cinema. The cinema also gives out a poster indicating the movies’ ratings and descriptions.

Please write a SQL query to output movies with an odd numbered ID and a description that is not 'boring'. Order the result by rating. 

For example, table cinema:

+---------+-----------+--------------+-----------+
|   id    | movie     |  description |  rating   |
+---------+-----------+--------------+-----------+
|   1     | War       |   great 3D   |   8.9     |
|   2     | Science   |   fiction    |   8.5     |
|   3     | irish     |   boring     |   6.2     |
|   4     | Ice song  |   Fantacy    |   8.6     |
|   5     | House card|   Interesting|   9.1     |
+---------+-----------+--------------+-----------+

For the example above, the output should be:

+---------+-----------+--------------+-----------+
|   id    | movie     |  description |  rating   |
+---------+-----------+--------------+-----------+
|   5     | House card|   Interesting|   9.1     |
|   1     | War       |   great 3D   |   8.9     |
+---------+-----------+--------------+-----------+

某城市开了一家新的电影院，吸引了很多人过来看电影。该电影院特别注意用户体验，专门有个 LED显示板做电影推荐，上面公布着影评和相关电影描述。

作为该电影院的信息部主管，您需要编写一个 SQL查询，找出所有影片描述为非 boring (不无聊) 的并且 id 为奇数 的影片，结果请按等级 rating 排列。 

例如，下表 cinema:

+---------+-----------+--------------+-----------+
|   id    | movie     |  description |  rating   |
+---------+-----------+--------------+-----------+
|   1     | War       |   great 3D   |   8.9     |
|   2     | Science   |   fiction    |   8.5     |
|   3     | irish     |   boring     |   6.2     |
|   4     | Ice song  |   Fantacy    |   8.6     |
|   5     | House card|   Interesting|   9.1     |
+---------+-----------+--------------+-----------+

对于上面的例子，则正确的输出是为：

+---------+-----------+--------------+-----------+
|   id    | movie     |  description |  rating   |
+---------+-----------+--------------+-----------+
|   5     | House card|   Interesting|   9.1     |
|   1     | War       |   great 3D   |   8.9     |
+---------+-----------+--------------+-----------+

108ms

1 # Write your MySQL query statement below
2 select id, movie, description, rating 
3 from cinema
4 where mod(id,2) = 1 AND description <> "boring"
5 order by rating desc

109ms

1 # Write your MySQL query statement below
2 select * from cinema where description != 'boring' and  id % 2 = 1 order by rating desc

110ms

1 # Write your MySQL query statement below
2 select a.* from cinema as a where mod(a.id,2)=1 and a.description != 'boring' order by rating desc;

111ms

1 # Write your MySQL query statement below
2 select *
3 from cinema
4 where not description = "boring" and mod(id, 2) = 1 
5 order by rating desc

115ms

1 # Write your MySQL query statement below
2 select * from cinema
3 where id%2 <> 0 
4 and description not like "%boring%"
5 order by rating desc