[Swift]LeetCode620. 有趣的电影 | Not Boring Movies

概述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 cinema3 insert into cinema (id, movie, description, rating) val

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 cinema3 insert into cinema (ID,movIE,description,rating) values (‘1‘,‘War‘,‘great 3D‘,‘8.9‘)4 insert into cinema (ID,rating) values (‘2‘,‘ScIEnce‘,‘fiction‘,‘8.5‘)5 insert into cinema (ID,rating) values (‘3‘,‘irish‘,‘boring‘,‘6.2‘)6 insert into cinema (ID,rating) values (‘4‘,‘Ice song‘,‘Fantacy‘,‘8.6‘)7 insert into cinema (ID,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 below2 select ID,rating 3 from cinema4 where mod(ID,2) = 1 AND description <> "boring"5 order by rating desc

109ms

1 # Write your MysqL query statement below2 select * from cinema where description != ‘boring‘ and  ID % 2 = 1 order by rating desc

110ms

1 # Write your MysqL query statement below2 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 below2 select *3 from cinema4 where not description = "boring" and mod(ID,2) = 1 5 order by rating desc

115ms

1 # Write your MysqL query statement below2 select * from cinema3 where ID%2 <> 0 4 and description not like "%boring%"5 order by rating desc

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