[Swift]LeetCode620. 有趣的电影 | Not Boring Movies
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,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? +---------+-----------+--------------+-----------+ | 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? 例如,下表? +---------+-----------+--------------+-----------+ | 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,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 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |