[Swift]LeetCode620. 有趣的電影 | Not Boring Movies

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SQL架構git

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.github

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. sql

For example, table cinema:微信

+---------+-----------+--------------+-----------+
|   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:架構

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

某城市開了一家新的電影院,吸引了不少人過來看電影。該電影院特別注意用戶體驗,專門有個 LED顯示板作電影推薦,上面公佈着影評和相關電影描述。post

做爲該電影院的信息部主管,您須要編寫一個 SQL查詢,找出全部影片描述爲非 boring (不無聊) 的而且 id 爲奇數 的影片,結果請按等級 rating 排列。 this

例如,下表 cinema:spa

+---------+-----------+--------------+-----------+
|   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     |
+---------+-----------+--------------+-----------+

對於上面的例子,則正確的輸出是爲:rest

+---------+-----------+--------------+-----------+
|   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
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