LeetCode--SQL 查询：有趣的电影

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

select * from cinema 
where mod(id,2)=1 and description<>'boring'
order by rating desc

延伸：mysql中判断奇数偶数（注意效率，千万级数据时索引使用情况等）

-- 按位与
select * from cinema WHERE id&1; 

-- id先除以2然后乘2 如果与原来的相等就是偶数
select * from cinema WHERE id=(id>>1)<<1; 

-- 正则匹配最后一位
select * from cinema WHERE id regexp '[13579]$';
select * from cinema WHERE id regexp '[02468]$';

-- id计算
select * from cinema WHERE id%2 = 1;
select * from cinema WHERE id%2 = 0;

-- 与上面的一样
select * from cinema WHERE mod(id, 2) = 1;
select * from cinema WHERE mod(id, 2) = 0;

-- -1的奇数次方和偶数次方
select * from cinema WHERE POWER(-1, id) = -1;
select * from cinema WHERE POWER(-1, id) = 1;