SQL编写一般思路：
1）复杂的查询，先划分为小任务，以降低难度。分别实现各个小任务后，再进行汇总；
2）涉及多表时，先进行联表查询；
3）简单分组，一般只需要group by即可；
4）组内TopN问题，使用row_number，rank，dense_rank；
5）熟练掌握常用函数；
1. 常用函数
1）字符串
split，分割字符串为数组，split(“a|b|c”, “|”) => 返回数组 [a, b, c]
参数1：待分割到字符串；
参数2：分割字符，因为"|“在Java中是特殊字符，所以需要进行转义，转义使用两个”"；
substr，取子字符串，substr(moviename, -5, 4)
参数1：原字符串；
参数2：截取的开始位置，如果是负数，则从右往左计数，如-1表示最后一个字符，-2表示倒数第二个字符；
参数3：截取长度；
示例：
0: jdbc:hive2://master:10000> select *, substr(moviename, -5, 4) as year from t_movie limit 5;
+------------------+-------------------------------------+-------------------------------+-------+
| t_movie.movieid | t_movie.moviename | t_movie.movietype | year |
+------------------+-------------------------------------+-------------------------------+-------+
| 1 | Toy Story (1995) | Animation|Children's|Comedy | 1995 |
| 2 | Jumanji (1995) | Adventure|Children's|Fantasy | 1995 |
| 3 | Grumpier Old Men (1995) | Comedy|Romance | 1995 |
| 4 | Waiting to Exhale (1995) | Comedy|Drama | 1995 |
| 5 | Father of the Bride Part II (1995) | Comedy | 1995 |
+------------------+-------------------------------------+-------------------------------+-------+
2）时间函数
year，获取时间的年份；
month，获取时间的月份；
from_unixtime，将时间戳转换为时间；
unix_timestamp()：获取当前时间戳；
unix_timestamp(string date)：时间转换为时间戳；
获取当前日期 & 时间：
当前日期：
0: jdbc:hive2://master:10000> select current_date();
+-------------+
| _c0 |
+-------------+
| 2019-09-21 |
+-------------+
当前时间：
0: jdbc:hive2://master:10000> select current_timestamp();
+--------------------------+
| _c0 |
+--------------------------+
| 2019-09-21 18:05:27.768 |
+--------------------------+
当前时间戳：
0: jdbc:hive2://master:10000> select unix_timestamp();
+-------------+
| _c0 |
+-------------+
| 1569060416 |
+-------------+
从时间中获取年份：
0: jdbc:hive2://master:10000> select year("2019-09-21 18:05:27.768 ") as year;
+-------+
| year |
+-------+
| 2019 |
+-------+
从时间戳中获取月份：
0: jdbc:hive2://master:10000> select month(from_unixtime(1569060416)) as month;
+--------+
| month |
+--------+
| 9 |
+--------+
3）聚合函数
sum、avg等；
4）explode
将数组等拆分为多行
0: jdbc:hive2://master:10000> select m.*, t.type from t_movie m lateral view explode(split(movietype, "\|")) t as type limit 10;
+------------+---------------------------+-------------------------------+-------------+
| m.movieid | m.moviename | m.movietype | t.type |
+------------+---------------------------+-------------------------------+-------------+
| 1 | Toy Story (1995) | Animation|Children's|Comedy | Animation |
| 1 | Toy Story (1995) | Animation|Children's|Comedy | Children's |
| 1 | Toy Story (1995) | Animation|Children's|Comedy | Comedy |
| 2 | Jumanji (1995) | Adventure|Children's|Fantasy | Adventure |
| 2 | Jumanji (1995) | Adventure|Children's|Fantasy | Children's |
| 2 | Jumanji (1995) | Adventure|Children's|Fantasy | Fantasy |
| 3 | Grumpier Old Men (1995) | Comedy|Romance | Comedy |
| 3 | Grumpier Old Men (1995) | Comedy|Romance | Romance |
| 4 | Waiting to Exhale (1995) | Comedy|Drama | Comedy |
| 4 | Waiting to Exhale (1995) | Comedy|Drama | Drama |
+------------+---------------------------+-------------------------------+-------------+
5）collect_set，可以理解为该函数实现了explode相反到功能;
collect_list：可以包含重复数据；collect_set：去重；
将多行某字段到数据，合并为一个数组，需要结合group by进行分组，以确定合并到行到范围。
0: jdbc:hive2://master:10000> select moviename, collect_set(type) as types from (select m.*, t.type from t_movie m lateral view explode(split(movietype, "\|")) t as type limit 10) t group by moviename;
+---------------------------+---------------------------------------+
| moviename | types |
+---------------------------+---------------------------------------+
| Jumanji (1995) | ["Children's","Adventure","Fantasy"] |
| Toy Story (1995) | ["Comedy","Children's","Animation"] |
| Grumpier Old Men (1995) | ["Comedy","Romance"] |
| Waiting to Exhale (1995) | ["Drama","Comedy"] |
+---------------------------+---------------------------------------+
2. 常见场景及方法
2. 常见场景及方法
1）简单条件过滤；
使用where，显示movieid为1到电影名：
0: jdbc:hive2://master:10000> select moviename from t_movie where movieid = 1;
+-------------------+
| moviename |
+-------------------+
| Toy Story (1995) |
+-------------------+
2）联表条件过滤；
使用join on，获取评分为5的电影名：
select moviename, rate from t_rating r join t_movie m on r.movieid=m.movieid where rate=5 limit 5;
+-----------------------------------------+-------+
| moviename | rate |
+-----------------------------------------+-------+
| One Flew Over the Cuckoo's Nest (1975) | 5.0 |
| Bug's Life, A (1998) | 5.0 |
| Ben-Hur (1959) | 5.0 |
| Christmas Story, A (1983) | 5.0 |
| Beauty and the Beast (1991) | 5.0 |
+-----------------------------------------+-------+
3）分组统计；
使用group by和聚合函数
获取电影的平均评分：
select movieid, avg(rate) avg_rate from t_rating group by movieid order by movieid limit 5;
+----------+---------------------+
| movieid | avg_rate |
+----------+---------------------+
| 1 | 4.146846413095811 |
| 2 | 3.20114122681883 |
| 3 | 3.01673640167364 |
| 4 | 2.7294117647058824 |
| 5 | 3.0067567567567566 |
+----------+---------------------+
4）组内TopN；
使用row_number，rank，dense_rank；
获取各部门工资最高的三名员工：
select * from (select deptid, name, (salary+nvl(bonus, 0)) salary, dense_rank() over(partition by deptid order by salary desc) as rank from emp) t where rank<=3;
+-----------+---------+-----------+---------+
| t.deptid | t.name | t.salary | t.rank |
+-----------+---------+-----------+---------+
| 10 | KING | 5000.0 | 1 |
| 10 | CLARK | 2450.0 | 2 |
| 10 | MILLER | 1300.0 | 3 |
| 20 | FORD | 3000.0 | 1 |
| 20 | SCOTT | 3000.0 | 1 |
| 20 | JONES | 2975.0 | 2 |
| 20 | ADAMS | 1100.0 | 3 |
| 30 | BLAKE | 2850.0 | 1 |
| 30 | ALLEN | 1900.0 | 2 |
| 30 | TURNER | 1500.0 | 3 |
+-----------+---------+-----------+---------+
5）累加；
使用sum() over(partition by order by)
获取每个人按月累计消费：
select name, dt, cost, sum(cost) over(partition by name, month(dt) order by cost) as sum from t_order order by name;
+-------+-------------+-------+------+
| name | dt | cost | sum |
+-------+-------------+-------+------+
| jack | 2015-01-01 | 10 | 10 |
| jack | 2015-01-05 | 46 | 56 |
| jack | 2015-01-08 | 55 | 111 |
| jack | 2015-02-03 | 23 | 23 |
| jack | 2015-04-06 | 42 | 42 |
| mart | 2015-04-08 | 62 | 62 |
| mart | 2015-04-09 | 68 | 130 |
| mart | 2015-04-11 | 75 | 205 |
| mart | 2015-04-13 | 94 | 299 |
注意，如果over语句中没有order by，则求取的是该月份的消费总金额，而不会出现累加。
select name, dt, cost, sum(cost) over(partition by name, month(dt)) as sum from t_order order by name;
+-------+-------------+-------+------+
| name | dt | cost | sum |
+-------+-------------+-------+------+
| jack | 2015-01-01 | 10 | 111 |
| jack | 2015-01-05 | 46 | 111 |
| jack | 2015-01-08 | 55 | 111 |
| jack | 2015-02-03 | 23 | 23 |
| jack | 2015-04-06 | 42 | 42 |
| mart | 2015-04-08 | 62 | 299 |
| mart | 2015-04-09 | 68 | 299 |
| mart | 2015-04-11 | 75 | 299 |
| mart | 2015-04-13 | 94 | 299 |