hive官方函数解释
hive官网函数大全地址： hive官网函数大全地址
示例
1、map(key1, value1, key2, value2, …)
SELECT map('name', '张三', 'age', 20, 'gender', '男') AS student;
---结果：
student
{"age":"20","gender":"男","name":"张三"}
2、map_values(Map<K.V>)
SELECT map_keys(map('name', '张三', 'age', 20, 'gender', '男')) AS keys;
---结果：
keys
["name","age","gender"]
3、map_values(Map<K.V>)
SELECT map_values(map('name', '张三', 'age', 20, 'gender', '男')) AS values;
---结果：
values
["张三","20","男"]
4、str_to_map(str, delimiter1, delimiter2)
str_to_map 函数用于将一个字符串转换为 Map 对象。具体来说，str_to_map 函数会将一个由键值对组成的字符串解析成一个 Map 对象，其中键和值之间使用指定的分隔符进行分隔。其中，str 是要转换的字符串，delimiter1 是键值对之间的分隔符，delimiter2 是键和值之间的分隔符。默认情况下，delimiter1 的值是 ‘,’，delimiter2 的值是 ‘:’。
SELECT str_to_map('name:张三,age:20,gender:男', ',', ':') AS student;
---结果：
student
{"age":"20","gender":"男","name":"张三"}
SELECT str_to_map('name=张三,age=20,gender=男', ',', '=') AS student;
---结果：
student
{"age":"20","gender":"男","name":"张三"}
5、explode (map)
select explode(map('A',10,'B',20,'C',30));
select explode(map('A',10,'B',20,'C',30)) as (key,value);
select tf.* from (select 0) t lateral view explode(map('A',10,'B',20,'C',30)) tf;
select tf.* from (select 0) t lateral view explode(map('A',10,'B',20,'C',30)) tf as key,value;
---上述四个结果均为：
key value
A 10
B 20
C 30
实战
给出一组学生数据，有名字，课程，等级，分数等字段，现在求每门课的情况，包含平均成绩，及这门课包含哪些学生及学生的等级
with stud as
( select 'zhang3' as name ,'优' as grade ,'math' as course ,'88' as score
union all
select 'li4' as name ,'良' as grade ,'math' as course ,'72' as score
union all
select 'zhao6' as name ,'差' as grade ,'math' as course ,'44' as score
union all
select 'wang5' as name ,'优' as grade ,'chinese' as course ,'80' as score
union all
select 'zhao6' as name ,'优' as grade ,'chinese' as course ,'55' as score
union all
select 'tian7' as name ,'优' as grade ,'chinese' as course ,'75' as score
)
--sql1
select course, collect_set(concat(name,':',grade)) as collect , avg(score) from stud group by course;
---结果：
course collect avg(score)
math ["li4:良","zhao6:差","zhang3:优"] 68.0
chinese ["wang5:优","tian7:优","zhao6:优"] 70.0
----sql2
select course, concat_ws(',',collect_set(concat(name,':',grade))) as strings , avg(score) from stud group by course;
---结果：
course strings avg(score)
math li4:良,zhao6:差,zhang3:优 68.0
chinese wang5:优,tian7:优,zhao6:优 70.0
----sql3
select course, str_to_map(concat_ws(',',collect_set(concat(name,':',grade))),',',':') as maps , avg(score) from stud group by course;
---结果：
course maps avg(score)
math {"li4":"良","zhang3":"优","zhao6":"差"} 68.0
chinese {"tian7":"优","wang5":"优","zhao6":"优"} 70.0
注意：
第一种sql,collect 字段的类型是array;第二种sql,strings字段的类型是string;第三种sql,maps字段的类型是map;
问题来了，能否在第二种的基础上，实现第一种和第三种的结果，且字段类型是string;
下面实现第二种转化为第三种，实际上就是map格式转换成json字符串；
with stud as
( select 'zhang3' as name ,'优' as grade ,'math' as course ,'88' as score
union all
select 'li4' as name ,'良' as grade ,'math' as course ,'72' as score
union all
select 'zhao6' as name ,'差' as grade ,'math' as course ,'44' as score
union all
select 'wang5' as name ,'优' as grade ,'chinese' as course ,'80' as score
union all
select 'zhao6' as name ,'优' as grade ,'chinese' as course ,'55' as score
union all
select 'tian7' as name ,'优' as grade ,'chinese' as course ,'75' as score
)
select
course
,concat('{"',string2,'"}') as string3
from
(select
course
,regexp_replace(string1,'\\,','\\"\\,\\"') as string2
from
(
select
course,
concat_ws(',', collect_list(concat_ws('":"', k,v) ) ) as string1
from (
select course, str_to_map(concat_ws(',',collect_set(concat(name,':',grade))),',',':') as maps , avg(score)
from stud group by course
)test_map_1
lateral view outer explode(maps) kv as k,v
group by course
) tt
) tm
---结果：
course string3
math {"li4":"良","zhang3":"优","zhao6":"差"}
chinese {"tian7":"优","wang5":"优","zhao6":"优"}