hive详细笔记(七)-Hive之case when语法详解(附带讲解视频 )
1 基本语法
语法一
CASE 字段
WHEN 值1 THEN 值1
[WHEN 值2 THEN 值2]*
[ELSE 值]
END 语法二
CASE
WHEN 条件表达式 THEN 值1
[WHEN 条件表达式 [and or] 条件表达式THEN 值2]*
[ELSE 值]
END 2 实例一
2.1 数据和需求
数据
悟空 A 男
娜娜 A 男
宋宋 B 男
凤姐 A 女
热巴 B 女
慧慧 B 女
求出不同部门男女各多少人。结果如下：
dpt 男 女
A 2 1
B 1 2
2.2 建表导入数据
create table tb_case_when_demo(
name string ,
dname string ,
gender string
)
row format delimited fields terminated by "\t" ;
load data local inpath "/hive/data/casedemo.txt" into table tb_case_when_demo ;
0: jdbc:hive2://linux01:10000> select * from tb_case_when_demo ;
OK
+-------------------------+--------------------------+---------------------------+
| tb_case_when_demo.name | tb_case_when_demo.dname | tb_case_when_demo.gender |
+-------------------------+--------------------------+---------------------------+
| 悟空 | A | 男 |
| 娜娜 | A | 男 |
| 宋宋 | B | 男 |
| 凤姐 | A | 女 |
| 热巴 | B | 女 |
| 慧慧 | B | 女 |
+-------------------------+--------------------------+---------------------------+2.3 case when实现一
select
dname ,
sum(case gender when '男' then 1 else 0 end) as m_cnts ,
sum(case gender when '女' then 1 else 0 end) as f_cnts
from
tb_case_when_demo
group by dname ;
+--------+---------+---------+
| dname | m_cnts | f_cnts |
+--------+---------+---------+
| A | 2 | 1 |
| B | 1 | 2 |
+--------+---------+---------+2.4 case when实现二
select
dname ,
sum(case when gender='男' then 1 else 0 end) as m_cnts ,
sum(case when gender='女' then 1 else 0 end) as f_cnts ,
case when dname='A' then '教学部' else '后勤部' end as ch
from
tb_case_when_demo
group by dname ;
+--------+---------+---------+------+
| dname | m_cnts | f_cnts | ch |
+--------+---------+---------+------+
| A | 2 | 1 | 教学部 |
| B | 1 | 2 | 后勤部 |
+--------+---------+---------+------+2.5 补充if语法
2) if 参数一条件判断 参数二 成立的结果 参数三不成立的结果
0: jdbc:hive2://linux01:10000> select if(1=1,1,0);
OK
+------+
| _c0 |
+------+
| 1 |
+------+
1 row selected (0.142 seconds)
0: jdbc:hive2://linux01:10000> select if(1=2,1,0);
OK
+------+
| _c0 |
+------+
| 0 |
+------+2.6 if实现上面的需求
select
dname,
sum(1) cnts ,
sum(if(gender='男', 1 , 0)) as M_CNTS ,
sum(if(gender='女', 1 , 0)) as F_CNTS
from
tb_case_when_demo
group by dname ;
+--------+-------+---------+---------+
| dname | cnts | m_cnts | f_cnts |
+--------+-------+---------+---------+
| A | 3 | 2 | 1 |
| B | 3 | 1 | 2 |
+--------+-------+---------+---------+3 练习
3.1 数据
gz.txt 用户工资组成表
uid jb jj tc deptno
1,2000,3000,1500,1
2,5000,500,1000,2
3,1500,1000,3000,2
4,3000,6000,8000,3
5,1500,2000,1800,1
6,2500,1000,1900,1
bm.txt 部门表
1,销售
2,技术
3,行政
yg.txt 员工信息表
uid name gender age
1,zs,M,28
2,ww,F,36
3,zl,F,48
4,pp,M,44
5,wb,M,32
6,TQ,F,32
3.2 建表导入数据
drop table gz ;
create table gz(
uid int,
jb int,
jj int,
tc int,
deptno int
)
row format delimited fields terminated by ",";
load data local inpath "/hive/data/gz.txt" into table gz;
drop table bm ;
create table bm(
deptno string ,
name string
)
row format delimited fields terminated by ",";
load data local inpath "/hive/data/bm.txt" into table bm;
create table yg(
uid int,
name string,
gender string,
age int
)
row format delimited fields terminated by ",";
load data local inpath "/hive/data/yg.txt" into table yg;
0: jdbc:hive2://linux01:10000> select * from gz ;
OK
+---------+--------+--------+--------+------------+
| gz.uid | gz.jb | gz.jj | gz.tc | gz.deptno |
+---------+--------+--------+--------+------------+
| 1 | 2000 | 3000 | 1500 | 1 |
| 2 | 5000 | 500 | 1000 | 2 |
| 3 | 1500 | 1000 | 3000 | 2 |
| 4 | 3000 | 6000 | 8000 | 3 |
| 5 | 1500 | 2000 | 1800 | 1 |
| 6 | 2500 | 1000 | 1900 | 1 |
+---------+--------+--------+--------+------------+
6 rows selected (0.223 seconds)
0: jdbc:hive2://linux01:10000> select * from bm ;
OK
+------------+----------+
| bm.deptno | bm.name |
+------------+----------+
| 1 | 销售 |
| 2 | 技术 |
| 3 | 行政 |
+------------+----------+
3 rows selected (0.2 seconds)
0: jdbc:hive2://linux01:10000> select * from yg ;
OK
+---------+----------+------------+---------+
| yg.uid | yg.name | yg.gender | yg.age |
+---------+----------+------------+---------+
| 1 | zs | M | 28 |
| 2 | ww | F | 36 |
| 3 | zl | F | 48 |
| 4 | pp | M | 44 |
| 5 | wb | M | 32 |
| 6 | TQ | F | 32 |
+---------+----------+------------+---------+3.3 练习
3.3.1求出公司中每个员工的姓名 和 三类收入中最高的那种收入的类型 greatest
方式一
select
name ,
greatest(jb, jj,tc) max_gz ,
case when greatest(jb, jj,tc) = jb then 'jb'
when greatest(jb, jj,tc) = tc then 'tc'
when greatest(jb, jj,tc) = jj then 'jj'
else "_"
end as gz_category
from
(select
yg.name ,
gz.jb ,
gz.jj ,
gz.tc
from
yg
join
gz
on
yg.uid = gz.uid)t ;
+-------+---------+--------------+
| name | max_gz | gz_category |
+-------+---------+--------------+
| zs | 3000 | jj |
| ww | 5000 | jb |
| zl | 3000 | tc |
| pp | 8000 | tc |
| wb | 2000 | jj |
| TQ | 2500 | jb |
+-------+---------+--------------+方式二
select
yg.uid ,
yg.name ,
t.max_gz ,
t.category
from
yg
join
(select
uid ,
greatest(jb , jj ,tc) as max_gz ,
case when greatest(jb , jj ,tc) = jb then 'jb'
when greatest(jb , jj ,tc) = jj then 'jj'
when greatest(jb , jj ,tc) = tc then 'tc'
end as category
from
gz) t
on t.uid = yg.uid ;
+---------+----------+-----------+-------------+
| yg.uid | yg.name | t.max_gz | t.category |
+---------+----------+-----------+-------------+
| 1 | zs | 3000 | jj |
| 2 | ww | 5000 | jb |
| 3 | zl | 3000 | tc |
| 4 | pp | 8000 | tc |
| 5 | wb | 2000 | jj |
| 6 | TQ | 2500 | jb |
+---------+----------+-----------+-------------+3.3.2求出公司中每个岗位的薪资总和
select
bm.deptno ,
bm.name ,
t.sum_gz
from
bm
join
(select
deptno,
sum(jj+tc+jb) sum_gz
from
gz
group by deptno)t
on t.deptno = bm.deptno ;
+------------+----------+-----------+
| bm.deptno | bm.name | t.sum_gz |
+------------+----------+-----------+
| 2 | 技术 | 12000 |
| 3 | 行政 | 17000 |
| 1 | 销售 | 17200 |
+------------+----------+-----------+3.3.3求出公司中每个岗位不同性别员工薪资总和
select
bm_name ,
gender ,
sum(jb+jj+tc) bm_gender_gz
from
(select
yg.uid ,
yg.name yg_name,
yg.gender ,
gz.jb ,
gz.jj ,
gz.tc ,
bm.name as bm_name
from
yg
join
gz
join
bm
on yg.uid = gz.uid
and gz.deptno = bm.deptno) t
group by bm_name , gender ;
+----------+---------+---------------+
| bm_name | gender | bm_gender_gz |
+----------+---------+---------------+
| 销售 | F | 5400 |
| 行政 | M | 17000 |
| 销售 | M | 11800 |
| 技术 | F | 12000 |
+----------+---------+---------------+3.3.4求出公司中不同性别、不同年龄阶段（20-30,31-40,41-50）的员工薪资总和
with t1 as (
select
*,
case
when age >=20 and age<30 then '20~30'
when age >=30 and age <40 then '30~40'
when age >=40 and age <50 then '40~50'
else '退休'
end as stage
from
yg
)
select
t1.stage ,
t1.gender,
sum(jb+jj+tc) stage_zg
from
t1
join
gz
on
gz.uid = t1.uid
group by t1.stage , t1.gender ;
+-----------+------------+-----------+
| t1.stage | t1.gender | stage_zg |
+-----------+------------+-----------+
| 30~40 | F | 11900 |
| 40~50 | M | 17000 |
| 30~40 | M | 5300 |
| 20~30 | M | 6500 |
| 40~50 | F | 5500 |
+-----------+------------+-----------+补充 : with 用法
WITH
t1 as (SELECT * FROM Table1) ,
t2 as (SELECT * FROM Table2)
SELECT t1.*
FROM t1
LEFT JOIN t2 ON t1.id=t2.id ;例如
with t1 as (select
yg.uid ,
yg.name yg_name,
yg.gender ,
gz.jb ,
gz.jj ,
gz.tc ,
bm.name as bm_name
from
yg
join
gz
join
bm
on yg.uid = gz.uid
and gz.deptno = bm.deptno)
t2 as (select .......)
t3 as (select ......)
select t1 join t2 ...