hiveSQL基本语句三-----连接查询--inner join、left join、full join、union all、union

一、(inner) join

连接2个表，取出公共部分

• 必须重命名
• on后连接条件键值唯一
• 连接前注意去重，提高效率
• inner可省略

select * from user_list_1 as a 
inner join user_list_2 as b
on a.user_id=b.user_id;

三表连接

select a.user_name 
from
(select distinct user_name from data1)as a 
inner join (select distinct user_name from data2) as b
on a.user_name=b.user_name
inner join (select distinct user_name from data3) as c 
on b.user_name=c.user_name;

二、left /right join

left join：以左表为全集，返回能匹配上的匹配结果，没匹配上的显示NULL；
应用：取出表1中存在，表2中不存在的项

select *
from user_list_1 a
left join user_list_2 b
on a.user_id=b.user_id;

三、full join全连接

并列将2个表的信息展示

select *
from user_list_1 a
full join user_list_2 b
on a.user_id=b.user_id;

coalesce函数
coalesce（expression_1，expression_2,…,expression_n)
依次取值，遇到非空立即返回并停止对后面参数取值。若所有参数都为空，最终返回一个空值

select coalesce(a.user_name,b.user_name)
from user_list_1 a
full join user_list_2 b
on a.user_id=b.user_id;

四、union all/union联合

增加行

• 字段名称必须一致
• 字段顺序必须一致
• 无连接条件

== union all和union的区别：==
union会去重且排序

• union all

select a.user_id,a.user_name
from user_list_1 a
union all
select b.user_id,b.user_name
from user_list_2 b;

• union

select a.user_id,a.user_name
from user_list_1 a
union
select b.user_id,b.user_name
from user_list_2 b;

练习
练习1:2019年购买后又退款的用户性别分布（user_trade/user_refund/user_info)

select c.sex,count(c.user_name)
from (select distinct user_name from user_trade where year(dt) =2019) a
join (select distinct user_name from user_refund where year(dt)=2019) b
on a.user_name=b.user_name
join user_info c
on b.user_name=c.user_name
group by c.sex;

练习2：在2018年购买，但是没有在2019年购买的用户城市分布（user_trade、user_info）

select d.city,count(d.user_name)
from
(select a.user_name
from
(select distinct user_name from user_trade where year(dt) =2018) a
left join (select distinct user_name from user_trade where year(dt) =2019) b
on a.user_name=b.user_name
where b.user_name is NULL) c
join user_info d
on c.user_name=d.user_name
group by d.city;

练习3:2017-2019年，有交易但是没有退款的手机品牌分布（trade_2017/trade_2018/trade_2019/user_refund/user_info)

select d.phone_brand,count(d.user_name)
from 
(select a.user_name
from
(select user_name from trade_2017
union
select user_name from trade_2018
union
select user_name from trade_2019) a
left join
(select distinct user_name from user_refund where dt > '0') b
on a.user_name=b.user_name
where b.user_name is NULL) c
join
(select user_name,get_json_object(extra1,'$.phonebrand') phone_brand from user_info) d
on c.user_name=d.user_name
group by d.phone_brand;

最开始的思路，user_info不用提前处理，由于’ . p h o n e b r a n d ′ 中 .phonebrand'中 .phonebrand′中后没加点失败，加了后成功，注意注意

select get_json_object(d.extra1,'$.phonebrand'),count(d.user_name)
from 
(select a.user_name
from
(select user_name from trade_2017
union
select user_name from trade_2018
union
select user_name from trade_2019) a
left join
(select distinct user_name from user_refund where year(dt) >= 2017 and year(dt)<= 2019) b
on a.user_name=b.user_name
where b.user_name is NULL) c
join user_info d
on c.user_name=d.user_name
group by get_json_object(d.extra1,'$.phonebrand');