Hive综合应用案例——用户学历查询

1 查询每一个用户从出生到现在的总天数

---------- 禁止修改 ----------
 drop database if exists mydb cascade;
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---------- begin ----------
---创建mydb数据库
create database if not exists mydb;
---使用mydb数据库
use mydb;



---创建表user
create table usertab(
    id string,
    sex string,
    time string,
    education string,
    occupation string,
    income string,
    area string,
    desired_area string,
    city_countryside string
)
row format delimited fields terminated by ',';



---导入数据：/root/data.txt
load data local inpath '/root/data.txt' into table usertab;



--查询每一个用户从出生到2019-06-10的总天数
select id, datediff('2019-06-10',regexp_replace(time, '/', '-')) from usertab;

2 同一个地区相同的教育程度的最高收入

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 drop database if exists mydb cascade;
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---------- begin ----------



--创建mydb数据库
create database if not exists mydb;



---使用mydb数据库
use mydb;



---创建表user
create table usertab1(
    id int,
    sex string,
    time string,
    education string,
    occupation string,
    income string,
    area string,
    desired_area string,
    city_countryside string
)
row format delimited fields terminated by ',';



---导入数据：/root/data.txt
load data local inpath '/root/data1.txt' into table usertab1;



--同一个地区相同的教育程度的最高收入
select area,education,income
from(
    select area,education,income,
    row_number() over(
        partition by area, education order by income desc
    ) as t1
    from usertab1
) as t2
where t2.t1 = 1;



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3 统计各级学历所占总人数百分比

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 drop database if exists mydb cascade;
 set hive.mapred.mode=nonstrict;
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---------- begin ----------



--创建mydb数据库
create database if not exists mydb;



---使用mydb数据库
use mydb;



---创建表user
create table usertab2(
    id int,
    sex string,
    time string,
    education string,
    occupation string,
    income string,
    area string,
    desired_area string,
    city_countryside string
)
row format delimited fields terminated by ',';



---导入数据：/root/data.txt
load data local inpath '/root/data.txt' into table usertab2;



--统计各级学历所占总人数百分比(对结果保留两位小数)
select concat(round(t1.cnted * 100 / t2.cnt, 2),'%'), t1.education
from
    (
        select count(*) as cnted,education
        from usertab2
        group by education
    ) as t1,

    (
        select count(*) as cnt from usertab2
    ) as t2
order by t1.education;



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