一、连续7天登录用户

1.数据准备ulogin.dat

1 2019-07-11 1
1 2019-07-12 1
1 2019-07-13 1
1 2019-07-14 1
1 2019-07-15 1
1 2019-07-16 1
1 2019-07-17 1
1 2019-07-18 1
2 2019-07-11 1
2 2019-07-12 1
2 2019-07-13 0
2 2019-07-14 1
2 2019-07-15 1
2 2019-07-16 0
2 2019-07-17 1
2 2019-07-18 0
3 2019-07-11 1
3 2019-07-12 1
3 2019-07-13 1
3 2019-07-14 0
3 2019-07-15 1
3 2019-07-16 1
3 2019-07-17 1
3 2019-07-18 1

2.建表与加载数据

create table ulogin
(
    uid    int,
    dt     date,
    status int
) row format delimited fields terminated by " ";
load data local inpath "/opt/stufile/ulogin.dat" into table ulogin;

select * from ulogin;

3. 使用 row_number 在组内给数据编号(rownum)

select uid,
       dt,
       row_number() over (partition by uid order by dt) rownum
from ulogin
where status = 1;

4.某个值 - rownum = gid，得到结果可以作为后面分组计算的依据

select uid,
       date_sub(dt, row_number() over (partition by uid order by dt)) gid
from ulogin
where status = 1;

5.根据求得的gid，作为分组条件，求最终结果

select tmp.uid,
       count(*) contlogin
from (
         select uid,
                date_sub(dt, row_number() over (partition by uid order by dt)) gid
         from ulogin
         where status = 1) tmp
group by tmp.uid, tmp.gid
having contlogin >= 7;

二、求TopN

1.需求

编写 sql 语句实现每班前三名，分数一样并列，同时求出前三名按名次排序的分差。

2.数据准备

3.建表与加载数据

create table stu
(
    sno   int,
    class string,
    score int
) row format delimited fields terminated by " ";

load data local inpath '/opt/stufile/stu.dat' into table stu;

4、上排名函数，分数一样并列，所以用dense_rank

select class,
       score,
       dense_rank() over (partition by class order by score desc ) rank
from stu;

5、将上一行数据下移，相减即得到分数差

select t.class,
       t.score,
       t.rank,
       t.score - lag(t.score) over (partition by class order by score desc ) lagscore
from (
         select class,
                score,
                dense_rank() over (partition by class order by score desc ) rank
         from stu) t
where t.rank <= 3;

6、处理 NULL

select t.class,
       t.score,
       t.rank,
       nvl(t.score - lag(t.score) over (partition by class order by score desc ), 0) lagscore
from (
         select class,
                score,
                dense_rank() over (partition by class order by score desc ) rank
         from stu) t
where t.rank <= 3;