连续活跃登陆的用户指至少连续2天都活跃登录的用户
解决类似场景的问题
创建数据
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CREATE
TABLE
test5active(
dt string,
user_id string,
age
int
)
ROW format delimited fields terminated
BY
','
;
INSERT
INTO
TABLE
test5active
VALUES
(
'2019-02-11'
,
'user_1'
,23),(
'2019-02-11'
,
'user_2'
,19),
(
'2019-02-11'
,
'user_3'
,39),(
'2019-02-11'
,
'user_1'
,23),
(
'2019-02-11'
,
'user_3'
,39),(
'2019-02-11'
,
'user_1'
,23),
(
'2019-02-12'
,
'user_2'
,19),(
'2019-02-13'
,
'user_1'
,23),
(
'2019-02-15'
,
'user_2'
,19),(
'2019-02-16'
,
'user_2'
,19);
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思路一:
1、因为每天用户登录次数可能不止一次，所以需要先将用户每天的登录日期去重。
2、再用row_number() over(partition by _ order by _)函数将用户id分组，按照登陆时间进行排序。
3、计算登录日期减去第二步骤得到的结果值，用户连续登陆情况下，每次相减的结果都相同。
4、按照id和日期分组并求和，筛选大于等于2的即为连续活跃登陆的用户。
第一步：用户登录日期去重
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select
DISTINCT
dt,user_id
from
test5active;
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第二步：用row_number() over()函数计数
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select
t1.user_id,t1.dt,
row_number() over(partition
by
t1.user_id
order
by
t1.dt) day_rank
from
(
select
DISTINCT
dt,user_id
from
test5active
)t1;
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第三步：日期减去计数值得到结果
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select
t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank)
as
dis
from
(
select
t1.user_id,t1.dt,
row_number() over(partition
by
t1.user_id
order
by
t1.dt) day_rank
from
(
select
DISTINCT
dt,user_id
from
test5active
)t1)t2;
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第四步：根据id和结果分组并计算总和，大于等于2的即为连续登陆的用户，得到 用户id，开始日期，结束日期，连续登录天数
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select
t3.user_id,
min
(t3.dt),
max
(t3.dt),
count
(1)
from
(
select
t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank)
as
dis
from
(
select
t1.user_id,t1.dt,
row_number() over(partition
by
t1.user_id
order
by
t1.dt) day_rank
from
(
select
DISTINCT
dt,user_id
from
test5active
)t1
)t2
)t3
group
by
t3.user_id,t3.dis
having
count
(1)>1;
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用户id 开始日期 结束日期 连续登录天数
最后：连续登陆的用户
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select
distinct
t4.user_id
from
(
select
t3.user_id,
min
(t3.dt),
max
(t3.dt),
count
(1)
from
(
select
t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank)
as
dis
from
(
select
t1.user_id,t1.dt,
row_number() over(partition
by
t1.user_id
order
by
t1.dt) day_rank
from
(
select
DISTINCT
dt,user_id
from
test5active
)t1
)t2
)t3
group
by
t3.user_id,t3.dis
having
count
(1)>1
)t4;
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思路二：使用lag(向后)或者lead(向前)
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select
user_id,t1.dt,
lead(t1.dt) over(partition
by
user_id
order
by
t1.dt)
as
last_date_id
from
(
select
DISTINCT
dt,user_id
from
test5active
)t1;
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select
distinct
t2.user_id
from
(
select
user_id,t1.dt,
lead(t1.dt) over(partition
by
user_id
order
by
t1.dt)
as
last_date_id
from
(
select
DISTINCT
dt,user_id
from
test5active
)t1
)t2
where
datediff(last_date_id,t2.dt)=1;
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