連續活躍登陸的用戶指至少連續2天都活躍登錄的用戶
解決類似場景的問題
創建數據
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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、因為每天用戶登錄次數可能不止一次，所以需要先將用戶每天的登錄日期去重。
2、再用row_number() over(partition by _ order by _)函數將用戶id分組，按照登陸時間進行排序。
3、計算登錄日期減去第二步驟得到的結果值，用戶連續登陸情況下，每次相減的結果都相同。
4、按照id和日期分組並求和，篩選大於等於2的即為連續活躍登陸的用戶。
第一步：用戶登錄日期去重
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select
DISTINCT
dt,user_id
from
test5active;
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第二步：用row_number() over()函數計數
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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：日期減去計數值得到結果
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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：根據id和結果分組並計算總和，大於等於2的即為連續登陸的用戶，得到 用戶id，開始日期，結束日期，連續登錄天數
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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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用戶id 開始日期 結束日期 連續登錄天數
最后：連續登陸的用戶
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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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