| 数据准备 | |
|---|---|
| id | login_date |
| 01 | 2021-02-28 |
| 01 | 2021-03-01 |
| 01 | 2021-03-02 |
| 01 | 2021-03-04 |
| 01 | 2021-03-05 |
| 01 | 2021-03-06 |
| 01 | 2021-03-08 |
| 02 | 2021-03-01 |
| 02 | 2021-03-02 |
| 02 | 2021-03-03 |
| 02 | 2021-03-06 |
| 03 | 2021-03-06 |
方案一
1.先把数据按照用户id分组，根据登录日期排序
SQL：
SELECT
id,
login_date,
row_number() over(partition by id order by login_date asc) as rn
FROM data;
| 结果： | ||
|---|---|---|
| id | login_date | rn |
| 01 | 2021-02-28 | 1 |
| 01 | 2021-03-01 | 2 |
| 01 | 2021-03-02 | 3 |
| 01 | 2021-03-04 | 4 |
| 01 | 2021-03-05 | 5 |
| 01 | 2021-03-06 | 6 |
| 01 | 2021-03-08 | 7 |
| 02 | 2021-03-01 | 1 |
| 02 | 2021-03-02 | 2 |
| 02 | 2021-03-03 | 3 |
| 02 | 2021-03-06 | 4 |
| 03 | 2021-03-06 | 1 |
2.用登录日期与rn求date_sub，得到的差值日期如果是相等大数据培训的，则说明这两天肯定是连续的
SQL：
SELECT
t1.id,
t1.login_date,
date_sub(t1.login_date, rn) as diff_dateFROM
(
SELECT
id,
login_date,
row_number() over(partition by id order by login_date asc) as rn
FROM data
) t1;| 结果： | ||
|---|---|---|
| id | login_date | diff_date |
| 01 | 2021-02-28 | 2021-02-27 |
| 01 | 2021-03-01 | 2021-02-27 |
| 01 | 2021-03-02 | 2021-02-27 |
| 01 | 2021-03-04 | 2021-02-28 |
| 01 | 2021-03-05 | 2021-02-28 |
| 01 | 2021-03-06 | 2021-02-28 |
| 01 | 2021-03-08 | 2021-03-01 |
| 02 | 2021-03-01 | 2021-02-28 |
| 02 | 2021-03-02 | 2021-02-28 |
| 02 | 2021-03-03 | 2021-02-28 |
| 02 | 2021-03-06 | 2021-03-02 |
| 03 | 2021-03-06 | 2021-03-05 |
3.根据id和日期差date_diff分组，登录次数即为分组后的count(1)
SQL：
SELECT
t2.id,
count(1) as login_times,
min(t2.login_date) as start_date,
max(t2.login_date) as end_date
FROM
(
SELECT
t1.id,
t1.login_date,
date_sub(t1.login_date,rn) as diff_date
FROM
(
SELECT
id,
login_date,
row_number() over(partition by id order by login_date asc) as rn
FROM data
) t1) t2
group by t2.id, t2.diff_date
having login_times >= 3;
| 结果： | |||
|---|---|---|---|
| id | login_times | start_date | end_date |
| 01 | 3 | 2021-02-28 | 2021-03-02 |
| 01 | 3 | 2021-03-04 | 2021-03-06 |
| 02 | 3 | 2021-03-01 | 2021-03-03 |
方案二
方案二利用lag和lead函数进行处理，思路类似。
SQL：
SELECT
id,
lag_login_date,
login_date,lead_login_date
FROM
(SELECT
id,
login_date,
lag(login_date,1,login_date) over(partition by id order by login_date) as lag_login_date,
lead(login_date,1,login_date) over(partition by id order by login_date) as lead_login_date
FROM data
) t1| where datediff(login_date,lag_login_date) =1 and datediff(lead_login_date,login_date) =1;结果： | |||
|---|---|---|---|
| id | lag_login_date | login_date | lead_login_date |
| 01 | 2018-02-28 | 2018-03-01 | 2018-03-02 |
| 01 | 2018-03-04 | 2018-03-05 | 2018-03-06 |
| 02 | 2018-03-01 | 2018-03-02 | 2018-03-03 |
行转列和列转行
以"连续登录"中的数据为例：
select id,
concat_ws(',',collect_list(login_date)) cwfrom data
| group by id;结果： | |
|---|---|
| id | cw |
| 01 | 2018-02-28,2018-03-01,2018-03-02,2018-03-04,2018-03-05,2018-03-06,2018-03-08 |
| 02 | 2018-03-01,2018-03-02,2018-03-03,2018-03-06 |
| 03 | 2018-03-06 |
以上面SQL生成的数据为基准，执行下列SQL：
select id, login_date
from t
| lateral view explode(split(cw,',')) b AS login_date;结果： | |
|---|---|
| id | login_date |
| 01 | 2018-02-28 |
| 01 | 2018-03-01 |
| 01 | 2018-03-02 |
| 01 | 2018-03-04 |
| 01 | 2018-03-05 |
| 01 | 2018-03-06 |
| 01 | 2018-03-08 |
| 02 | 2018-03-01 |
| 02 | 2018-03-02 |
| 02 | 2018-03-03 |
| 02 | 2018-03-06 |
| 03 | 2018-03-06 |