How does SQL query continuously logged-in users?

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Take 3 consecutive days as an example, use the tool: MySQL.

1. Create the SQL table:

Create table if not exists orde (id varchar (10), date datetime,orders varchar (10)); insert into orde values ('1x, '2019); insert into orde values (' 1, '2019); insert into orde values (' 1, '20119); insert into orde values (' 1, '20119) Insert into orde values ('1hammer,' 20119); insert into orde values ('1, '20119); insert into orde values (' 4', '20119); insert into orde values (' 2mm, '20119) Insert into orde values ('3hammer,' 20119); insert into orde values ('4yen,' 20119); insert into orde values ('5yen,' 20119); insert into orde values ('6here,' 20119); insert into orde values ('7march,' 20119); insert into orde values ('858,' 20119) Insert into orde values ('999,' 20119); insert into orde values ('9, '20119); insert into orde values (' 9, '20119) Insert into orde values ('9', '2019); insert into orde values (' 9', '2019)

View the data sheet:

two。 Use the row_number () over () sort function to calculate the ranking of each id, SQL as follows:

Select *, row_number () over (partition by id order by date) 'rank'from ordewhere orders is not NULL

View the data sheet:

3. Subtract the date date field from the rank ranking field, SQL as follows:

Select *, DATE_SUB (a.dategery interval a.rank day) 'date_sub'from (select *, row_number () over (partition by id order by date)' rank'from ordewhere orders is not NULL) a

View the data:

4. According to id and date grouping and calculating the number of grouped (count), the earliest login time and the latest login time, the SQL is as follows:

Select b.idjimmin (date) 'start_time',max (date)' end_time',count (*) 'date_count'from (select *, DATE_SUB (a.datejure interval a.rank day)' date_sub'from (select *, row_number () over (partition by id order by date) 'rank'from ordewhere orders is not NULL) a) bgroup by b.datekeeper subjournal count (*) > = 3

View the data:

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