number of visitors gained and lost comparing with previous day - sql - mysql

Question

CREATE TABLE visitors (
Date DATE,
visitor VARCHAR(20));
INSERT INTO visitors VALUES
("2019-10-01", "v1"),
("2019-10-01", "v2"),
("2019-10-01", "v3"),
("2019-10-02", "v2"),
("2019-10-03", "v2"),
("2019-10-03", "v4"),
("2019-10-03", "v5");
please find Find number_of_visitors_per_date, number_of_visitors gained comparing with previous day, and number_of_visitors lost comparing with previous day. The expected results for the above table should be:
Date | number_of_visitors | number_of_visitors_gained | number_of_visitors_lost
2019-10-01 | 3 | 3 | 0
2019-10-02 | 1 | 0 | 2
2019-10-03 | 3 | 2 | 0
Obviously, the most challenging part is how to get the last two columns. Note, since there is no previous day for the first day, the number_of_visitors_gained is the total number of visitors in the first day, and the number_of_visitors_lost is 0.

Answer 1

If you RDBMS supports window functions, you can aggregate, and then use lag():
select
date,
number_of_visitors,
case
when lag(number_of_visitors) over(order by date) is null
then number_of_visitors
when lag(number_of_visitors) over(order by date) < number_of_visitors
then number_of_visitors - lag(number_of_visitors) over(order by date)
else 0
end number_of_visitors_gained,
case when lag(number_of_visitors) over(order by date) > number_of_visitors
then lag(number_of_visitors) over(order by date) - number_of_visitors
else 0
end number_of_visitors_lost
from (
select date, count(*) number_of_visitors
from visitors
group by date
) t
order by date
Demo on DB Fiddle:
date | number_of_visitors | number_of_visitors_gained | number_of_visitors_lost
:--------- | -----------------: | ------------------------: | ----------------------:
2019-10-01 | 3 | 3 | 0
2019-10-02 | 1 | 0 | 2
2019-10-03 | 3 | 2 | 0

Answer 2

You are looking for aggregation using lead() and lag() window functions:
select date, count(*) as visitors_on_day,
sum(case when prev_vdate is null or prev_vdate <> date - interval 1 day
then 1 else 0
end) as gained_visitors,
sum(case when next_vdate <> date + interval 1 day
then 1 else 0
end) as lost_visitors
from (select v.*,
lag(date) over (partition by visitor) as prev_vdate,
lead(date) over (partition by visitor) as next_vdate
from visitors v
) t
group by date
order by date;