get totals each day based on a given timestamp - mysql

Question

I have a simple table:
user | timestamp
===================
Foo | 1440358805
Bar | 1440558805
BarFoo | 1440559805
FooBar | 1440758805
I would like to get a view with total number of users each day:
date | total
===================
...
2015-08-23 | 1 //Foo
2015-08-24 | 1
2015-08-25 | 1
2015-08-26 | 3 //+Bar +BarFoo
2015-08-27 | 3
2015-08-28 | 4 //+FooBar
...
What I currently have is
SELECT From_unixtime(a.timestamp, '%Y-%m-%d') AS date,
Count(From_unixtime(a.timestamp, '%Y-%m-%d')) AS total
FROM thetable AS a
GROUP BY From_unixtime(a.timestamp, '%Y-%m-%d')
ORDER BY a.timestamp ASC
which counts only the user of a certain day:
date | total
===================
2015-08-23 | 1 //Foo
2015-08-26 | 2 //Bar +BarFoo
2015-08-28 | 1 //FooBar
I've prepared a sqlfiddle
EDIT
The solution by #splash58 returns this result:
date | #t:=coalesce(total, #t)
==================================
2015-08-23 | 1
2015-08-26 | 3
2015-08-28 | 4
2015-08-21 | 4
2015-08-22 | 4
2015-08-24 | 4
2015-08-25 | 4
2015-08-27 | 4
2015-08-29 | 4
2015-08-30 | 4

Answer 1

You can get the cumulative values by using variables:
SELECT date, total, (#cume := #cume + total) as cume_total
FROM (SELECT From_unixtime(a.timestamp, '%Y-%m-%d') as date, Count(*) AS total
FROM thetable AS a
GROUP BY From_unixtime(a.timestamp, '%Y-%m-%d')
) a CROSS JOIN
(SELECT #cume := 0) params
ORDER BY date;
This gives you the dates that are in your data. If you want additional dates (where no users start), then one way is a calendar table:
SELECT c.date, a.total, (#cume := #cume + coalesce(a.total, 0)) as cume_total
FROM Calendar c JOIN
(SELECT From_unixtime(a.timestamp, '%Y-%m-%d') as date, Count(*) AS total
FROM thetable AS a
GROUP BY From_unixtime(a.timestamp, '%Y-%m-%d')
) a
ON a.date = c.date CROSS JOIN
(SELECT #cume := 0) params
WHERE c.date BETWEEN '2015-08-23' AND '2015-08-28'
ORDER BY c.date;
You can also put the dates explicitly in the query (using a subquery), if you don't have a calendar table.

Answer 2

To save order of dates, i think, we need to wrap query in one more select
select date, #n:=#n + ifnull(total,0) total
from
(select Calendar.date, total
from Calendar
left join
(select From_unixtime(timestamp, '%Y-%m-%d') date, count(*) total
from thetable
group by date) t2
on Calendar.date= t2.date
order by date) t3
cross join (select #n:=0) n
Demo on sqlfiddle

Answer 3

You can use function
TIMESTAMPDIFF(DAY,`timestamp_field`, CURDATE())
You will not have to convert timestamp to other field dypes.

Answer 4

drop table if exists thetable;
create table thetable (user text, timestamp int);
insert into thetable values
('Foo', 1440358805),
('Bar', 1440558805),
('BarFoo', 1440559805),
('FooBar', 1440758805);
DROP PROCEDURE IF EXISTS insertTEMP;
DELIMITER //
CREATE PROCEDURE insertTEMP (first date, last date) begin
drop table if exists Calendar;
CREATE TEMPORARY TABLE Calendar (date date);
WHILE first <= last DO
INSERT INTO Calendar Values (first);
SET first = first + interval 1 day;
END WHILE;
END //
DELIMITER ;
call insertTEMP('2015-08-23', '2015-08-28');
select Calendar.date, #t:=coalesce(total, #t)
from Calendar
left join
(select date, max(total) total
from (select From_unixtime(a.timestamp, '%Y-%m-%d') AS date,
#n:=#n+1 AS total
from thetable AS a, (select #n:=0) n
order by a.timestamp ASC) t1
group by date ) t2
on Calendar.date= t2.date,
(select #t:=0) t
result
date, #t:=coalesce(total, #t)
2015-08-23 1
2015-08-24 1
2015-08-25 1
2015-08-26 3
2015-08-27 3
2015-08-28 4