Calculate open cases per day - mysql

Question

I am working on queries which are based on time and I would like to get the optimal way to calculate opened cases during the day. I do have table task_interval which has 2 columns start and end.
JSON example:
[
{
"start" : "2019-10-15 20:41:38",
"end" : "2019-10-16 01:44:03"
},
{
"start" : "2019-10-15 20:43:52",
"end" : "2019-10-15 22:18:54"
},
{
"start" : "2019-10-16 20:21:38",
"end" : null,
},
{
"start" : "2019-10-17 01:42:35",
"end" : null
},
{
"create_time" : "2019-10-17 03:15:57",
"end_time" : "2019-10-17 04:14:17"
},
{
"start" : "2019-10-17 03:16:44",
"end" : "2019-10-17 04:14:31"
},
{
"start" : "2019-10-17 04:15:23",
"end" : "2019-10-17 04:53:28"
},
{
"start" : "2019-10-17 04:15:23",
"end" : null,
},
]
The result of query should return:
[
{ time: '2019-10-15', value: 1 },
{ time: '2019-10-16', value: 1 }, // Not 2! One task from 15th has ended
{ time: '2019-10-17', value: 3 }, // We take 1 continues task from 16th and add 2 from 17th which has no end in same day
]
I have written the query which will return cumulative sum of started tasks which end date is not same as started date:
SELECT
time,
#running_total:=#running_total + tickets_number AS cumulative_sum
FROM
(SELECT
CAST(ti.start AS DATE) start,
COUNT(*) AS tickets_number
FROM
ticket_interval ti
WHERE
DATEDIFF(ti.start, ti.end) != 0
OR ti.end IS NULL
GROUP BY CAST(ti.start AS DATE)) X
JOIN
(SELECT #running_total:=0) total;

Answer 1

If you are running MySQL 8.0, one option is to unpivot, then aggregate and perform a window sum to compute the running count:
select
date(dt) dt_day,
sum(sum(no_tasks)) over(order by date(dt)) no_tasks
from (
select start_dt dt, 1 no_tasks from mytable
union all select end_dt, -1 from mytable where end_dt is not null
) t
group by date(dt)
order by dt_day
Side note: start and end are reserved words, hence not good choices for column names. I renamed those to start_dt and end_dt.
In earlier versions, we can emulate the window sum with a user variable, like so:
select
dt_day,
#no_tasks := #no_tasks + no_tasks no_tasks
from (
select date(dt) dt_day, sum(no_tasks) no_tasks
from (
select start_dt dt, 1 no_tasks from mytable
union all select end_dt, -1 from mytable where end_dt is not null
) t
group by dt_day
order by dt_day
) t
cross join (select #no_tasks := 0) x
order by dt_day
Demo on DB Fiddle - both queries yield:
dt_day | no_tasks
:--------- | -------:
2019-10-15 | 1
2019-10-16 | 1
2019-10-17 | 3