Find period since last pause in MySQL - php

Question

I need to find the period without pause in days since last pause.
I have a next table:
id | user | date
-----------------------
1 | 1 | 16.02.2017
1 | 1 | 15.02.2017
1 | 1 | 14.02.2017
1 | 1 | 13.02.2017
1 | 1 | 10.02.2017
Last pause: 10-13 February
Last period without pause: 4 days
I tried to found the difference between day like in this question, but as result it was always NULL. And this is only first part. For second part I thought to use something like ranking, but don't know if it will work.
I plan to use it with PHP 7 + MySQL 5.6.

Answer 1

I've used this sample:
create table if not exists myt(id int, dd date);
insert into myt values
(1, '2017-01-01'),
(1, '2017-01-02'),
(1, '2017-01-03'),
(1, '2017-01-04'),
(1, '2017-01-08'),
(1, '2017-01-09'),
(1, '2017-01-10');
First you should set a partition by consecutive days:
select id, dd,
if(#last_date = '1900-01-01' or datediff(dd, #last_date) = -1, #cn := #cn, #cn := +1) consecutive,
#last_date := dd
from
(select #last_date := '1900-01-01', #cn := 0) x,
(select id, dd
from myt
order by dd desc) y
;
This returns:
+----+---------------------+-------------+
| id | dd | consecutive |
+----+---------------------+-------------+
| 1 | 10.01.2017 00:00:00 | 0 |
| 1 | 09.01.2017 00:00:00 | 0 |
| 1 | 08.01.2017 00:00:00 | 0 |
+----+---------------------+-------------+
| 1 | 04.01.2017 00:00:00 | 1 |
| 1 | 03.01.2017 00:00:00 | 1 |
| 1 | 02.01.2017 00:00:00 | 1 |
| 1 | 01.01.2017 00:00:00 | 1 |
+----+---------------------+-------------+
After you set a partition, then get MAX and MIN date for each partition:
select id, min(dd) as ini, max(dd) as fin, datediff(max(dd), min(dd)) as Days
from (
select id, dd,
if(#last_date = '1900-01-01' or datediff(dd, #last_date) = -1, #cn := #cn, #cn := +1) consecutive,
#last_date := dd
from
(select #last_date := '1900-01-01', #cn := 0) x,
(select id, dd
from myt
order by dd desc) y
) z
group by consecutive
;
Result:
+----+---------------------+---------------------+------+
| id | ini | fin | Days |
+----+---------------------+---------------------+------+
| 1 | 08.01.2017 00:00:00 | 10.01.2017 00:00:00 | 2 |
+----+---------------------+---------------------+------+
| 1 | 01.01.2017 00:00:00 | 04.01.2017 00:00:00 | 3 |
+----+---------------------+---------------------+------+
Check it: http://rextester.com/XMIX80360

Answer 2

Try this query. It will find all pauses -
SELECT curr_date, prev_date FROM (
SELECT t1.date curr_date, MAX(t2.date) prev_date FROM periods t1
LEFT JOIN periods t2
ON t1.date > t2.date
GROUP BY t1.date) t
WHERE DATEDIFF(curr_date, prev_date) > 1
The result is:
13-Feb-17 10-Feb-17
Then add condition/LIMIT to get only one row.

Answer 3

I am trying to write general query like,
select count(distinct t1.`date`) from period t1 left join period t2 ON t2.user = t1.user and t1.`date` - INTERVAL 1 DAY = t2.date where t2.id is null
Can you try this query once,
it should work, I have used it in my almost same case.

Answer 4

This query will return the latest hole:
select
m.`date`,
min(m1.`date`) as next_date,
datediff(min(m1.`date`), m.`date`)+1 as diff
from
mytable m left join mytable m1
on m.`date`<m1.`date`
group by
m.`date`
having
datediff(min(m1.`date`), m.`date`)>1
order by
m.`date` desc
limit 1

Answer 5

The following query:
SELECT MIN(`date`) AS date_start,
MAX(`date`) AS date_end,
MAX(days_diff) AS pause_days,
COUNT(*) + 1 AS period_without_pays
FROM (
SELECT id, user, `date`,
DATE_SUB(`date`, INTERVAL rn DAY) AS group_date,
DATEDIFF(`date`, COALESCE(prevDate, `date`)) AS days_diff
FROM (
SELECT t1.id, t1.user, t1.`date`,
#rn := #rn + 1 AS rn,
(SELECT t2.`date`
FROM mytable AS t2
WHERE t1.id = t2.id AND t1.user = t2.user AND t1.`date` > t2.`date`
ORDER BY `date` DESC LIMIT 1) AS prevDate
FROM mytable AS t1
CROSS JOIN (SELECT #rn := 0) AS v
ORDER BY `date`) AS t) AS x
GROUP BY id, user, group_date, days_diff
HAVING SUM(days_diff) > 0
returns:
date_start date_end pause_days period_without_pays
-----------------------------------------------------
2017-02-14 2017-02-16 1 4
2017-02-13 2017-02-13 3 2
The row with pause_days > 1 returns the start date of the pause along with the number of days.
The row with pause_days = 1 returns the start / end dates of an island of consecutive records having consecutive dates along with the count of these dates.
Note: The above query works with the sample data provided. You may have to tweak the query a little bit so as to adjust it to the complexity of the actual data.

Answer 6

Try this one:
SET #dateDiff=NULL;SET #dateDiff2='';
SELECT diff.secondDate AS fromDate,diff.initialDate AS toDate, diff.dateDiffR FROM (
SELECT d.date AS initialDate,#dateDiff AS secondDate,IF(#dateDiff IS NULL,#dateDiff:=d.date,0) AS try,
IF(DATE(#dateDiff)!=DATE(d.date),DATEDIFF(d.date,#dateDiff),NULL) AS dateDiffR,
IF(#dateDiff!=#dateDiff2,#dateDiff2:=#dateDiff,0) AS try1,
IF(DATE(#dateDiff)!=DATE(d.date),#dateDiff:=d.date,NULL) AS assign FROM
(SELECT b.date FROM mytable b)d ) diff WHERE diff.dateDiffR>0
it will give you the date difference with its date range. If you get negative count then swap the dates on parameters' for DATEDIFF;

Answer 7

(Posted on behalf of the OP).
I adapted little bit query from #McNets:
select user, min(dd) as ini, max(dd) as fin, datediff(max(dd), min(dd))+1 as Days, consecutive
from (
select user,dd,
if(#last_date = curdate() or datediff(dd, #last_date) >= -1, #cn := #cn, #cn := #cn+1) consecutive,
#last_date := dd
from
(select #last_date := curdate(), #cn := 0) x,
(select user, date as dd
from myt
where user = %id
order by dd desc) y
) z
group by consecutive
order by CAST(consecutive AS UNSIGNED)
I added filter by user, changed cause to '>= -1' to accept time usage, added number of the serie and changed initial date from '1900-01-01' to CURDATE() function (i don't see any influence to query result from this action).
Now using number of the series can find the longest series and its dates.