My SQL Finding a span of dates accross rows - mysql

Question

I am looking for some help with even knowing where to start. Essentially we have a table for clients that hold employment start dates and end dates. For annual reports we have to calculate "continuous employment" which is defined as earliest start date to last end date as long as there is not more than 21 days between one end date and the next start date.
here is an example
employee | Start Date | End Date
1 | 2012-10-1 | 2012-11-05
1 | 2012-11-08 | 2013-1-25
2 | 2012-10-1 | 2012-11-05
2 | 2012-11-30 | 2013-1-02
in the above, i would like to see employee 1 as continuously employed from 2012-10-1 to 2013-1-25
but employee 2 would have 2 separate employment lines showing continuous employment from 2012-10-1 to 2012-11-05 and a different from 012-11-30 to 2013-1-02
Thanks for the help!

Answer 1

The theory is similar to #mellamokb's answer, but somewhat more concise:
SELECT employee, MIN(start) start, end
FROM (
SELECT #end:=IF(employee<=>#emp AND #stt<=end+INTERVAL 21 DAY,#end,end) end,
#stt:=start start,
#emp:=employee AS employee
FROM my_table, (SELECT #emp:=NULL, #stt:=0, #end:=0) init
ORDER BY employee, start DESC
) t
GROUP BY employee, end
See it on sqlfiddle.

Answer 2

One way to find "continuous groups" among a set of records is to use variables to track the difference between each line and develop groupings that combine continuous ranges together. In the example below, I use three variables to track enough information for generating the groups:
#curEmployee - tracks the current employee from the previous record, and is compared with the employee on the current record to know when we've switched to a different employee, which automatically becomes another grouping
#curEndDate - tracks the last end date from the previous record, so it can be compared with the start date of the current record to see if the current record belongs in the same "group" as the previous record - that is to say, it is part of continuous employment with the previous record
#curGroup - this is the key variable which segregates the rows into separate "groups" that represent continuous employment. The logic is that a row should be considered as continuous with the previous row if and only if the following two conditions are true: the two rows have the same employee number, and the end date of the previous row is less than 21 days from the current row.
NOTE: You may want to validate the edge conditions, i.e., whether exactly 20/21/22 days apart will be considered continuous employment or not, and tweak the logic below.
Here is the sample query that calculates those groups. A couple things to take note of: the order of variable assignment matters, because they are assigned from top to bottom in the select list. We need to assign #curGroup first, so that it still has the values of #curEmployee and #curEndDate from the previous record to draw on. Secondly, the order by clause is very important to ensure that when we are comparing the previous and current record, they are the two records that are the closest to each other. If we looked at the records in a random order, they would likely end up all as separate groups.
select
e.employee, e.`start date`, e.`end date`
,#curGroup :=
case when employee = #curEmployee
and #curEndDate + INTERVAL 21 DAY >= e.`start date`
then #curGroup
else #curGroup + 1
end as curGroup
,#curEmployee := employee as curEmployee
,#curEndDate := e.`end date` as curEndDate
from
employment e
JOIN (SELECT #curEmployee := 0, #curEndDate := NULL, #curGroup := 0) r
order by e.employee, e.`start date`
Sample Result (DEMO) - notice how CURGROUP stays at 1 for the first two lines, because they are within 21 days of each other and represent continuous employment, while the last two lines get identified as separate group numbers:
| EMPLOYEE | START DATE | END DATE | CURGROUP | CUREMPLOYEE | CURENDDATE |
-------------------------------------------------------------------------------------------------------------------------------
| 1 | October, 01 2012 00:00:00+0000 | November, 05 2012 00:00:00+0000 | 1 | 1 | 2012-11-05 00:00:00 |
| 1 | November, 08 2012 00:00:00+0000 | January, 25 2013 00:00:00+0000 | 1 | 1 | 2013-01-25 00:00:00 |
| 2 | October, 01 2012 00:00:00+0000 | November, 05 2012 00:00:00+0000 | 2 | 2 | 2012-11-05 00:00:00 |
| 2 | November, 30 2012 00:00:00+0000 | January, 02 2013 00:00:00+0000 | 3 | 2 | 2013-01-02 00:00:00 |
Now that we've established groups of records that are part of continuous employment, we merely need to group by those group numbers and find the minimum and maximum date range for the output:
select
employee,
min(`start date`) as `start date`,
max(`end date`) as `end date`
from (
select
e.employee, e.`start date`, e.`end date`
,#curGroup :=
case when employee = #curEmployee
and #curEndDate + INTERVAL 21 DAY >= e.`start date`
then #curGroup
else #curGroup + 1
end as curGroup
,#curEmployee := employee as curEmployee
,#curEndDate := e.`end date` as curEndDate
from
employment e
JOIN (SELECT #curEmployee := 0, #curEndDate := NULL, #curGroup := 0) r
order by e.employee, e.`start date`
) as T
group by curGroup
Sample Result (DEMO):
| EMPLOYEE | START DATE | END DATE |
--------------------------------------------------------------------------------
| 1 | October, 01 2012 00:00:00+0000 | January, 25 2013 00:00:00+0000 |
| 2 | October, 01 2012 00:00:00+0000 | November, 05 2012 00:00:00+0000 |
| 2 | November, 30 2012 00:00:00+0000 | January, 02 2013 00:00:00+0000 |