SQL Group By Date Conflicts - mysql

Question

I have a table with columns start_date and end_date. What we need to do is Select everything and group them by date conflicts for each Object_ID.
A date conflict is when a row's start date and/or end date pass through another rows'. For instance, here are some examples of conflicts:
Row 1 has dates 1st through the 5th, Row 2 has dates 2nd through the 3rd.
Row 1 has dates 2nd through the 5th, Row 2 has dates 1st through the 3rd.
Row 1 has dates 2nd through the 5th, Row 2 has dates 3rd through the 6th.
Row 1 has dates 2nd through the 5th, Row 2 has dates 1st through the 7th.
So for example, if we have some sample data (assume the numbers are just days of the month for simplicity):
id | object_id | start_date | end_date
1 | 1 | 1 | 5
2 | 1 | 2 | 4
3 | 1 | 6 | 8
4 | 2 | 2 | 3
What i would expect to see is this:
object_id | start_date | end_date | numconflicts
1 | <na> | <na> | 2
1 | 6 | 8 | 0 or null
2 | 2 | 3 | 0 or null
And for a Second Test Case, Here is some sample data:
id | object_id | start_date | end_date
1 | 1 | 1 | 5
2 | 1 | 2 | 4
3 | 1 | 6 | 8
4 | 2 | 2 | 3
5 | 2 | 4 | 5
6 | 1 | 2 | 3
7 | 1 | 10 | 12
8 | 1 | 11 | 13
And for the second Test Case, what I would expect to see as output:
object_id | start_date | end_date | numconflicts
1 | <na> | <na> | 3
1 | 6 | 8 | 0 or null
2 | 2 | 3 | 0 or null
2 | 4 | 5 | 0 or null
1 | <na> | <na> | 2
Yes, I will need some way of differentiating the first and the second grouping (the first and last rows) but I haven't quite figured that out. The goal is to view this list, and then when you click on a group of conflicts you can view all of the conflicts in that group.
My first thought was to attempt some GROUP BY CASE ... clause but I just wrapped by head around itself.
The language I am using to call mysql is php. So if someone knows of a php-loop solution rather than a large mysql query i am all ears.
Thanks in advance.
Edit: Added in primary Keys to provide a little less confusion.
Edit: Added in a Test case 2 to provide some more reasoning.

Answer 1

This query finds the number of duplicates:
select od1.object_id, od1.start_date, od1.end_date, sum(od2.id is not null) as dups
from object_date od1
left join object_date od2
on od2.object_id = od1.object_id
and od2.end_date >= od1.start_date
and od2.start_date <= od1.end_date
and od2.id != od1.id
group by 1,2,3;
You can use this query as the basis of a query that gives you exactly what you asked for (see below for output).
select
object_id,
case dups when 0 then start_date else '<na>' end as start_date,
case dups when 0 then end_date else '<na>' end as end_date,
sum(dups) as dups
from (
select od1.object_id, od1.start_date, od1.end_date, sum(od2.id is not null) as dups
from object_date od1
left join object_date od2
on od2.object_id = od1.object_id
and od2.end_date >= od1.start_date
and od2.start_date <= od1.end_date
and od2.id != od1.id
group by 1,2,3) x
group by 1,2,3;
Note that I have used an id column to distinguish the rows. However, you could replace the test of id's not matching with comparisons on every column, ie replace od2.id != od1.id with tests that every other column is not equal, but that would require a unique index on all the other columns to make sense, and having an id column is a good idea anyway.
Here's a test using your data:
create table object_date (
id int primary key auto_increment,
object_id int,
start_date int,
end_date int
);
insert into object_date (object_id, start_date, end_date)
values (1,1,5),(1,2,4),(1,6,8),(2,2,3);
Output of first query when run against this sample data:
+-----------+------------+----------+------+
| object_id | start_date | end_date | dups |
+-----------+------------+----------+------+
| 1 | 1 | 5 | 1 |
| 1 | 2 | 4 | 1 |
| 1 | 6 | 8 | 0 |
| 2 | 2 | 3 | 0 |
+-----------+------------+----------+------+
Output of second query when run against this sample data:
+-----------+------------+----------+------+
| object_id | start_date | end_date | dups |
+-----------+------------+----------+------+
| 1 | 6 | 8 | 0 |
| 1 | <na> | <na> | 2 |
| 2 | 2 | 3 | 0 |
+-----------+------------+----------+------+

Answer 2

Oracle : This could be done with a subquery in a group by CASE statement.
https://forums.oracle.com/forums/thread.jspa?threadID=2131172
Mysql : You could have a view which had all the conflicts .
select distinct a1.appt, a2.appt from appointment a1, appointment a2 where a1.start < a2.end and a1.end > a2.start.
and then simply do a count(*) on that table.

Answer 3

Something like the following should work:
select T1.object_id, T1.start_date, T1.end_date, count(T1.object_id) as numconflicts
from T1
inner join T2 on T1.start_date between T2.start_date and T2.end_date
inner join T3 on T1.end_date between T2.start_date and T2.end_date
group by T1.object_id
I might be off a little bit, but it should help you get started.
Edit: Indented it properly