CODEWITHSUNDEEP
mysqlsqlgroup-by
Peanut
Peanut

Reputation: 2136

SQL Group By Date Conflicts

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.

Upvotes: 8

Views: 585

Answers (3)

Bohemian
Bohemian

Reputation: 425003

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 |
+-----------+------------+----------+------+

Upvotes: 2

cabanaboy
cabanaboy

Reputation: 71

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

Upvotes: 0

jayunit100
jayunit100

Reputation: 17648

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.

Upvotes: 0

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