在 Mysql 中未排序的具有多个日期范围的给定月份的日期范围间隙查询

Question

I have a pretty tricky calendar gap project that I could use some assistance. I hope I have outlined my issue well enough and really appreciate any assistance in solving this.

I need to find calendar gaps between multiple date ranges. I have a mysql DB with car rentals. I need a query that shows all cars that have a gap during a given month. I am using mysql 5.7, so I CANNOT use recursive queries.

Here is what I have in my DB (mysql):

TABLE:

Car   ||  from_date   || to_date
Ford  ||  2021-05-01  || 2021-05-09 (10 day gap)
Ford  ||  2021-05-20  || 2021-05-30 (0 day gap till next)
Ford  ||  2021-06-01  || 2021-06-07 (7 day gap till next)
Ford  ||  2021-06-15  || 2021-06-20 (5 day gap till next)
Ford  ||  2021-06-26  || 2021-06-30 (0 day gap till next)
Chevy ||  2021-04-20  || 2021-04-29 (7 day gap till next)
Chevy ||  2021-05-07  || 2021-05-12 (7 day gap till next)
Chevy ||  2021-05-20  || 2021-05-23 (8 day gap till next)
Chevy ||  2021-06-02  || 2021-06-10 (13 day gap till next)
Tesla ||  2021-04-15  || 2021-04-30 
Tesla ||  2021-05-01  || 2021-05-30 

SCENARIOS: In my table above I have created the following scenarios: Ford: has gaps completely within May and June. Chevy: has overlapping gaps at start and end of month Tesla: has all June open

WARNING: These records are sorted by from_date in this example, but they may not be in real life, so you cannot assume their order. This is why the "show gaps" example referenced below did not work for me.

EXPECTED/ACCEPTABLE RESULTS: The tables below shows what I expect, but I am fine with just knowing if a car has ANY gap during the given month. So, for 05, I do not need Chevy listed 3 times. If we can build this faster by just pulling the first occurrence. Just showing the first match of each car would be fine.

TRUE RESULT FOR: month = 05, gap = 7

+------------+------------+------------+------+
|    Car     |  gap_start |  gap_end   | gap  |
+------------+------------+------------+------+
|    Ford    | 2021-05-09 | 2021-05-20 |  10  |
|    Chevy   | 2021-04-29 | 2014-05-07 |   7  |
|    Chevy   | 2021-05-12 | 2014-05-20 |   7  |
|    Chevy   | 2021-05-23 | 2014-06-02 |   8  |
+------------+------------+------------+------+

ACCEPTABLE RESULTS FOR: month = 05, gap = 7

+------------+------------+
|    Car     |  has_gap   |
+------------+------------+
|    Ford    |     1      |
|    Chevy   |     1      |
+------------+------------+

TRUE RESULT FOR: month = 06, gap = 5

+------------+------------+------------+------+
|    Car     |  gap_start |  gap_end   | gap  |
+------------+------------+------------+------+
|    Ford    | 2021-06-07 | 2021-06-15 |   7  |
|    Ford    | 2021-06-20 | 2021-06-26 |   5  |
|    Chevy   | 2021-06-10 | 2021-06-31 |  20  |
|    Tesla   | 2021-06-01 | 2021-06-30 |  30  |
+------------+------------+------------+------+

ACCEPTABLE RESULTS FOR: month = 06, gap = 5

+------------+----------+
|    Car     |  has_gap |
+------------+----------+
|    Ford    |     1    |
|    Chevy   |     1    |
|    Tesla   |     1    |
+------------+----------+

NOTE: I realize my math is off for the Tesla because I have had to fake the dates. I am not as concerned about how that one is returned, as long as Tesla is included in the results.

NEXT STEPS: I will probably build a calendar (horizontal strip display) for each car with their booked dates, so I am thinking once I get the results I would loop through them and pull all the reservations for that given month to build that calendar. Unless someone can figure how to do all that in one query and keep it fast.

I have spend some time trying to cover all my bases in asking this question and hope I haven't made any mistakes. Please feel free to ask questions or for clarification. AND, THANKS IN ADVANCE FOR ANY HELP.

Similar question, but not close enough to work. Show gaps between dates in MySQL

Answer 1

Basically, I think you can use not exists and aggregation. To get the cars during a time period that have a gap:

select car, min(from_date)
from t
where from_date >= $month_start and from_date < $month_end and
      not exists (select 1
                  from t t2
                  where t2.car = t.car and
                        t2.from_date > t.to_date and
                        t2.from_date < t.to_date + interval $gap day
                 )
group by car;

Here is a db<>fiddle.