Checking for date range conflicts in MySQL

Question

I am writing a hotel booking system. after lots of studying (including stack overflow) i wrote this sql to find out free rooms:

SELECT
*
FROM room
WHERE
    room         


        

Answer 1

Your original logic was very close, you just need to swap the '$check_in' and '$check_out' values. I.e.:

SELECT *
FROM room
WHERE room_id NOT IN
(
    SELECT room_id
    FROM bookings
    WHERE checkin <= '$check_out' AND checkout >= '$check_in'
)

Brian Driscoll's answer focusses on the scenarios that constitute booking conflicts, as so:

---------------|-----Booked-----|---------------
       |----A1----|
                             |----A2----|
                    |--A3--|
            |----------A4----------|

Case A2 & A3: checkin <= '$check_in' AND checkout >= '$check_in'
Case A1 & A3: checkin <= '$check_out' AND checkout >= '$check_out'
Case A4:      checkin >= '$check_in' AND checkout <= '$check_out'

However the senarios that constitute no conflict are much simpler. There are only two:

---------------|-----Booked-----|---------------
  |----B1----|                             
                                  |----B2----|

Case B1: checkin > '$check_out'
Case B2: checkout < '$check_in'

So the situation where there is no conflict between a booking and a potential booking can be expressed with this SQL:

checkin > '$check_out' OR checkout < '$check_in'

To check for conflicts instead, we just need to negate this. So, using DeMorgans Law, the negation is:

checkin <= '$check_out' AND checkout >= '$check_in'

...which arrives at the solution given above.