Check if date range conflicts with current documents MongoDB - node.js

Question

This problem has a layer of complexity and I'm not sure best practices to work around.
Attempting to create a booking calendar that serves the next availability date for their selected range. We have had cancellations that open up gaps.
Current functionality just takes the furthest end date and serves the next day, this has become very inaccurate and is leaving gaps due to cancellations.
Ex. User is looking to book for 10 days at the nearest available date. In the database there is a gap between 2 documents that would suit this user, how do I query to verify this availability?
Database document example
{
"start_date" : "2020-02-06T19:58:25.430Z",
"end_date" : "2020-02-16T19:58:25.430Z"
},
{
"start_date" : "2020-02-17T19:58:25.430Z",
"end_date" : "2020-02-27T19:58:25.430Z"
},
{
"start_date" : "2020-03-21T19:58:25.430Z",
"end_date" : "2020-03-31T19:58:25.430Z"
},
In between the second and third document there is a 2 week period that the user could claim. Currently this user would receive 2020-04-01 as the next availability. I would like to give them the option to claim a cancellation date by sending 2020-02-28 as the next availability.
How do I query mongodb to see if there is an x day amount availability in between document ranges before defaulting to the furthest end_date?

Answer 1

One idea: create a separate cancelled collection, and move any cancelled appointments into this collection.
If a new appointment is to be made, search this cancelled collection first for any open spots due to cancellations. If nothing suitable is found, then proceed to the furthest end date as per your current method.
An illustration:
> db.appointments.find()
{ "_id" : 0, "start" : 1, "end" : 2, "duration" : 1, "name": "John" }
{ "_id" : 1, "start" : 2, "end" : 4, "duration" : 2, "name": "Jane" }
{ "_id" : 2, "start" : 4, "end" : 5, "duration" : 1, "name": "Jack" }
// Jane cancelled the appt with _id 1
> c = db.appointments.findOne({_id: 1})
> db.cancelled.insert(c)
> db.appointments.deleteOne({_id: 1})
// current state of appointments
> db.appointments.find()
{ "_id" : 0, "start" : 1, "end" : 2, "duration" : 1, "name": "John" }
{ "_id" : 2, "start" : 4, "end" : 5, "duration" : 1, "name": "Jack" }
// Bob wanted to make an appointment spanning two "time units". Is it available?
> db.cancelled.findOne({duration: 2})
{ "_id" : 1, "start" : 2, "end" : 4, "duration" : 2, "name" : "Jane" }
// If Bob agreed to take over Jane's time slot
// reinsert the document from "cancelled" to "appointments", changing the name
> a = db.cancelled.findOne({duration: 2})
> a.name = "Bob"
> db.appointments.insert(a)
> db.cancelled.deleteOne({_id:1})
// end state of appointments
> db.appointments.find().sort({start: 1})
{ "_id" : 0, "start" : 1, "end" : 2, "duration" : 1, "name": "John" }
{ "_id" : 1, "start" : 2, "end" : 4, "duration" : 2, "name": "Bob" }
{ "_id" : 2, "start" : 4, "end" : 5, "duration" : 1, "name": "Jack" }
Note in this illustration, I also recorded the duration of each appointment so they can be searched efficiently. The end state is _id: 1 from Jane was replaced by Bob, with nothing in the cancelled collection.
To make sure that both appointments and cancelled collections are both updated correctly, you might want to use Transactions that's available since MongoDB 4.0.

Answer 2

I figured out a solution after letting it ferment in my brain for a week. I couldn't get the wanted results out of querying mongodb. I had to use javascript to compare some of the documents. Not sure about how the run time will hold up, for me that's a problem for a different day.
Thought I would add my solution as I came across my own question in google searches for the last week.
//Find all the documents that have an end date greater than today, sort them based // on the start date to order documents
let sort = await db.collection(COLLECTION)
.find({end_date : { $gte:new Date() }
})
.sort({start_date: 1 })
.toArray();
//Set the nextDate variable
let nextDate;
if(sort.length >= 2){
//Javascript iteration if there are more than 2 documents returned from sorting
for (let i = 0; i < sort.length; i++) {
var startDate = Date.parse(sort[i].end_date);
//endDate is ternary incase the amount of documents is odd number
var endDate = sort[i + 1] ? Date.parse(sort[i + 1].start_date) : new Date()
var timeDiff = endDate - startDate;
var daysDiff = Math.floor(timeDiff / (1000 * 60 * 60 * 24))
//Get the day difference between the date if the difference is greater than the //days alloted by user return the end date
if(daysDiff > days) {
return nextDate = sort[i].end_date
}
}
//If no days have room for the decided days, return the last document in array
return nextDate = sort[sort.length - 1].end_date
//If only one document, return end date
} else if (sort.length === 1) {
nextDate = sort[0].endDate
//No documents found, return todays date as it is available.
} else {
nextDate = new Date()
}
This actually works perfect for what I need it to do. Hopefully will get a good refactor and can be simplified.