How to sort items in arrays meeting three conditions?

Question

I am trying since several hours to make it work. That is why I finally decided to ask you if it is even possible to achieve this by using one single sort-method.

I have an array of objects with the following type:

{
  seats: number
  area: string
}[]

1. Now I would like to have all the objects with 5 seats on the top of the array.

2. Of those the ones that have "garden" as an area should be on the top.

3. The ones that have more than 5 seats should follow in an ascending order.

4. The rest should follow in a descending order.

I am very frustrated right now. I did not think that this is that difficult...

Answer 1

Let's say this is the array you have

const data = [
  { seats: 6, area: 'garden' },
  { seats: 3, area: 'garden' },
  { seats: 5, area: 'patio' },
  { seats: 7, area: 'patio' },
  { seats: 2, area: 'indoor' },
  { seats: 5, area: 'garden' },
  { seats: 4, area: 'patio' },
];

you can sort it like this:

data.sort((a, b) => {

  // if one object has five seats, prioritize it
  if (a.seats === 5 && b.seats !== 5) {
    return -1; // a comes first
  } else if (b.seats === 5 && a.seats !== 5) {
    return 1; // b comes first
  } else if (a.seats === b.seats) {

    // if they both have the same number of seats, prioritize the 'garden'
    if (a.area === 'garden' && b.area !== 'garden') {
      return -1; // a comes first
    } else if (b.area === 'garden' && a.area !== 'garden') {
      return 1; // b comes first
    } else {
      return 0; // no change in order
    }

  // if it's more than 5 seats, sort it by ascending. otherwise, descending.
  } else if (a.seats > 5 && b.seats > 5) {
    return a.seats - b.seats; // ascending order
  } else {
    return b.seats - a.seats; // descending order
  }
});