sort keeping 2 items on top

user_277696 0
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I have an array object like

const d = [
    { id: 20, text: 'deaf' },
    { id: 30, text: 'acta', },
    { id: 0, text: 'deema1' },
    { id: -1, text: 'deema2' },
]

I want to sort the array by text but want to keep id -1 and 0 always on top, so the result would be

// sorted
[
    { id: -1, text: 'deema2' },
    { id: 0, text: 'deema1' },
    { id: 30, text: 'acta' },
    { id: 20, text: 'deaf' },
]

I tried to sort like d.sort((a, b) => (a.text).localeCompare(b.text)) but not sure how to handle case -1 and 0

const d = [
    { id: 20, text: 'deaf' },
    { id: 30, text: 'acta', },
    { id: 0, text: 'deema1' },
    { id: -1, text: 'deema2' },
];

d.sort((a, b) => (a.text).localeCompare(b.text))

console.log(d);

algorithm javascript
3 Answers

You can work with indexOf:

const d = [
    { id: 20, text: 'deaf' },
    { id: 30, text: 'acta', },
    { id: 0, text: 'deema1' },
    { id: -1, text: 'deema2' },
    { id: 0, text: 'ace' },
    { id: -999, text: 'hello' },
    { id: 800, text: 'game' },
]

d.sort((a, b) => [0,-1].indexOf(b.id) - [0,-1].indexOf(a.id) 
                 || a.text.localeCompare(b.text));

console.log(d);

This will order all objects with id -1 first, and among those, the text property will define the order, then all objects with id 0, (again relatively ordered by text) and finally all other objects by text.

In case you want there to be no distinction between 0 and -1, but sort all those by text among themselves, then use includes instead of indexOf:

const d = [
    { id: 20, text: 'deaf' },
    { id: 30, text: 'acta', },
    { id: 0, text: 'deema1' },
    { id: -1, text: 'deema2' },
    { id: 0, text: 'ace' },
    { id: -999, text: 'hello' },
    { id: 800, text: 'game' },
]

d.sort((a, b) => [0,-1].includes(b.id) - [0,-1].includes(a.id) 
                 || a.text.localeCompare(b.text));

console.log(d);

user_54598390

You can put multiple conditions inside of sort to get the desired result. The first condition being more important than the next and so on. This is how it would look like:

const d = [{
    id: 20,
    text: 'deaf'
  },
  {
    id: 30,
    text: 'acta',
  },
  {
    id: 0,
    text: 'deema1'
  },
  {
    id: -1,
    text: 'deema2'
  },
]

const sorted = d.sort((a, b) => {
  if (a.id === -1) {
    return -1;
  }
  if (a.id === 0) {
    return -1;
  }
  return a.text.localeCompare(b.text);
});

console.log(sorted)

user_120666770

Slightly modified from one of the answer above to use the pipe operator, this way you can add any other numbers you wish just by using more pipes.

const d = [
    { id: 20, text: 'deaf' },
    { id: 30, text: 'acta', },
    { id: 0, text: 'deema1' },
    { id: -1, text: 'deema2' },
    { id: 0, text: 'ace' },
    { id: -999, text: 'hello' },
    { id: 800, text: 'game' },
]

const sorted = d.sort((a, b) => {
  if (a.id === -1 || a.id === 0) {
    return -1;
  }
  return a.text.localeCompare(b.text);
});

console.log(sorted)
user_90901950
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