Sort an array first by boolean column then by string column

Question

I'm trying to so sort an array by 2 fields. I have a boolean: isFavorite and a string: name. All the booleans who are true have to be the first items. But I want the array to be alphabetic. This is my code so far (tried multiple things):

data.sort(function (x,y) {
    if (x.isFavorite){
        return -1;
    }
    if (x.isFavorite && !y.isFavorite && (x.name < y.name)){
        return -1;
    }  else if ((x.isFavorite === y.isFavorite) && (x.name === y.name)){
        return 0;
    } else if (x.isFavorite && y.isFavorite && (x.name < y.name)){
        return -1;
    } else if (!x.isFavorite && !y.isFavorite && (x.name > y.name)){
        return 1;
    }
}

Possible duplicate of Javascript sort array by two fields

Muhammet Enginar
– Muhammet Enginar

2017-12-21 08:44:04 +00:00
Commented Dec 21, 2017 at 8:44 — Muhammet Enginar
– Muhammet Enginar, Commented Dec 21, 2017 at 8:44

Mihai Alexandru-Ionut · Accepted Answer · 2017-12-21 08:30:16Z

16

Why so much conditions and overlapping ?

Just use logical OR operator in order to sort array by two fields.

Also, I used + (Boolean) in order to force converting to Number.

grouperArray.sort((a, b) => (+a.isFavorite) - (+b.isFavorite) || a.name.localeCompare(b.name));

let data = [{ isFavorite:false, name:'A' }, { isFavorite:true, name:'C' }, { isFavorite:true, name:'B' }];
data.sort((a,b) => (+b.isFavorite) - (+a.isFavorite) || a.name.localeCompare(b.name));
console.log(data);

answered Dec 21, 2017 at 8:30

Mihai Alexandru-Ionut

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Comments

klugjo · Accepted Answer · 2017-12-21 08:33:38Z

3

const data = [
  {isFavorite: false, name: 'b'},
  {isFavorite: false, name: 'f'},
  {isFavorite: true, name: 'c'},
  {isFavorite: true, name: 'a'},
  {isFavorite: false, name: 'd'},
  {isFavorite: true, name: 'g'}
];

const sortedData = data.sort((a, b) => {
  if (a.isFavorite !== b.isFavorite) {
    return a.isFavorite ? -1 : 1;
  } else {
    return a.name > b.name ? 1 : -1;
  }
});
  
console.log(sortedData);

answered Dec 21, 2017 at 8:33

klugjo

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Comments

Muhammet Enginar · Accepted Answer · 2017-12-21 10:11:18Z

3

Try this one:

const data = [
  {isFavorite: false, name: 'B'}, 
  {isFavorite: false, name: 'A'}, 
  {isFavorite: true, name: 'C'}, 
  {isFavorite: true, name: 'D'},
  {isFavorite: true, name: 'A'},
  {isFavorite: false, name: 'A'},
  {isFavorite: false, name: 'D'},
  {isFavorite: true, name: 'Z'},
];

const compareLoc = (a, b) => a.name.localeCompare(b.name)
const result = [...data.filter(d => d.isFavorite).sort(compareLoc),
...data.filter(d => !d.isFavorite).sort(compareLoc)];
console.log(result);

Following snippet works.

const data = [
  {isFavorite: false, name: 'B'}, 
  {isFavorite: false, name: 'A'}, 
  {isFavorite: true, name: 'C'}, 
  {isFavorite: true, name: 'D'},
  {isFavorite: true, name: 'A'},
  {isFavorite: false, name: 'A'},
  {isFavorite: false, name: 'D'},
  {isFavorite: true, name: 'Z'},
];

data.sort(function (x, y) {
    // if both are true or false, we should compare name attributes
    if (x.isFavorite === y.isFavorite) {
        return x.name.localeCompare(y.name);
    } return x.isFavorite ? -1 : 1
}
)
console.log(data)

edited Dec 21, 2017 at 10:11

answered Dec 21, 2017 at 8:37

Muhammet Enginar

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9 Comments

Pieter Over a year ago

This almost works for me, the strange thing is that everything works fine except at the bottom of my dropdown (array), I have a few items that didn't get sorted like this: a-z, a, c, f, g , w => so the first part is perfectly fine, but after Z comes like 10 items that starts again from "a"

user4676340 Over a year ago

That's not weird, he doesn't sort the end of the array. You're supposed to sort every case of a condition, and he only does it in the if statement. Copy/paste the code of the if in the else, and below it, and it should work

Pieter Over a year ago

Can you post the snippet for clarification? I did what I think you meant but doesn't work

Muhammet Enginar Over a year ago

I didn't get the exact point, don't you need isFavorite(a-z) then !isFavorite(a-z)? Why a-z comes after a-z is I order isFavorite = true ones before false ones in all scenarios

Muhammet Enginar Over a year ago

can you check the upper snippet I just shared

|

gurvinder372 · Accepted Answer · 2017-12-21 08:29:36Z

0

You problem is that first two if's are overlapping and second if is not reachable if x.isFavorite is true then first if will return.

Also, you are not checking some scenarios like y.isFavorite is true and x.isFavorite is not.

You can write it as

if ( x.isFavorite != y.isFavorite )
{
   return x.isFavorite ? -1 : 1;
}
else
{
   return x.name.localeCompare( y.name );
}

answered Dec 21, 2017 at 8:29

gurvinder372

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Comments

Zenoo · Accepted Answer · 2017-12-21 08:32:38Z

0

Here you go :)

var arr = [
  {
    isFavorite: false,
    name: 'a'
  },
  {
    isFavorite: true,
    name: 'b'
  },
  {
    isFavorite: true,
    name: 'c'
  },
  {
    isFavorite: false,
    name: 'c'
  }
];

arr.sort(function (x,y) {
    if (x.isFavorite > y.isFavorite){
        return -1;
    }else if(x.isFavorite == y.isFavorite){
        return x.name > y.name;
    }else{
        return 1;
    }
});

console.log(arr);

answered Dec 21, 2017 at 8:32

Zenoo

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