What am I doing wrong while trying to sort this array?

Question

Need to sort an array of objects based on a particular value in that object.

const obj = [
    {
        field: "FULL_NAME",
        required: "Y"
    },
    {
        name: "EMAIL",
        required: "N"
    },
    {
        name: "ADDRESS",
        required: "N"
    },
    {
        name: "NUMBER",
        required: "Y"
    },
]

I want to sort this array in a way that fields with required 'Y' come first.

Tried to write a comparison function like this :

const obj = [
    {
        field: "FULL_NAME",
        required: "Y"
    },
    {
        name: "EMAIL",
        required: "N"
    },
    {
        name: "ADDRESS",
        required: "N"
    },
    {
        name: "NUMBER",
        required: "Y"
    },
];

const compare = (a, b) => {
    if(a.required === "Y" && b.required === "N"){
        return 1
    }

    if(a.required === "N" && b.required === "Y"){
        return -1
    }

    return 0;
}

console.log(obj.sort(compare));

How do I fix it so it works?

Especially before the code is runnable, it would be helpful to include exactly what went wrong. Here it looks like it's just sorting it backwards, and you just needed to reverse your condition. Did you not try that? — Brian Thompson, Dec 13 '21 at 16:12
I would've understood the minor mistake if it run properly, made a small error from my side because of which the sort function wasn't even being executed. Fixed that. — Parth, Dec 13 '21 at 16:33

jsejcksn · Accepted Answer · 2021-12-13T16:19:55.370

You have the 1 and -1 backwards. -1 means the first element argument is sorted first, and 1 means the second element argument is sorted first. See the compareFunction/ sort order table at the Array.prototype.sort() article:

`compareFunction(a, b)` return value	sort order
`> 0`	sort `b` before `a`
`< 0`	sort `a` before `b`
`=== 0`	keep original order of `a` and `b`

const obj = [
    {
        field: "FULL_NAME",
        required: "Y"
    },
    {
        name: "EMAIL",
        required: "N"
    },
    {
        name: "ADDRESS",
        required: "N"
    },
    {
        name: "NUMBER",
        required: "Y"
    },
];

const compare = (a, b) => {
    if(a.required === "Y" && b.required !== "Y"){
        return -1
    }

    if(a.required !== "Y" && b.required === "Y"){
        return 1
    }

    return 0;
}

console.log(obj.sort(compare));

score 1 · Answer 2 · answered Dec 13 '21 at 16:28

1

Shortest version using ES6

obj.sort((a,b) => a.required > b.required ? -1: 1);

Reference: Sort objects in an array alphabetically on one property of the array

answered Dec 13 '21 at 16:28

rphsantos

111
1
4

What am I doing wrong while trying to sort this array?

2 Answers2