Sort array by firstname (alphabetically) in JavaScript

Question

I got an array (see below for one object in the array) that I need to sort by firstname using JavaScript. How can I do it?

var user = {
   bio: null,
   email:  "user@domain.example",
   firstname: "Anna",
   id: 318,
   lastAvatar: null,
   lastMessage: null,
   lastname: "Nickson",
   nickname: "anny"
};

Answer 1

Shortest possible code with ES6!

users.sort((a, b) => a.firstname.localeCompare(b.firstname))

String.prototype.localeCompare() basic support is universal!

Answer 2

Suppose you have an array users. You may use users.sort and pass a function that takes two arguments and compare them (comparator)

It should return

• something negative if first argument is less than second (should be placed before the second in resulting array)
• something positive if first argument is greater (should be placed after second one)
• 0 if those two elements are equal.

In our case if two elements are a and b we want to compare a.firstname and b.firstname

Example:

users.sort(function(a, b){
    if(a.firstname < b.firstname) { return -1; }
    if(a.firstname > b.firstname) { return 1; }
    return 0;
})

This code is going to work with any type.

Note that in "real life"™ you often want to ignore case, correctly sort diacritics, weird symbols like ß, etc. when you compare strings, so you may want to use localeCompare. See other answers for clarity.

Answer 3

Something like this:

array.sort(function(a, b){
 var nameA = a.name.toLowerCase(), nameB = b.name.toLowerCase();
 if (nameA < nameB) //sort string ascending
  return -1;
 if (nameA > nameB)
  return 1;
 return 0; //default return value (no sorting)
});

Answer 4

If compared strings contain unicode characters you can use localeCompare function of String class like the following:

users.sort(function(a,b){
    return a.firstname.localeCompare(b.firstname);
})

Answer 5

Nice little ES6 one liner:

users.sort((a, b) => a.firstname !== b.firstname ? a.firstname < b.firstname ? -1 : 1 : 0);

Answer 6

We can use localeCompare but need to check the keys as well for falsey values

The code below will not work if one entry has missing lname.

obj.sort((a, b) => a.lname.localeCompare(b.lname))

So we need to check for falsey value like below

let obj=[
{name:'john',lname:'doe',address:'Alaska'},
{name:'tom',lname:'hopes',address:'California'},
{name:'harry',address:'Texas'}
]
let field='lname';
console.log(obj.sort((a, b) => (a[field] || "").toString().localeCompare((b[field] || "").toString())));

OR

we can use lodash , its very simple. It will detect the returned values i.e whether number or string and do sorting accordingly .

import sortBy from 'lodash/sortBy';
sortBy(obj,'name')

https://lodash.com/docs/4.17.5#sortBy

Answer 7

underscorejs offers the very nice _.sortBy function:

_.sortBy([{a:1},{a:3},{a:2}], "a")

or you can use a custom sort function:

_.sortBy([{a:"b"},{a:"c"},{a:"a"}], function(i) {return i.a.toLowerCase()})

Answer 8

In case we are sorting names or something with special characters, like ñ or áéíóú (commons in Spanish) we could use the params locales (es for spanish in this case ) and options like this:

let user = [{'firstname': 'Az'},{'firstname': 'Áb'},{'firstname':'ay'},{'firstname': 'Ña'},{'firstname': 'Nz'},{'firstname': 'ny'}];


user.sort((a, b) => a.firstname.localeCompare(b.firstname, 'es', {sensitivity: 'base'}))


console.log(user)

The oficial locale options could be found here in iana, es (spanish), de (German), fr (French). About sensitivity base means:

Only strings that differ in base letters compare as unequal. Examples: a ≠ b, a = á, a = A.

Answer 9

A more compact notation:

user.sort(function(a, b){
    return a.firstname === b.firstname ? 0 : a.firstname < b.firstname ? -1 : 1;
})

Answer 10

I'm surprised no one mentioned Collators. You shouldn't use localeCompare unless you have to as it has significantly worse performance

const collator = new Intl.Collator('zh-CN'); // Chinese Simplified for example

function sortAsc(a: string, b: string) {
  return collator.compare(a, b)
}

function sortDesc(a: string, b: string) {
  return collator.compare(b, a);
}

Answer 11

Basically you can sort arrays with method sort, but if you want to sort objects then you have to pass function to sort method of array, so I will give you an example using your array

user = [
  {
    bio: "<null>",
    email: "user@domain.example",
    firstname: "Anna",
    id: 318,
    last_avatar: "<null>",
    last_message: "<null>",
    lastname: "Nickson",
    nickname: "anny",
  },
  {
    bio: "<null>",
    email: "user@domain.example",
    firstname: "Senad",
    id: 318,
    last_avatar: "<null>",
    last_message: "<null>",
    lastname: "Nickson",
    nickname: "anny",
  },
  {
    bio: "<null>",
    email: "user@domain.example",
    firstname: "Muhamed",
    id: 318,
    last_avatar: "<null>",
    last_message: "<null>",
    lastname: "Nickson",
    nickname: "anny",
  },
];

var ar = user.sort(function (a, b) {
  var nA = a.firstname.toLowerCase();
  var nB = b.firstname.toLowerCase();

  if (nA < nB) return -1;
  else if (nA > nB) return 1;
  return 0;
});

Answer 12

try

users.sort((a,b)=> (a.firstname>b.firstname)*2-1)

var users = [
  { firstname: "Kate", id: 318, /*...*/ },
  { firstname: "Anna", id: 319, /*...*/ },
  { firstname: "Cristine", id: 317, /*...*/ },
]

console.log(users.sort((a,b)=> (a.firstname>b.firstname)*2-1) );

Answer 13

also for both asec and desc sort, u can use this : suppose we have a variable SortType that specify ascending sort or descending sort you want:

 users.sort(function(a,b){
            return   sortType==="asc"? a.firstName.localeCompare( b.firstName): -( a.firstName.localeCompare(  b.firstName));
        })

Answer 14

A generalized function can be written like below

    function getSortedData(data, prop, isAsc) {
        return data.sort((a, b) => (a[prop] < b[prop] ? -1 : 1) * (isAsc ? 1 : -1));
   }

you can pass the below parameters

1. The data which you want to sort
2. The property in the data by it should be sorted
3. The last parameter is of boolean type. It checks if you want to sort by ascending or by descending

Answer 15

Inspired from this answer,

users.sort((a,b) => (a.firstname  - b.firstname));

Answer 16

in simply words you can use this method

users.sort(function(a,b){return a.firstname < b.firstname ? -1 : 1});

Answer 17

Just for the record, if you want to have a named sort-function, the syntax is as follows:

let sortFunction = (a, b) => {
 if(a.firstname < b.firstname) { return -1; }
 if(a.firstname > b.firstname) { return 1; }
 return 0;
})
users.sort(sortFunction)

Note that the following does NOT work:

users.sort(sortFunction(a,b))

Answer 18

You can use this for objects

transform(array: any[], field: string): any[] {
return array.sort((a, b) => a[field].toLowerCase() !== b[field].toLowerCase() ? a[field].toLowerCase() < b[field].toLowerCase() ? -1 : 1 : 0);}

Answer 19

Pushed the top answers into a prototype to sort by key.

Array.prototype.alphaSortByKey= function (key) {
    this.sort(function (a, b) {
        if (a[key] < b[key])
            return -1;
        if (a[key] > b[key])
            return 1;
        return 0;
    });
    return this;
};

Answer 20

You can use the in-built array method - sort. This method takes a callback method as a param

    // custom sort function to be passed as param/callback to the Array's sort method
    function myCustomSort(a, b) {
        return (a.toLowerCase() > b.toLowerCase()) ? 1 : -1;
    }

    // Actual method to be called by entity that needs sorting feature
    function sortStrings() {
        var op = Array.prototype.sort.call(arguments, myCustomSort);
    }

    // Testing the implementation
    var sortedArray = sortStrings("Burger", "Mayo1", "Pizza", "boxes", "Apples", "Mayo");
    console.log(sortedArray); //["Apples", "boxes", "Burger", "Mayo", "Mayo1", "Pizza"]

Key Points to be noted for understanding this code.

1. The custom method, in this case, myCustomSort, should return +1 or -1 for each element pair(from the input array) comparison.
2. Use toLowerCase()/toUpperCase() in the custom sorting callback method so that case difference does not affect the correctness of the sorting process.

I hope this is clear enough explanation. Feel free to comment if you think, more info is needed.

Cheers!

Answer 21

My implementation, works great in older ES versions:

sortObject = function(data) {
    var keys = Object.keys(data);
    var result = {};

    keys.sort();

    for(var i = 0; i < keys.length; i++) {
        var key = keys[i];

        result[key] = data[key];
    }

    return result;
};

Answer 22

for a two factors sort (name and lastname):

users.sort((a, b) => a.name.toLowerCase() < b.name.toLowerCase() ? -1 : a.name.toLowerCase() > b.name.toLowerCase() ? 1 : a.lastname.toLowerCase() < b.lastname.toLowerCase() ? -1 : a.lastname.toLowerCase() > b.lastname.toLowerCase() ? 1 : 0)

Answer 23

You can use something similar, to get rid of case sensitive

users.sort(function(a, b){

  //compare two values
  if(a.firstname.toLowerCase() < b.firstname.toLowerCase()) return -1;
  if(a.firstname.toLowerCase() > b.firstname.toLowerCase()) return 1;
  return 0;

})