Sort objects in an array alphabetically on one property of the array [duplicate]

Question

Let's say you have a JavaScript class like this

var DepartmentFactory = function(data) {
    this.id = data.Id;
    this.name = data.DepartmentName;
    this.active = data.Active;
}

Let's say you then create a number of instances of that class and store them in an array

var objArray = [];
objArray.push(DepartmentFactory({Id: 1, DepartmentName: 'Marketing', Active: true}));
objArray.push(DepartmentFactory({Id: 2, DepartmentName: 'Sales', Active: true}));
objArray.push(DepartmentFactory({Id: 3, DepartmentName: 'Development', Active: true}));
objArray.push(DepartmentFactory({Id: 4, DepartmentName: 'Accounting', Active: true}));

So I now would have an array of objects created by DepartmentFactory. How would I go about using the array.sort() method to sort this array of objects by the DepartmentName property of each object?

The array.sort() method works just fine when sorting an array of strings

var myarray=["Bob", "Bully", "Amy"];
myarray.sort(); //Array now becomes ["Amy", "Bob", "Bully"]

But how do I make it work with a list of objects?

You can pass a sort function as the first argument to .sort(). — Paul Tomblin
– Paul Tomblin, Commented Jan 17, 2012 at 19:47
Since you're using DepartmentFactory as a constructor, create its objects using new DepartmentFactory, otherwise the array will be filled with a bunch of undefined values. — Anurag
– Anurag, Commented Jan 17, 2012 at 19:49

Omer Bokhari · Accepted Answer · 2012-01-17 20:03:52Z

537

you would have to do something like this:

objArray.sort(function(a, b) {
    var textA = a.DepartmentName.toUpperCase();
    var textB = b.DepartmentName.toUpperCase();
    return (textA < textB) ? -1 : (textA > textB) ? 1 : 0;
});

note: changing the case (to upper or lower) ensures a case insensitive sort.

edited Jan 17, 2012 at 20:03

answered Jan 17, 2012 at 19:52

Omer Bokhari

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3 Comments

MikeT Over a year ago

I've come to your answer 50 gazillion times because i can never remember how to do this. FWIW, my Linter always informs me of "gratuitous parentheses around expression": return (textA < textB) ? -1 : (textA > textB) ? 1 : 0; Not gonna edit your answer, holding out hope I'll at least remember the linting thing :)

Kedar Over a year ago

Found an issue with this code. Use a combination of lower and capital letters. ex. credit_card_no and City. The code does sort the list but words starting with 'c' are not grouped together.

Taran Over a year ago

we can simplify it to return textA.localeCompare(textB);

Oliver Baumann · Accepted Answer · 2025-06-16 14:51:30Z

281

To support unicode:

objArray.sort(function(a, b) {
   return a.DepartmentName.localeCompare(b.DepartmentName);
});

Example

edited Jun 16 at 14:51

Oliver Baumann

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answered Jan 17, 2012 at 20:02

ron tornambe

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4 Comments

Ben Barreth Over a year ago

For a case-insensitive version, use the following on line 2: return a.DepartmentName.toLowerCase().localeCompare(b.DepartmentName.toLowerCase());

obzenner Over a year ago

learn something new every day - localeCompare is cool and has all browser support for first arg. Not bad!

jrz Over a year ago

@BenBarreth there is no reason to lower-case the strings. the whole point of localeCompare is to shunt the work of managing sort logic and locale quirks to the system. If doing a case-insensitive sort is normal for the locale, as it is in english, this will be done for you: "Z" > 'a' // false "Z".localeCompare('a') // 1 If you wish to deviate from the locale's default, you can send along overrides in the locales and options parameters: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

Kris Boyd Over a year ago

Just remember that this isn't null safe, need to check to see if "DepartmentName" is an actual string before running this otherwise the sort will fail on a null pointer exception.

Ludovic · Accepted Answer · 2020-08-21 14:42:57Z

57

Shorter code with ES6

objArray.sort((a, b) => a.DepartmentName.toLowerCase().localeCompare(b.DepartmentName.toLowerCase()))

answered Aug 21, 2020 at 14:42

Ludovic

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Comments

Neil · Accepted Answer · 2020-04-04 18:56:01Z

25

objArray.sort((a, b) => a.DepartmentName.localeCompare(b.DepartmentName))

answered Apr 4, 2020 at 18:56

Neil

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Comments

Diode · Accepted Answer · 2012-01-17 20:37:10Z

var DepartmentFactory = function(data) {
    this.id = data.Id;
    this.name = data.DepartmentName;
    this.active = data.Active;
}

// use `new DepartmentFactory` as given below. `new` is imporatant

var objArray = [];
objArray.push(new DepartmentFactory({Id: 1, DepartmentName: 'Marketing', Active: true}));
objArray.push(new DepartmentFactory({Id: 2, DepartmentName: 'Sales', Active: true}));
objArray.push(new DepartmentFactory({Id: 3, DepartmentName: 'Development', Active: true}));
objArray.push(new DepartmentFactory({Id: 4, DepartmentName: 'Accounting', Active: true}));

function sortOn(property){
    return function(a, b){
        if(a[property] < b[property]){
            return -1;
        }else if(a[property] > b[property]){
            return 1;
        }else{
            return 0;   
        }
    }
}

//objArray.sort(sortOn("id")); // because `this.id = data.Id;`
objArray.sort(sortOn("name")); // because `this.name = data.DepartmentName;`
console.log(objArray);

demo: http://jsfiddle.net/diode/hdgeH/

Alissa · Accepted Answer · 2019-11-20 15:50:20Z

11

objArray.sort( (a, b) => a.id.localeCompare(b.id, 'en', {'sensitivity': 'base'}));

This sorts them alphabetically AND is case insensitive. It's also super clean and easy to read :D

answered Nov 20, 2019 at 15:50

Alissa

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2 Comments

Jules Over a year ago

This is also the only answer which mentions 'sensitivity': 'base', which makes a big difference depending on your application according to MDN.

bilelz Over a year ago

thanks developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

bob · Accepted Answer · 2013-01-09 08:23:21Z

// Sorts an array of objects "in place". (Meaning that the original array will be modified and nothing gets returned.)
function sortOn (arr, prop) {
    arr.sort (
        function (a, b) {
            if (a[prop] < b[prop]){
                return -1;
            } else if (a[prop] > b[prop]){
                return 1;
            } else {
                return 0;   
            }
        }
    );
}

//Usage example:

var cars = [
        {make:"AMC",        model:"Pacer",  year:1978},
        {make:"Koenigsegg", model:"CCGT",   year:2011},
        {make:"Pagani",     model:"Zonda",  year:2006},
        ];

// ------- make -------
sortOn(cars, "make");
console.log(cars);

/* OUTPUT:
AMC         : Pacer : 1978
Koenigsegg  : CCGT  : 2011
Pagani      : Zonda : 2006
*/



// ------- model -------
sortOn(cars, "model");
console.log(cars);

/* OUTPUT:
Koenigsegg  : CCGT  : 2011
AMC         : Pacer : 1978
Pagani      : Zonda : 2006
*/



// ------- year -------
sortOn(cars, "year");
console.log(cars);

/* OUTPUT:
AMC         : Pacer : 1978
Pagani      : Zonda : 2006
Koenigsegg  : CCGT  : 2011
*/

Kris Boyd · Accepted Answer · 2023-04-06 19:57:13Z

7

Because all of the solutions here were presented without null/undefined safe operations, I handle that this way (you can handle nulls as you see fit):

ES5

objArray.sort(
  function(a, b) {
    var departmentNameA = a.DepartmentName ? a.DepartmentName : '';
    var departmentNameB = b.DepartmentName ? b.DepartmentName : '';

    return departmentNameA.localeCompare(departmentNameB);
  }
);

ES6+

objArray.sort(
 (a: DepartmentFactory, b: DepartmentFactory): number => {
   const departmentNameA = a.DepartmentName ? a.DepartmentName : '';
   const departmentNameB = b.DepartmentName ? b.DepartmentName : '';

   return departmentNameA.localeCompare(departmentNameB);
 }
);

I have also removed the toLowerCase that others used since localeCompare is case insensitive. Also I prefer to be a bit more explicit on parameters when using Typescript or ES6+ to make it more explicit for future developers.

edited Apr 6, 2023 at 19:57

answered Jun 14, 2021 at 20:58

Kris Boyd

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2 Comments

RogerSmith Over a year ago

Neither sort actually returns anything so no sorting is happening. Also you can shorten the ternary to use a null coalescing operator (??).

Kris Boyd Over a year ago

@RogerSmith thanks for the comment! You are correct, I mistakenly forgot to return the localeCompare result. The ?? operator was introduced in ES2020, so in 2021 I would have hesitated using anything that fresh due to browser compatibility issues with raw JS. Also quick note, ?? is the "nullish coalescing operator" the null coalescing operator is xxx?.yyy... Its a minor semantic difference in names, but big implication in implementation. Nullish works just as weirdly as Falsey and has a slightly different precedence than ternary. I lean towards ternary for compatibility and explicit syntax.

qwertymk · Accepted Answer · 2012-01-17 20:02:54Z

5

DEMO

var DepartmentFactory = function(data) {
    this.id = data.Id;
    this.name = data.DepartmentName;
    this.active = data.Active;
}

var objArray = [];
objArray.push(new DepartmentFactory({Id: 1, DepartmentName: 'Marketing', Active: true}));
objArray.push(new DepartmentFactory({Id: 2, DepartmentName: 'Sales', Active: true}));
objArray.push(new DepartmentFactory({Id: 3, DepartmentName: 'Development', Active: true}));
objArray.push(new DepartmentFactory({Id: 4, DepartmentName: 'Accounting', Active: true}));

console.log(objArray.sort(function(a, b) { return a.name > b.name}));

answered Jan 17, 2012 at 20:02

qwertymk

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Comments

nickleefly · Accepted Answer · 2013-01-04 07:11:31Z

2

do it like this

objArrayy.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)
});
console.log(objArray)

edited Jan 4, 2013 at 7:11

answered Jan 4, 2013 at 6:55

nickleefly

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Comments

Michael Norward · Accepted Answer · 2020-04-19 21:01:20Z

Here is a simple function you can use to sort array of objects through their properties; it doesn't matter if the property is a type of string or integer, it will work.

    var cars = [
        {make:"AMC",        model:"Pacer",  year:1978},
        {make:"Koenigsegg", model:"CCGT",   year:2011},
        {make:"Pagani",     model:"Zonda",  year:2006},
    ];

    function sortObjectsByProp(objectsArr, prop, ascending = true) {
        let objectsHaveProp = objectsArr.every(object => object.hasOwnProperty(prop));
        if(objectsHaveProp)    {
            let newObjectsArr = objectsArr.slice();
            newObjectsArr.sort((a, b) => {
                if(isNaN(Number(a[prop])))  {
                    let textA = a[prop].toUpperCase(),
                        textB = b[prop].toUpperCase();
                    if(ascending)   {
                        return textA < textB ? -1 : textA > textB ? 1 : 0;
                    } else {
                        return textB < textA ? -1 : textB > textA ? 1 : 0;
                    }
                } else {
                    return ascending ? a[prop] - b[prop] : b[prop] - a[prop];
                }
            });
            return newObjectsArr;
        }
        return objectsArr;
    }

    let sortedByMake = sortObjectsByProp(cars, "make"); // returns ascending order by its make;
    let sortedByYear = sortObjectsByProp(cars, "year", false); // returns descending order by its year,since we put false as a third argument;
    console.log(sortedByMake);
    console.log(sortedByYear);

Alexandre Magno Teles Zimerer · Accepted Answer · 2018-08-17 11:16:20Z

After try a little bit on this, and trying to make less loops as possible, I ended up with this solution:

Demo on codepen

const items = [
      {
        name: 'One'
      },
      {
        name: 'Maria is here'
      },
      {
        name: 'Another'
      },
      {
        name: 'Z with a z'
      },
      {
        name: '1 number'
      },
      {
        name: 'Two not a number'
      },
      {
        name: 'Third'
      },
      {
        name: 'Giant'
      }
    ];

    const sorted = items.sort((a, b) => {
      return a[name] > b[name];
    });

    let sortedAlphabetically = {};

    for(var item in sorted) {
      const firstLetter = sorted[item].name[0];
      if(sortedAlphabetically[firstLetter]) {
        sortedAlphabetically[firstLetter].push(sorted[item]);
      } else {
        sortedAlphabetically[firstLetter] = [sorted[item]]; 
      }
    }

    console.log('sorted', sortedAlphabetically);

zconnelly13 · Accepted Answer · 2012-01-17 19:51:49Z

-4

You have to pass a function that accepts two parameters, compares them, and returns a number, so assuming you wanted to sort them by ID you would write...

objArray.sort(function(a,b) {
    return a.id-b.id;
});
// objArray is now sorted by Id

answered Jan 17, 2012 at 19:51

zconnelly13

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2 Comments

Paul Tomblin Over a year ago

He asked about sorting by DepartmentName, not Id.

tsando Over a year ago

I tried this and this doesn't seem to work on string columns... it worked on a date column. What worked was the solution from @omer-bokhari

Mar · Accepted Answer · 2016-07-29 01:15:53Z

-4

A simple answer:

objArray.sort(function(obj1, obj2) {
   return obj1.DepartmentName > obj2.DepartmentName;
});

ES6 way:

objArray.sort((obj1, obj2) => {return obj1.DepartmentName > obj2.DepartmentName};

If you need to make it lowercase/uppercase etc, just do that and store that result in a variable than compare that variable. Example:

objArray.sort((obj1, obj2) => {
   var firstObj = obj1.toLowerCase();
   var secondObj = obj2.toLowerCase();
   return firstObj.DepartmentName > secondObj.DepartmentName;
});

answered Jul 29, 2016 at 1:15

Mar

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