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

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?

This question is related to javascript

The answer is


Shorter code with ES6

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

To support unicode:

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

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

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/


// 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
*/

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);

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}));

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.

_x000D_
_x000D_
    var cars = [_x000D_
        {make:"AMC",        model:"Pacer",  year:1978},_x000D_
        {make:"Koenigsegg", model:"CCGT",   year:2011},_x000D_
        {make:"Pagani",     model:"Zonda",  year:2006},_x000D_
    ];_x000D_
_x000D_
    function sortObjectsByProp(objectsArr, prop, ascending = true) {_x000D_
        let objectsHaveProp = objectsArr.every(object => object.hasOwnProperty(prop));_x000D_
        if(objectsHaveProp)    {_x000D_
            let newObjectsArr = objectsArr.slice();_x000D_
            newObjectsArr.sort((a, b) => {_x000D_
                if(isNaN(Number(a[prop])))  {_x000D_
                    let textA = a[prop].toUpperCase(),_x000D_
                        textB = b[prop].toUpperCase();_x000D_
                    if(ascending)   {_x000D_
                        return textA < textB ? -1 : textA > textB ? 1 : 0;_x000D_
                    } else {_x000D_
                        return textB < textA ? -1 : textB > textA ? 1 : 0;_x000D_
                    }_x000D_
                } else {_x000D_
                    return ascending ? a[prop] - b[prop] : b[prop] - a[prop];_x000D_
                }_x000D_
            });_x000D_
            return newObjectsArr;_x000D_
        }_x000D_
        return objectsArr;_x000D_
    }_x000D_
_x000D_
    let sortedByMake = sortObjectsByProp(cars, "make"); // returns ascending order by its make;_x000D_
    let sortedByYear = sortObjectsByProp(cars, "year", false); // returns descending order by its year,since we put false as a third argument;_x000D_
    console.log(sortedByMake);_x000D_
    console.log(sortedByYear);
_x000D_
_x000D_
_x000D_


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

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)

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


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;
});