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?

Answers:

Answer

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.

Answer

To support unicode:

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

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}));
Answer

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)
Answer
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

Answer

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
Answer

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);
Answer

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;
});
Answer
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/

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

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);

Answer

Shorter code with ES6

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

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