Modern Javascript

modern-javascript.js

// Lets start with an array
var array = ['hello', 'world'];

// forEach is supported from Javascript 1.6 which means it works in new browsers incl IE9 (not IE8 and back)
array.forEach(function(item){
	console.log(item)
});


// other new array features is like LINQ i C#. filter() returns a new list filtered by the function
var filteredArray = array.filter(function(item){
	return item == 'hello';
});

// returns a new array changed by the function. In this case will return a new array with the contents ['HELLO', 'WORLD'].
array.map(function(item){return item.toUpperCase()});

// every returns true if the whole array returns true.
array.every(function(item){ return item == 'hello' });

// some returns true if any of the array items returns true.
array.some(function(item){ return item == 'hello' });

// returns a part of the array. In this case the first item in the array (same as .pop()).
array.splice(0, 1);

// sort in alphabetic order
var sortedArray = array.sort(function(a,b){
	return a-b;
})

// sort descending alpabetic order
var sortedArray = array.sort(function(a,b){
	return b-a;
})



// now we will show how to use the new functions, but first we must have an array.
var playHistory = [
    {
        artist:'Madonna',
        title:'Frozen'
    },
    {
        artist:'Madonna',
        title:'La Isla Bonita'
    },
    {
        artist:'Madonna',
        title:'American Pie'
    },
    {
        artist:'Madonna',
        title:'Frozen'
    },
    {
        artist:'Madonna',
        title:'Frozen'
    }
];

var group = [];

// OK, now we want to get the artist with most albums
var topSongs = playHistory

// we go trough the array and return a reduced list of top albums
.map(function(item){
    
    // we'll initialize a temporary variable
    if (song = group[item.title]) {
        song.count++;
        return null; // by returning null we'll return a smaller array than we had before
    }
    else
        return group[item.title] = {artist: item.artist, title : item.title, count : 1};
}, topSongs)

// sort on artists with most albums
.sort(function(a,b){return a.count > b.count})

// we'll take top 3, similar to LINQ .take(3);
.slice(0, 3);