KimSarabia

var userChoice = "";
do
{
    var targetNumber = Number(prompt("Please enter your target number", ""));
    var start = 0;
    while (start <= targetNumber)
    {
        document.write(start + "<br/>");
        start = start + 2;
    }

KimSarabia

var swapCase = function(string){
    var newString = " ";
    for(var i = 0; i< string.length; i++){
        if(string[i] === string[i].toLowerCase()){
            newString += string[i].toUpperCase();
        }else {
            newString += string[i].toLowerCase();
        }
    }
    console.log(newString);

KimSarabia

function triangular(N) {
  var total = 0;
    for(var i = 1; i <= N; i++){
      total += i;
    }
    return total;
}
console.log(triangular(100));

//5050

KimSarabia

/*Recursion
We’ve seen that % (the remainder operator) can be used to test whether a number 
is even or odd by using % 2 to check whether it’s divisible by two. 
Here’s another way to define whether a positive whole number is even or odd:
Zero is even.
One is odd.
For any other number N, its evenness is the same as N - 2.
Define a recursive function isEven corresponding to this description. 
The function should accept a number parameter and return a Boolean.
Test it on 50 and 75. See how it behaves on -1. Why? Can you think of a way to fix this?

KimSarabia

//www.lsauer.com 2012
//Answer to:
//http://stackoverflow.com/questions/881085/count-the-number-of-occurances-of-a-character-in-a-string-in-javascript/10671743#10671743
//There are at least four ways. The best option, which should also be the fastest -owing to the native RegEx engine -, is placed at the top. //jsperf.com is currently down, otherwise I would provide you with performance statistics.

#1.
     ("this is foo bar".match(/o/g)||[]).length
     //>2
#2.
    "this is foo bar".split("o").length-1

KimSarabia

/*You can get the Nth character, or letter, from a string by writing "string".charAt(N)
similar to how you get its length with "s".length. 
The returned value will be a string containing only one character (for example, "b"). 
The first character has position zero, which causes the last one to be found at position string.length - 1. 
In other words, a two-character string has length 2, and its characters have positions 0 and 1.
Write a function countBs that takes a string as its only argument 
Returns a number that indicates how many uppercase “B” characters are in the string.
Next, write a function called countChar that behaves like countBs, 
except it takes a second argument that indicates the character that is to be counted 
(rather than counting only uppercase “B” characters). Rewrite countBs to make use of this new function.

KimSarabia

<div class='search'>
  <input type="text" id='query' required></input>
  <button id='search'>Search</button>
</div>
<div class="movieOptions">
  <button class="categories" onclick="topMovies()">Top Movies</button>
  <button class="categories" onclick="popular()">Popular</button>
  <button class="categories" onclick="nowPlaying()"> Now Playing </button>
  <button class="categories" onclick="upcoming()"> Upcoming </button>
  <button class="categories" onclick="tvPopular()"> TV Popular </button>

KimSarabia

// takes the form field value and returns true on valid number
function valid_credit_card(value) {
  // accept only digits, dashes or spaces
	if (/[^0-9-\s]+/.test(value)) return false;

	// The Luhn Algorithm. It's so pretty.
	var nCheck = 0, nDigit = 0, bEven = false;
	value = value.replace(/\D/g, "");

	for (var n = value.length - 1; n >= 0; n--) {

KimSarabia

Project Name

TODO: Write a project description

Installation

TODO: Describe the installation process

Usage

KimSarabia

//Source: https://stackoverflow.com/questions/33232823/javascript-compare-two-objects-and-get-key-value-pair
// compare objects with symmetrical keys
function diffObject(a, b) {
  return Object.keys(a).reduce(function(map, k) {
    if (a[k] !== b[k]) map[k] = b[k];
    return map;
  }, {});
}

console.log(diffObject({A: 10, B: 20, C: 30}, {A: 10, B: 22, C: 30}));