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how i get out from the loop if switch case implemented(there is a switch inside the loop)

How I get out from the loop if switch-case implemented (there is a switch inside the loop).
function playInbestPlace() {
console.log("hello from playInbestPlace ")
findEmptyarea();
for (var i = 0; i < indexOfEmpty.length; i++) {
var elem = indexOfEmpty[i];
switch (elem) {
case 0:
cells[elem].childNodes[0].append("o");
break;
case 2:
cells[elem].childNodes[0].append("o");
break;
case 4:
cells[elem].childNodes[0].append("o");
break;
case 6:
cells[elem].childNodes[0].append("o");
break;
case 8:
cells[elem].childNodes[0].append("o");
break;
}
}
}
I want it to get out if any case valid.

you can add a variable found and break out of the loop if it's true :
function playInbestPlace() {
console.log("hello from playInbestPlace ")
findEmptyarea();
for (var i = 0; i < indexOfEmpty.length; i++) {
var elem = indexOfEmpty[i];
var found = false; // initial found is false
switch (elem) {
case 0:
cells[elem].childNodes[0].append("o");
found = true;
break;
case 2:
cells[elem].childNodes[0].append("o");
found = true;
break;
case 4:
cells[elem].childNodes[0].append("o");
found = true;
break;
case 6:
cells[elem].childNodes[0].append("o");
found = true;
break;
case 8:
cells[elem].childNodes[0].append("o");
found = true;
break;
}
if(found) // break out if it's true
break;
}
}

You could use a flag variable to break from the loop when some condition is verified.
function playInbestPlace() {
console.log("hello from playInbestPlace ");
findEmptyarea();
var keepOnLooping = true;
for (var i = 0; keepOnLooping && i < indexOfEmpty.length; i++) {
if (elem % 2 === 0) {
cells[elem].childNodes[0].append("o");
keepOnLooping = false;
}
}
}
I've also added epascarello optimization in my answer.

Javascript switch statement?

I have some code like below:
for (var i = 0; i < $scope.Option.length; i++) {
var option = $scope.Option[i].Code;
if (option == "A") {
$scope.aSelected = true;
break;
}
}
for (var i = 0; i < $scope.Option.length; i++) {
var option = $scope.Option[i].Code;
if (option == "B") {
$scope.bSelected = true;
break;
}
}
Is it possible to right this in a switch statement like below:
for (var i = 0; i < $scope.Option.length; i++) {
var option = $scope.Option[i].Code;
switch (option) {
case "A":
$scope.aSelected = true;
break;
case "B":
$scope.bSelected = true;
break;
default:
console.log('unrecognized option');
}
}
Is this actually incorrect in the switch case because the first option may be A which will break out of the loop and then for example if 'B' was the option in a later position of the collection it would never get bSelected = true;

It's correct because the BREAK inside of SWITCH will break out of it, not the whole FOR loop, and therefore it will check for B too.

change html page according to day (mo, tu etc.)

window.onload=function(){
var day=new Date().getDay();
switch (day)
{
case 0:
window.document.location.href = 'su.html';
break;
case 1:
window.document.location.href = 'mo.html';
break;
case 2:
window.document.location.href = 'tu.html';
break;
case 3:
window.document.location.href = 'we.html';
break;
case 4:
window.document.location.href = 'th.html';
break;
case 5:
window.document.location.href = 'fr.html';
break;
case 6:
window.document.location.href = 'sa.html';
break;
}
};
it does not load the current html page. It loads the index.html file
can somebody pls tell me how to make this could work ?
thanks!
i've got an array whith week days

You can't put the javascript in every page, but only in a "loading" page that execute the code and navigate the user in the correct page.
Loading page (With your script) -> Go in the correct page

You will need to change your script to something like this if you want to have the same script in every page:
window.onload=function(){
var day=new Date().getDay();
var path = window.location.pathname;
var page = path.split("/").pop();
if(day == 0 && page != 'su.html')
window.document.location.href = 'su.html';
else if(day == 1 && page != 'mo.html')
window.document.location.href = 'mo.html';
else if(day == 2 && page != 'tu.html')
window.document.location.href = 'tu.html';
else if(day == 3 && page != 'we.html')
window.document.location.href = 'we.html';
else if(day == 4 && page != 'th.html')
window.document.location.href = 'th.html';
else if(day == 5 && page != 'fr.html')
window.document.location.href = 'fr.html';
else if(day == 6 && page != 'sa.html')
window.document.location.href = 'sa.html';
};

Calling a method is giving me code and not the value (Javascript)

basically I have created an object with a method which adds several attributes in the object together. But when I try and call the method to the console log it fires out the code (which is a if statement) to me instead of the value which I was hoping it would return, so I am confused!
Why is this happening ?
Code Below:
var Granite = function(ty, gr, th, wi, le, ed, ad){
this.type = ty;
this.group = gr;
this.thickness = th;
this.width = wi;
this.length = le;
this.edgeProfile = ed;
this.addOns = ad;
this.groupPrice = function(){
if (thickness === 20){
switch(group)
{
case 1:
return 160;
break;
case 2:
return 194;
break;
case 3:
return 244;
break;
case 4:
return 288;
break;
case 5:
return 336;
break;
case 6:
return 380;
break;
default:
return 380;
}
}else{
switch(group)
{
case 1:
return 200;
break;
case 2:
return 242;
break;
case 3:
return 305;
break;
case 4:
return 360;
break;
case 5:
return 420;
break;
case 6:
return 475;
break;
default:
return 475;
}
}
}
this.price = function(){
if(length <= 2000 && length > 1000){
return ((edgeProfile + groupPrice)*2) - addOns;
}else if(length <= 3000 && length > 2000){
return ((edgeProfile + groupPrice)*3) - addOns;
}else if(length <= 4000 && length > 3000){
return ((edgeProfile + groupPrice)*4) - addOns;
}else if(length <= 5000 && length > 4000){
return ((edgeProfile + groupPrice)*5) - addOns;
}
}
}
var granite1 = new Granite("Rosa Porrino", 1, 30, 400, 3200, 30.05, 86.18);
console.log(granite1.groupPrice);
It returns the full if statement within the groupPrice method to me

You are not calling the method but providing reference of function to console,log(). In JavaScript you need use '()' to call a function.
this will surely work console.log(granite1.groupPrice());
In side this.price
use this.groupPrice(). instead of groupPrice
Modified this,price Method
this.price = function(){
if(length <= 2000 && length > 1000){
return ((this.edgeProfile + this.groupPrice())*2) - addOns;
}else if(length <= 3000 && length > 2000){
return ((this.edgeProfile + this.groupPrice())*3) - addOns;
}else if(length <= 4000 && length > 3000){
return ((this.edgeProfile + this.groupPrice())*4) - addOns;
}else if(length <= 5000 && length > 4000){
return ((this.edgeProfile + this.groupPrice())*5) - addOns;
}
}

If you're calling the function, append with () otherwise you're just referencing the function.

Loop program again after method is executed

i guess this is a really simple problem, but i just can't get it to work! I want that my menu should show up again after that i have run a method. Thanks in advance.
function menu () {
var choice = prompt("0. Exit \n\n1. Fahrenheit to Celsius \n2. Celsius to Fahrenheit \n3. Guess a number");
choice = parseInt(choice);
if (choice > 4 || choice < 0) {
alert("FEL!!");
} else if (isNaN(choice)) {
alert("Måste vara en siffra");
}
switch (choice) {
case 0:
choice = false;
break;
case 1:
CelsiusToFarenheit();
break;
case 2:
FahrenheitToCelsius();
break;
case 3:
Guess();
break;
}
return choice;
}
do {
menu();
} while(choice == true);

you forgot to store the variable returned by the menu() function :
function menu () {
var choice = prompt("0. Exit \n\n1. Fahrenheit to Celsius \n2. Celsius to Fahrenheit \n3. Guess a number");
choice = parseInt(choice);
if (choice > 4 || choice < 0) {
alert("FEL!!");
} else if (isNaN(choice)) {
alert("Måste vara en siffra");
}
switch (choice) {
case 0:
choice = false;
break;
case 1:
CelsiusToFarenheit();
break;
case 2:
FahrenheitToCelsius();
break;
case 3:
Guess();
break;
}
return choice;
}
var choice;
do {
choice = menu();
} while(choice == true);

You are calling Menu() and after it executes, it is done. Therefore you need to run Menu() again.