Switch code will not work.

Question

You are given a program that takes
the number as input. Complete the program so that it will output to the console the theme according to input number.
function main() {
var themeNumber = parseInt(readLine(), 10)
switch(themeNumber){
case "Light":
console.log("Light");
break;
case "Dark":
console.log("Dark");
break;
case "Nocturne":
console.log("Nocturne
");
break;
case "Terminal":
console.log
("Terminal");
break;
case "Indigo":
console.log("Indigo
");
break;
default:
console.log("Another color");
}}

Answer 1

Emmanuel Riano

Any reason you are comparing string inside switch..🤔?

Answer 2

You read <themeNumber> as an integer by parseInt(), but you compare it to string inside the switch.

Answer 3

I honestly thought themenumber was the input expression that will compare the cases and give the output. what is wrong about it?

Answer 4

The problem here is, you can't compare numbers to strings, just like you can't compare apples to bananas.

Answer 5

Right but whats the number in this situation ive used color names and closest thing to a number is the value themenumber

Answer 6

If switch compares the expression to the cases but im trying to get user input by using the variable given. So how am i using a number comparing to a string?

Answer 7

function main() {
var themeNumber = parseInt(readLine(), 10)
If someone could explain what this part of the code does. Does it introduce the function and variable and what is parse int? This loads on every code coach i do automatically and has comments telling me where to add my code so i know its necessary but not what for and how it affects the rest of the code.

Answer 8

switch(themeNumber)
{
case 1:
console.log("Light");
break;
// more cases follows here
}

parseInt() tries to convert the given input into an integer. If given input is somehow not convertible, then a NaN (Not a Number) was returned instead. This is what I've come to know about Javascript.

Answer 9

So do i get rid of parse int for the code to be read as a string or is there another way of doing that?

Answer 10

Try to read the input with just

var themeNumber = readLine();

And provide one of the string options (Light, Dark etc.) as input.

But then again, aestethically, the variable was named <themeNumber>, I would expect it to be created for storing a number rather than string : )

Answer 11

I had an extra space in one of my strings! Sometimes its the smallest details