Javascript multi statement

Question

I am building little app with JS and I find difficult with that:

I want to get from users input an integer number from range 0-100. If it's not an int I am parsing it to int. Then I want to put that number to function checkNum, and there appears a problem. Let me paste my code:

$(function () {

initApp();

function getInput(){
    var userInput = window.prompt("Input number from range 0-100");
    var num = parseInt(userInput);
    return num;
};


function checkInput(num){
    if( num < 0 || num > 100){
        displayError();
    }
    else{
        alert('num is ok');
    }
};

function displayError(){
  alert("error");
};

function initApp(){
    getInput();
    checkInput();
};

});

As you can see even if I put number out of range there is always alert says num is ok - what is not true. I have no idea what I am doing wrong. I'll appreciate any help. What is more, I'd like to put here function Number.isNaN() to provide typing non-sense in the input. I thought about statement like this:

if(Number.isNaN(num) || num <0 || num > 100){...}

I guess that my mistake is obvious and trivial for someone more experienced than me.

This is unrelated to your question, but should always explicitly set the second parameter of parseInt(): parseInt(userInput, 10); — Arnauld
– Arnauld, Commented Jul 25, 2016 at 16:53

Madara's Ghost · Accepted Answer · 2016-07-25 16:53:35Z

1

To answer your question, you're passing nothing to checkInput() so it's always undefined, and undefined < 0 || undefined > 100 is always false. The solution is to pass the returned value from getInput() to checkInput() like so:

function initApp(){
    var value = getInput();
    checkInput(value);
}

To answer what you're aiming for: use <input type="number"> and have the browser do the work for you.

<form>
  <input type="number" min="0" max="100" step="1" />
  <button>Try to submit!</button>
</form>

No JavaScript required.

answered Jul 25, 2016 at 16:53

Madara's Ghost

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Comments

T.J. Crowder · Accepted Answer · 2016-07-25 16:53:02Z

0

You need to pass the return value from getInput into checkInput:

function initApp(){
    checkInput(getInput());
}

Side note: Function declarations (what you've used) don't need ; at the end.

answered Jul 25, 2016 at 16:53

community wiki

T.J. Crowder

Comments

R. Foubert · Accepted Answer · 2016-07-25 16:53:37Z

0

You just need to give the value returned by getInput like this ;)

initApp();

function getInput(){
    var userInput = window.prompt("Input number from range 0-100");
    var num = parseInt(userInput);
    return num;
};


function checkInput(num){
    if( num < 0 || num > 100){
        displayError();
    }
    else{
        alert('num is ok');
    }
};

function displayError(){
  alert("error");
};

function initApp(){
    var value = getInput();
    checkInput(value);
};

answered Jul 25, 2016 at 16:53

R. Foubert

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