MucaP MucaP - 1 year ago 38
Javascript Question

How to open different local files according to the result?

I'm very new to javascript and I'm having an issue with this script:

function processInp() {
var x = parseInt(document.querySelector('input[name="num"]').value);
if (!x) {
return false
} else if (x > 480) {
return false
}"../files/" + x + '.html');
return true

First, it checks if the input is a number, then if it is higher than 480, so throw an alert and an error. That's completly fine and working. What I really can't do is opening the url according to the input. Is it possible with numbers? How? Full example if possible. Thank you!

Stu Stu

Looking at your example, I think this might be due to scope.

As changing the line function processInp() { to window.processInp = function () { (i.e. seems to work correctly?

I think this is a pretty good post about scope if you want to read up further: What is the scope of variables in JavaScript?