JavaScript - textContent 实现

Question

I am currently working on a HTML Calculator as part of the Odin Project.

I have encountered a weird behavior of .textContent: Here is my Code snippet from JavaScript:

//---output is a HTML tag at which the input of the user should be entered-------------------------------
const output=document.querySelector('.Output');

//------I store the input (pressed keys) into an array
let input=[]
//-------------------------------------------------

//--------------keybord support--------------------
document.addEventListener('keydown', function(e) {
    if (e.key != "+" || e.key !="-" || e.key !="*" || e.key !="/") {
        let internalVariable = 0;
        input.push(parseInt(e.key));
        internalVariable=input.join('');
        output.innerHTML=internalVariable;
    }   

    if (e.key=="+") {
        console.log(typeof e.key,input)**-> Test if condition works**
    }
    

Problem is: Whenever I press + button, I still get an output (NaN) and I get an entry(NaN) into my input-array whoch should not happen.

Did I miss understood text.Content ?

Answer 1

The problem is this line:

if (e.key != "+" || e.key !="-" || e.key !="*" || e.key !="/") {

Let's simplify it to just two conditions:

if (e.key != "+" || e.key !="-") {
}

This will always be true . If the key is + , then it will not be - , satisfying the second part. If the key is - , then it will not be + , satisfying the first part.

Use a string of keys instead, and check if the pressed key is included in it.

document.addEventListener('keydown', function (e) {
    if ('+-*/'.includes(e.key)) {

Or, another option is to check if the key is a digit.

if (/\d/.test(e.key)) {

Demo:

 const output = document.querySelector('.Output'); const input = [] document.addEventListener('keydown', function(e) { if (/\\d/.test(e.key)) { input.push(parseInt(e.key)); internalVariable = input.join(''); output.innerHTML = internalVariable; } else { console.log('do something when a non-digit was pressed'); } });

 <div class="Output"></div>