Why cant I catch invalid input in this code?

Question

Why this code doesn't catch invalid input like "&, *, (, p, ]" with "while (!guess)"?

let guess= window.prompt('guess the number!');
const rand=Math.floor(Math.random()*10)+1;
let tries= 0;
while (!guess){

let guess=parseInt(prompt('enter a valid number'));

}

while (parseInt(guess)!==rand){
tries+=1;
if (guess==='q'){
console.log('you have exited');
break;
}
if (guess<rand)
guess=prompt('enter higher!');

else (guess>rand)
guess=prompt('enter lower!');

}

console.log(`correct number is : ${rand} and it took you ${tries} tries`);

Answer 1

If the first value you enter into prompt is "&", that won't be equated with true. If you enter an empty string in the first prompt, you'll get into the while-loop that you want.

The while-loop won't work the way you want because it isn't written very well. let game =... declares another variable with the same name in a scope that you don't want. The scope of that while-loop game variable is between the { } brackets of the loop so the while-loop condition will always be true.

The following will work more the way you want:
function getGuessedNumber(msg) {
let guess= prompt(msg);
if (guess === 'q')
return guess;
guess = parseInt(guess);
while (isNaN(guess)){
guess=prompt('enter a valid number');
if (guess === 'q')
return guess;
else
guess = parseInt(guess);
}
return guess;
}

const rand=Math.floor(Math.random()*10)+1;
let tries= 0;
let guess = getGuessedNumber('guess the number!');
while (guess !== rand){
tries+=1;
if (guess==='q'){
console.log('you have exited');
break;
}
if (guess<rand)
guess=getGuessedNumber('enter higher!');
else if (guess>rand)
guess=getGuessedNumber('enter lower!');
}

console.log(`correct number is : ${rand} and it took you ${tries} tries`);