How can I catch invalid user input in JavaScript?

Question

It checks if the number is bigger or smaller or not and also if the user has quit by entering "q". But it 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`);

Probably because you're not converting the initial guess to an integer. A value of "&" is a non-empty string and is therefore truthy, which means the first while loop would be skipped. — mario_sunny
– mario_sunny, Commented Apr 14, 2021 at 2:51

VikingGlen · Accepted Answer · 2021-04-14 04:12:30Z

0

You can use else if between each conditional. Your final else would be something like, 'Invalid key pressed. Try again.'

if (guess === 'q') {...} else
if (parseInt(guess) < rand) {...} else
if (parseInt(guess) > rand {...}
else guess = prompt('Invalid key entered. Please try again.');

answered Apr 14, 2021 at 4:12

VikingGlen

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fortunee · Accepted Answer · 2021-04-14 04:42:27Z

0

You can compare the value of guess to see if it matches any of the special characters using Regex with String.test.

while (!guess || /[&*()\[\]p]/g.test(guess)) {
  let guess = prompt('enter a valid number');
}

edited Apr 14, 2021 at 4:42

answered Apr 14, 2021 at 3:09

fortunee

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