How to check if one string ONLY contains items of an array?

Question

Yo guys, I have an array of the alphabet and I wanna check whether a string only contains the items of this alphabet. If not, my bot kick a player from a game. I managed to write this script:

var letters =
    "a b c d e f g h i j k l m n o p q r s t u v w x y z";
  var arrayLetters = letters.split(/ +/);
  var allLetters = arrayLetters.map((x) => x.toLowerCase());

  if (!allLetters.some((x) => player.name.includes(x)) {
    room.kickPlayer("Your name can't have strange fonts");
  }

Actually, this works but with names that doesn't contain a letter at all.

For example, 𝐏𝐚𝐧𝐝𝐚 would be kicked but 𝐏𝐚nda won't because there are some letters which are part of the array. How can I solve this? I wanna also keep players who only contains those letters in my array.

Thanks so much for the answers!

Does this answer your question? stackoverflow.com/questions/23476532/… — match
– match, Commented Sep 11, 2021 at 15:54
What you have checked is if the string contains at least one element of the array. For example player.name = ["2","a"] will always return` true` because there is at least to whom to check equality so your checking is not efficient. — segandiaye
– segandiaye, Commented Sep 11, 2021 at 15:58

vanowm · Accepted Answer · 2021-09-11 16:19:24Z

0

It sounds like there is a simple regex solution for this:

var letters = "abcdefghijklmnopqrstuvwxyz";
var regex = new RegExp("[^" + letters + "]", "i");
var players = ["𝐏𝐚𝐧𝐝𝐚", "𝐏𝐚nda", "Panda"];
players.forEach(p =>
{
  console.log(p, regex.test(p));
});

The basic idea is to match any characters that are NOT part of the letters

This particular case can also be shortned by using [^a-z] instead:

var regex = new RegExp("[^a-z]", "i");
var players = ["𝐏𝐚𝐧𝐝𝐚", "𝐏𝐚nda", "Panda"];
players.forEach(p =>
{
  console.log(p, regex.test(p));
});

P.S. You don't need separate your alphabet with a space in order to split it. "abcd".split("") would do same thing

edited Sep 11, 2021 at 16:19

answered Sep 11, 2021 at 16:06

vanowm

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Comments

Hai Thai · Accepted Answer · 2021-09-11 16:18:51Z

0

Regex is the best solution for you.

var letters = "a b c d e f g h i j k l m n o p q r s t u v w x y z";
var arrayLetters = letters.split(/ +/);
var allLetters = arrayLetters.map((x) => x.toLowerCase());
var regex = new RegExp(`[^${allLetters.join('')}]`, 'i');
if (regex.test(player.name)) {
  room.kickPlayer("Your name can't have strange fonts");
}

answered Sep 11, 2021 at 16:18

Hai Thai

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Comments

Andy · Accepted Answer · 2021-09-11 17:25:18Z

0

Do it the other way around. Use some to iterate over the name characters and test to see if they exist in letters.

const letters = 'a b c d e f g h i j k l m n o p q r s t u v w x y z';

function validLetters(letters, name) {
  const found = [...name].some(letter => letters.includes(letter));
  return found ? `${name}: ok` : `${name}: Your name can\'t have strange fonts`;
}

console.log(validLetters(letters, '𝐏𝐚𝐧𝐝𝐚'));
console.log(validLetters(letters, 'Bob'));
console.log(validLetters(letters, 'BOB'));
console.log(validLetters(letters, '123'));
console.log(validLetters(letters, '𝐏𝐚nda'));

answered Sep 11, 2021 at 17:25

Andy

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