Replace a letter with its alphabet position

Question

This looked fairly straightforward to me when I started, but for some reason I'm getting an empty array everytime I try to run the result on codewars. I'm hoping you can help me identify what the problem is.

function alphabetPosition(text) {
  text.split(' ').join('');
  var chari = "";
  var arr = [];
  var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
  for(var i = 0; i < text.len; i++){
    chari = text.charAt(i).toLowerCase();
    if(alphabet.indexOf(chari) > -1){
      arr.push(alphabet.indexOf(chari));
    }
  }
  return arr;
}
console.log(alphabetPosition("Hello World"));

My idea is to get the text from the parameter then strip out the spaces. I made a variable for my empty array and make an alphabet string that I can later search through. In the for loop, i make each character lowercase, and if the character is found in the alphabet string, its position gets pushed into the array (arr). I appreciate your time.

Answer 1

The Kata works with this code. Try with this one:

function alphabetPosition(text) {
  var result = "";
  for (var i = 0; i < text.length; i++) {
    var code = text.toUpperCase().charCodeAt(i)
    if (code > 64 && code < 91) result += (code - 64) + " ";
  }

  return result.slice(0, result.length - 1);
}
console.log(alphabetPosition("The sunset sets at twelve o' clock."));

Answer 2

You need the String#length property

text.length

instead of text.len.

function alphabetPosition(text) {
    var chari,
        arr = [],
        alphabet = "abcdefghijklmnopqrstuvwxyz",
        i;

    for (var i = 0; i < text.length; i++){
        chari = text[i].toLowerCase();
        if (alphabet.indexOf(chari) !== -1){
            arr.push(alphabet.indexOf(chari));
        }
    }
    return arr;
}
console.log(alphabetPosition("Hello World!!1"));

A solution with ES6

function alphabetPosition(text) {
    return [...text].map(a => parseInt(a, 36) - 10).filter(a => a >= 0);
}
console.log(alphabetPosition("Hello World!!1"));