Check for Palindromes - eyeforofeye

JavaScript
1

Tell us what’s happening:

Your code so far

function palindrome(str) {
  
  //string interpolating uses special single quotes and ${variable}
  console.log(`^\n : ${str}`);

  //get rid of special characters
  let s = str.replace(/[^a-zA-Z ]/g, "");
  
  //remove spaces 
  s = s.replace(/\s/g, '');
  
  //lowercase
  let sl = s.toLowerCase();
  console.log(`   compare : ${sl}`);
  
  //string to array 
  let a = sl.split("");
  
  
  //reverse the array
  let r = a.reverse();
  
  //create a string from array using join()
  let jstr = r.join("");
  
  console.log(`   Reverse : ${jstr}`);
  
  if(jstr == sl ){
    console.log(" :  true")
    return true;
  }  
  else{
    console.log(" :  false")
    return false;
  }
}


palindrome("eye");

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Link to the challenge:
https://www.freecodecamp.org/challenges/check-for-palindromes

One of the tests for Palindrome is :palindrome(“1 eye for of 1 eye.”) should return false.
once the '1’s are removed from the string it reads “eyeforofeye” which is a palindrome.
but the question indicates that it should not be a palindrome.

2

This question comes up a lot, but it’s because of something said in the question:

Blockquote
You’ll need to remove all non-alphanumeric characters

The number 1 then shouldn’t be removed, and of course then 1eyeforof1eye is not the same as eye1forofeye1