Search and Replace [0] of string

JavaScript
1

Tell us what’s happening:
I everyone, I have an issue here. This program is working well but I don’t understand why I need to split the ‘after’ string but not the before string. If I do not split the after string I have an error that after[0] doesn’t exist.

Do you know why ?
thank you.

Your code so far

function myReplace(str, before, after) {
  
  //before = before.split('');
  after = after.split('');
  
  if(before[0] === before[0].toUpperCase()){
     after[0] = after[0].toUpperCase();
  }
  
  //before = before.join('');
  after = after.join('');
  
  return str.replace(before, after);
}

myReplace("A quick brown fox jumped over the lazy dog", "jumped", "leaped");

Your browser information:

Your Browser User Agent is: Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36.

Link to the challenge:

2

Hello!

As with many languages, there are multiple ways to handle this without splitting/joining. I personally believe the easiest way is to just recreate the string. See my code sample below:

function myReplace(str, before, after) {
  
  if(before.charAt(0) == before.charAt(0).toUpperCase()) after = after[0].toUpperCase() + after.substring(1);
  return str.replace(before, after);
}

What my statement above does is use built in JS functions to locate the positions instead of arrays. After which I snag the values I need and rebuild the string without having to do any join/splits.

I don’t personally know the exact answer to yours but I felt I would share my solution as well. I believe yours has some sort of conversion to it, perhaps the split invokes a char type object?

1 Like
3

When I run your code, all tests pass. Try resetting the challenge and pasting your solution in again.

4

ArielLeslie,

They already know their code works. Their question was to why it doesn’t work if they remove a piece of code, see their example.

If they remove the .split() call on the after variable it throws an error. That is why I suggested in my post that perhaps it adds some sort of unicode or “char” type to the string which throws the error.