如何在javascript中遍历两个索引

Question

why it's giving me the first index and i need the ouput below which is "StOp mAkInG SpOnGeBoB MeMeS!"

 function spongeMeme(sentence) {
  let x = sentence.split("");
  for(let i = 0; i < sentence.length;i+=2){
    return x[i].toUpperCase()
  }
}
console.log(spongeMeme("stop Making spongebob Memes!")) // S

// output : StOp mAkInG SpOnGeBoB MeMeS!"

Answer 1

Try this:

        function spongeMeme(sentence) {
        let x = sentence.split("");
        let newSentence = '';
        for(let i = 0; i < sentence.length;i++){
            // If the letter is even
            if(i % 2 ==0) {
                newSentence += x[i].toUpperCase();
            } 
            // If the letter is odd
            else {
                newSentence += x[i].toLowerCase();
            }
        }
        return newSentence
      }
      console.log(spongeMeme("stop Making spongebob Memes!"))

First of all you can only return once per function, so you have to create a variable and store all the new letter inside, and, at the end, return this variable.

Answer 2

You're returning just the first letter of what is stored inside of x. Try to debug it and see what is going on.

The way I'd do it is like so:

function spongeMeme(sentence) {
return sentence.split("")
.map((s, i) => (i % 2 != 0 ? s.toUpperCase() : s))
.join("");
}
console.log(spongeMeme("stop Making spongebob Memes!"));

Answer 3

You're immediately returning on the first iteration of your for loop, hence the reason you're only getting back a single letter.

Let's start by fixing your existing code:

function spongeMeme(sentence) {
  let x = sentence.split("");

  for(let i = 0; i < sentence.length; i+=2) {

    // You can't return here. Update the value instead.
    return x[i].toUpperCase()
  }

  // Join your sentence back together before returning.
  return x.join("");
}

console.log(spongeMeme("stop Making spongebob Memes!"))

That'll work, but we can clean it up. How? By making it more declarative with map.

function spongeMeme(sentence) {
  return sentence
    .split('') // convert sentence to an array
    .map((char, i) => { // update every other value
      return (i % 2 === 0) ? char.toUpperCase() : char.toLowerCase(); 
    })
    .join(''); // convert back to a string
}

Answer 4

You can also do that:

 const spongeMeme = (()=> { const UpperLower = [ (l)=>l.toUpperCase(), (l)=>l.toLowerCase() ] return (sentence) => { let i = -1, s = '' for (l of sentence) s += UpperLower[i=++i&1](l) return s } })() console.log(spongeMeme("stop Making spongebob Memes!"))

 .as-console-wrapper {max-height: 100% !important;top: 0;} .as-console-row::after {display: none !important;}

OR

function spongeMemez(sentence)
  {
  let t = false, s = ''
  for (l of sentence) s += (t=!t) ? l.toUpperCase() : l.toLowerCase()
  return s
  }

but I think this version is slower, because using an index should be faster than a ternary expression