CODEWITHSUNDEEP
javascriptarrays
dadadodo
dadadodo

Reputation: 347

array.splice returining a value twice

I have this function that replaces an array element at indexOf(before) with a string called "after". This almost works fine except I'm getting two "after" values instead of just one

here is my code: 

function myReplace(str, before, after) {
  var strArr = []; 
  strArr = str.split(' ');
  for (var i = 0; i < strArr.length; i++) {
    strArr.splice(strArr.indexOf(before), 1, after);
  }
  return strArr;
}

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

and here what it returns:

["A", "quick", "brown", "fox", "leaped", "over", "the", "lazy", "leaped"]

what is wrong?

Upvotes: 0

Views: 829

Answers (4)

Richard Casetta
Richard Casetta

Reputation: 446

Array.prototype.splice returns -1 when no element is found : https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf

So in the first loop, it will replace "jumped" by "leaped" as expected, but for all other, it will do strArr(-1, 1, after) which means it will replace the last element with after.

The problem here is your for loop, it checks if before is present for every element in your table. You must use a while loop as in the example provided in the link above (look at Finding all the occurrences of an element)

Upvotes: 1

sidanmor
sidanmor

Reputation: 5189

Use this:

function myReplace(str, before, after) {
   var strArr = []; 
   strArr = str.split(' ');
   for (var i = 0; i < strArr.length; i++) {
      if(strArr[i] === before){
         strArr[i] = after;
      }
   }
   return strArr;
}

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

Upvotes: 0

gurvinder372
gurvinder372

Reputation: 68393

If you simply want to replace, then keep it simple

var regex = new RegExp("jumped", "gi");
var output = "A quick brown fox jumped over the lazy dog".replace( regex, "leaped" );
console.log(output);

You can read more about RegExp here

Upvotes: 1

Caspar Wylie
Caspar Wylie

Reputation: 2833

Ah the problem here is rather unlucky! This is because after it finds element 4, indexOf returns -1 (its way of saying not found), and replaces the last (-1) from the array. 

function myReplace(str, before, after) {
  var strArr = []; 
  strArr = str.split(' ');
  for (var i = 0; i < strArr.length; i++) {
    var index = strArr.indexOf(before);
    if(index!=-1){
        strArr.splice(index, 1, after);
    }
  }
  return strArr;
}

Hope this helps.

Upvotes: 2

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