CODEWITHSUNDEEP
javascriptjqueryloopsindexingindexof
Cory
Cory

Reputation: 135

Looping Through String To Find Multiple Indexes

I am trying to figure out the most efficient way of looping through a string and finding all indexes of a certain letter.

I have used $word_or_phrase.indexOf( $letter ); to find a single index of a letter, but the letter is in $word_or_phrase multiple times. Would the most efficient way to do this be to build an array of all the indexes until .indexOf returns -1? Or how would you suggest me finding all the the indexes?

I have already taken time and found this: Javascript str.search() multiple instances

This works, but to me doesn't seem efficient when dealing with more than 2 indexes. What if I had 10?

Thanks for the advice in advance!

Upvotes: 3

Views: 10539

Answers (5)

Ian
Ian

Reputation: 50933

As the answer in the StackOverflow link you posted shows, you can use the second parameter of indexOf to define where the search starts in the string. You can continue looping over the string, using this technique, to get the indexes of all matched substrings:

function getMatchIndexes(str, toMatch) {
    var toMatchLength = toMatch.length,
        indexMatches = [], match,
        i = 0;

    while ((match = str.indexOf(toMatch, i)) > -1) {
        indexMatches.push(match);
        i = match + toMatchLength;
    }

    return indexMatches;
}

console.log(getMatchIndexes("asdf asdf asdf", "as"));

DEMO: http://jsfiddle.net/qxERV/

Another option is to use a regular expression to find all matches:

function getMatchIndexes(str, toMatch) {
    var re = new RegExp(toMatch, "g"),
        indexMatches = [], match;

    while (match = re.exec(str)) {
        indexMatches.push(match.index);
    }

    return indexMatches;
}

console.log(getMatchIndexes("asdf asdf asdf", "as"));

DEMO: http://jsfiddle.net/UCpeY/

And yet another option is to manually loop through the string's characters and compare to the target:

function getMatchIndexes(str, toMatch) {
    var re = new RegExp(toMatch, "g"),
        toMatchLength = toMatch.length,
        indexMatches = [], match,
        i, j, cur;

    for (i = 0, j = str.length; i < j; i++) {
        if (str.substr(i, toMatchLength) === toMatch) {
            indexMatches.push(i);
        }
    }

    return indexMatches;
}

console.log(getMatchIndexes("asdf asdf asdf", "as"));

DEMO: http://jsfiddle.net/KfJ9H/

Upvotes: 5

Bilal Murtaza
Bilal Murtaza

Reputation: 795

check this ...

var data = 'asd 111 asd 222 asd 333';
var count = countOccurence('asd', data);
console.info(count);
function countOccurence(item, data, count){
    if (count == undefined) { count = 0; }
    if (data.indexOf(item) != -1)
    {
        count = count+1;
        data = data.substring(data.indexOf(item) + item.length);
        count = countOccurence(item, data, count);
    }
    return count;
}

Upvotes: 1

A. Wolff
A. Wolff

Reputation: 74410

Could be a solution:

http://jsfiddle.net/HkbpY/

var str = 'some kind of text with letter e in it',
    letter = 'e',
    indexes = [];

$.each(str.split(''),function(i,v){
    if(v === letter) indexes.push(i);
});

console.log(indexes);

Upvotes: 2

Dineshkani
Dineshkani

Reputation: 3015

Try like this

var str = "foodfoodfoodfooooodfooooooood";
for (var index = str.indexOf("o");index > 0; index = str.indexOf("o", index+1)){
console.log(index);
}

See Demo

Upvotes: 0

sjkm
sjkm

Reputation: 3937

var mystring = 'hello world';
var letterToCount = 'l';

var indexes = [];
for(var i=0; i<mystring.length; i++) {
    if(mystring[i] == letterToCount)
       indexes.push(i); 
}

alert(indexes.join(',')); //2,3,9

Upvotes: 0

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