# javascript – How to get the nth occurrence in a string?-ThrowExceptions

Exception or error:

I would like to get the starting position of the `2nd` occurrence of `ABC` with something like this:

``````var string = "XYZ 123 ABC 456 ABC 789 ABC";
getPosition(string, 'ABC', 2) // --> 16
``````

How would you do it?

How to solve:
``````function getPosition(string, subString, index) {
return string.split(subString, index).join(subString).length;
}
``````

###

You can also use the string indexOf without creating any arrays.

The second parameter is the index to start looking for the next match.

``````function nthIndex(str, pat, n){
var L= str.length, i= -1;
while(n-- && i++<L){
i= str.indexOf(pat, i);
if (i < 0) break;
}
return i;
}

var s= "XYZ 123 ABC 456 ABC 789 ABC";

nthIndex(s,'ABC',3)

/*  returned value: (Number)
24
*/
``````

###

Working off of kennebec’s answer, I created a prototype function which will return -1 if the nth occurence is not found rather than 0.

``````String.prototype.nthIndexOf = function(pattern, n) {
var i = -1;

while (n-- && i++ < this.length) {
i = this.indexOf(pattern, i);
if (i < 0) break;
}

return i;
}
``````

###

Because recursion is always the answer.

``````function getPosition(input, search, nth, curr, cnt) {
curr = curr || 0;
cnt = cnt || 0;
var index = input.indexOf(search);
if (curr === nth) {
if (~index) {
return cnt;
}
else {
return -1;
}
}
else {
if (~index) {
return getPosition(input.slice(index + search.length),
search,
nth,
++curr,
cnt + index + search.length);
}
else {
return -1;
}
}
}
``````

###

Here’s my solution, which just iterates over the string until `n` matches have been found:

``````String.prototype.nthIndexOf = function(searchElement, n, fromElement) {
n = n || 0;
fromElement = fromElement || 0;
while (n > 0) {
fromElement = this.indexOf(searchElement, fromElement);
if (fromElement < 0) {
return -1;
}
--n;
++fromElement;
}
return fromElement - 1;
};

var string = "XYZ 123 ABC 456 ABC 789 ABC";
console.log(string.nthIndexOf('ABC', 2));

>> 16
``````

###

This method creates a function that calls for the index of nth occurrences stored in an array

``````function nthIndexOf(search, n) {
var myArray = [];
for(var i = 0; i < myString.length; i++) { //loop thru string to check for occurrences
if(myStr.slice(i, i + search.length) === search) { //if match found...
myArray.push(i); //store index of each occurrence
}
}
return myArray[n - 1]; //first occurrence stored in index 0
}
``````

###

Using `indexOf` and Recursion:

First check if the nth position passed is greater than the total number of substring occurrences. If passed, recursively go through each index until the nth one is found.

``````var getNthPosition = function(str, sub, n) {
if (n > str.split(sub).length - 1) return -1;
var recursePosition = function(n) {
if (n === 0) return str.indexOf(sub);
return str.indexOf(sub, recursePosition(n - 1) + 1);
};
return recursePosition(n);
};
``````

###

Using `[String.indexOf][1]`

``````var stringToMatch = "XYZ 123 ABC 456 ABC 789 ABC";

function yetAnotherGetNthOccurance(string, seek, occurance) {
var index = 0, i = 1;

while (index !== -1) {
index = string.indexOf(seek, index + 1);
if (occurance === i) {
break;
}
i++;
}
if (index !== -1) {
console.log('Occurance found in ' + index + ' position');
}
else if (index === -1 && i !== occurance) {
}
else {
}
}

yetAnotherGetNthOccurance(stringToMatch, 'ABC', 2);

// Output: Occurance found in 16 position

yetAnotherGetNthOccurance(stringToMatch, 'ABC', 20);

yetAnotherGetNthOccurance(stringToMatch, 'ZAB', 1)

``````

###

``````function getStringReminder(str, substr, occ) {
let index = str.indexOf(substr);
let preindex = '';
let i = 1;
while (index !== -1) {
preIndex = index;
if (occ == i) {
break;
}
index = str.indexOf(substr, index + 1)
i++;
}
return preIndex;
``````

}
console.log(getStringReminder(‘bcdefgbcdbcd’, ‘bcd’, 3));

###

I was playing around with the following code for another question on StackOverflow and thought that it might be appropriate for here. The function printList2 allows the use of a regex and lists all the occurrences in order. (printList was an attempt at an earlier solution, but it failed in a number of cases.)

``````<html>
<title>Checking regex</title>
<script>
var string1 = "123xxx5yyy1234ABCxxxabc";
var search1 = /\d+/;
var search2 = /\d/;
var search3 = /abc/;
function printList(search) {
document.writeln("<p>Searching using regex: " + search + " (printList)</p>");
var list = string1.match(search);
if (list == null) {
document.writeln("<p>No matches</p>");
return;
}
// document.writeln("<p>" + list.toString() + "</p>");
// document.writeln("<p>" + typeof(list1) + "</p>");
// document.writeln("<p>" + Array.isArray(list1) + "</p>");
// document.writeln("<p>" + list1 + "</p>");
var count = list.length;
document.writeln("<ul>");
for (i = 0; i < count; i++) {
document.writeln("<li>" +  "  " + list[i] + "   length=" + list[i].length +
" first position=" + string1.indexOf(list[i]) + "</li>");
}
document.writeln("</ul>");
}
function printList2(search) {
document.writeln("<p>Searching using regex: " + search + " (printList2)</p>");
var index = 0;
var partial = string1;
document.writeln("<ol>");
for (j = 0; j < 100; j++) {
var found = partial.match(search);
if (found == null) {
break;
}
var size = found[0].length;
var loc = partial.search(search);
var actloc = loc + index;
document.writeln("<li>" + found[0] + "  length=" + size + "  first position=" + actloc);
// document.writeln("  " + partial + "  " + loc);
partial = partial.substring(loc + size);
index = index + loc + size;
document.writeln("</li>");
}
document.writeln("</ol>");

}
</script>
<body>
<p>Original string is <script>document.writeln(string1);</script></p>
<script>
printList(/\d+/g);
printList2(/\d+/);
printList(/\d/g);
printList2(/\d/);
printList(/abc/g);
printList2(/abc/);
printList(/ABC/gi);
printList2(/ABC/i);
</script>
</body>
</html>``````