How to get the nth occurrence in a string?
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?
Solution 1:
const string = "XYZ 123 ABC 456 ABC 789 ABC";
function getPosition(string, subString, index) {
return string.split(subString, index).join(subString).length;
}
console.log(
getPosition(string, 'ABC', 2) // --> 16
)
Solution 2:
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
*/
Solution 3:
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;
}
Solution 4:
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;
}
}
}
Solution 5:
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