How to get the nth occurrence in a string

Question

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     -- gt  16   How would you do it

User · Answer

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.)

_x000D_

<html>_x000D_
<head>_x000D_
<title>Checking regex</title>_x000D_
<script>_x000D_
var string1 = "123xxx5yyy1234ABCxxxabc";_x000D_
var search1 = /\d+/;_x000D_
var search2 = /\d/;_x000D_
var search3 = /abc/;_x000D_
function printList(search) {_x000D_
   document.writeln("<p>Searching using regex: " + search + " (printList)</p>");_x000D_
   var list = string1.match(search);_x000D_
   if (list == null) {_x000D_
      document.writeln("<p>No matches</p>");_x000D_
      return;_x000D_
   }_x000D_
   // document.writeln("<p>" + list.toString() + "</p>");_x000D_
   // document.writeln("<p>" + typeof(list1) + "</p>");_x000D_
   // document.writeln("<p>" + Array.isArray(list1) + "</p>");_x000D_
   // document.writeln("<p>" + list1 + "</p>");_x000D_
   var count = list.length;_x000D_
   document.writeln("<ul>");_x000D_
   for (i = 0; i < count; i++) {_x000D_
      document.writeln("<li>" +  "  " + list[i] + "   length=" + list[i].length + _x000D_
          " first position=" + string1.indexOf(list[i]) + "</li>");_x000D_
   }_x000D_
   document.writeln("</ul>");_x000D_
}_x000D_
function printList2(search) {_x000D_
   document.writeln("<p>Searching using regex: " + search + " (printList2)</p>");_x000D_
   var index = 0;_x000D_
   var partial = string1;_x000D_
   document.writeln("<ol>");_x000D_
   for (j = 0; j < 100; j++) {_x000D_
       var found = partial.match(search);_x000D_
       if (found == null) {_x000D_
          // document.writeln("<p>not found</p>");_x000D_
          break;_x000D_
       }_x000D_
       var size = found[0].length;_x000D_
       var loc = partial.search(search);_x000D_
       var actloc = loc + index;_x000D_
       document.writeln("<li>" + found[0] + "  length=" + size + "  first position=" + actloc);_x000D_
       // document.writeln("  " + partial + "  " + loc);_x000D_
       partial = partial.substring(loc + size);_x000D_
       index = index + loc + size;_x000D_
       document.writeln("</li>");_x000D_
   }_x000D_
   document.writeln("</ol>");_x000D_
_x000D_
}_x000D_
</script>_x000D_
</head>_x000D_
<body>_x000D_
<p>Original string is <script>document.writeln(string1);</script></p>_x000D_
<script>_x000D_
   printList(/\d+/g);_x000D_
   printList2(/\d+/);_x000D_
   printList(/\d/g);_x000D_
   printList2(/\d/);_x000D_
   printList(/abc/g);_x000D_
   printList2(/abc/);_x000D_
   printList(/ABC/gi);_x000D_
   printList2(/ABC/i);_x000D_
</script>_x000D_
</body>_x000D_
</html>

_x000D_

User · Answer

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  lt  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

User · Answer

Shorter way and I think easier  without creating unnecessary strings   const findNthOccurence    string  nth  char    gt      let index   0   for  let i   0  i  lt  nth  i    1        if  index     -1  index   string indexOf char  index   1        return index

User · Answer

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  gt  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

User · Answer

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--  amp  amp  i    lt  this length            i   this indexOf pattern  i           if  i  lt  0  break             return i

User · Answer

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--  amp  amp  i   lt L           i  str indexOf pat  i           if  i  lt  0  break            return i     var s   XYZ 123 ABC 456 ABC 789 ABC    nthIndex s  ABC  3       returned value   Number  24

User · Answer

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

User · Answer

a simple solution just add string  character and idx  function getCharIdx str char n     let r   0   for  let i   0  i lt str length  i         if  str i      char         r         if  r     n           return i

User · Answer

x000D   x000D  const string    XYZ 123 ABC 456 ABC 789 ABC   x000D   x000D  function getPosition string  subString  index    x000D    return string split subString  index  join subString  length  x000D    x000D   x000D  console log  x000D    getPosition string   ABC   2     -- gt  16 x000D    x000D   x000D   x000D

User · Answer

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

User · Answer

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  gt  0            fromElement   this indexOf searchElement  fromElement           if  fromElement  lt  0                return -1                    --n            fromElement            return fromElement - 1      var string    XYZ 123 ABC 456 ABC 789 ABC   console log string nthIndexOf  ABC   2      gt  gt  16

User · Answer

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  amp  amp  i     occurance            console log  Occurance not found in     occurance     position              else           console log  Occurance not found             yetAnotherGetNthOccurance stringToMatch   ABC   2       Output  Occurance found in 16 position  yetAnotherGetNthOccurance stringToMatch   ABC   20       Output  Occurance not found in 20 position  yetAnotherGetNthOccurance stringToMatch   ZAB   1      Output  Occurance not found

[javascript] How to get the nth occurrence in a string?

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