How to find indices of all occurrences of one string in another in JavaScript

Question

I m trying to find the positions of all occurrences of a string in another string  case-insensitive   For example  given the string   I learned to play the Ukulele in Lebanon   and the search string le  I want to obtain the array    2  25  27  33    Both strings will be variables - i e   I can t hard-code their values   I figured that this was an easy task for regular expressions  but after struggling for a while to find one that would work  I ve had no luck   I found this example of how to accomplish this using  indexOf    but surely there has to be a more concise way to do it

User · Answer

Thanks for all the replies  I went through all of them and came up with a function that gives the first an last index of each occurrence of the  needle  substring   I am posting it here in case it will help someone   Please note  it is not the same as the original request for only the beginning of each occurrence  It suits my usecase better because you don t need to keep the needle length   function findRegexIndices text  needle  caseSensitive     var needleLen   needle length      reg   new RegExp needle  caseSensitive    gi     g        indices           result     while    result   reg exec text           indices push  result index  result index   needleLen          return indices

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Here is a simple code snippet   x000D   x000D  function getIndexOfSubStr str  searchToken  preIndex  output        var result   str match searchToken       if  result            output push result index  preIndex           str str substring result index searchToken length           getIndexOfSubStr str  searchToken  preIndex  output            return output     var str    my name is  xyz  and my school name is  xyz  and my area name is  xyz     var searchToken   my   var preIndex   0   console log getIndexOfSubStr str  searchToken  preIndex        x000D   x000D   x000D

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One liner using String protype matchAll  ES2020        sourceStr matchAll new RegExp searchStr   gi     map a   gt  a index    Using your values   const sourceStr    I learned to play the Ukulele in Lebanon    const searchStr    le   const indexes       sourceStr matchAll new RegExp searchStr   gi     map a   gt  a index   console log indexes       2  25  27  33    If you re worried about doing a spread and a map   in one line  I ran it with a for   of loop for a million iterations  using your strings   The one liner averages 1420ms while the for   of averages 1150ms on my machine  That s not an insignificant difference  but the one liner will work fine if you re only doing a handful of matches   See matchAll on caniuse

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Follow the answer of  jcubic  his solution caused a small confusion for my case For example var result   indexes  aaaa    aa   will return  0  1  2  instead of  0  2   So I updated a bit his solution as below to match my case function indexes text  subText  caseSensitive        var  source   text      var  find   subText      if  caseSensitive    true             source    source toLowerCase             find    find toLowerCase              var result           for  var i   0  i  lt   source length             if   source substring i  i    find length      find                result push i               i     find length      found a subText  skip to next position           else               i    1                      return result

User · Answer

function countInString searchFor searchIn     var results 0   var a searchIn indexOf searchFor    while a  -1      searchIn searchIn slice a 1 searchFor length      results       a searchIn indexOf searchFor       return results

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If you just want to find the position of all matches I d like to point you to a little hack   x000D   x000D  var haystack    I learned to play the Ukulele in Lebanon        needle    le       splitOnFound   haystack split needle  map function  culm                return this pos    culm length   needle length         pos  -needle length   slice 0  -1       pos           Object wich is used as this  console log splitOnFound   x000D   x000D   x000D   It might not be applikable if you have a RegExp with variable length but for some it might be helpful  This is case sensitive  For case insensitivity use String toLowerCase function before

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Here s my code  using search and slice methods   x000D   x000D      let s    I learned to play the Ukulele in Lebanon      let sub   0      let matchingIndex          let index   s search  le i      while  index  gt   0           matchingIndex push index sub          sub   sub     s length - s slice  index 1   length          s   s slice  index 1          index   s search  le i             console log matchingIndex  x000D   x000D   x000D

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Here is regex free version   function indexes source  find      if   source        return              if find is empty string return all indexes    if   find           or shorter arrow function         return source split     map    i    gt  i       return source split     map function    i    return i            var result         for  i   0  i  lt  source length    i           If you want to search case insensitive use         if  source substring i  i   find length  toLowerCase      find        if  source substring i  i   find length     find          result push i               return result     indexes  I learned to play the Ukulele in Lebanon     le     EDIT  and if you want to match strings like  aaaa  and  aa  to find  0  2  use this version   function indexes source  find      if   source        return           if   find          return source split     map function    i    return i            var result         var i   0    while i  lt  source length        if  source substring i  i   find length     find          result push i         i    find length        else         i                return result

User · Answer

You sure can do this     make a regular expression out of your needle var needle    le  var re   new RegExp needle  gi    var haystack    I learned to play the Ukulele    var results   new Array     this is the results you want while  re exec haystack      results push re lastIndex       Edit  learn to spell RegExp  Also  I realized this isn t exactly what you want  as lastIndex tells us the end of the needle not the beginning  but it s close - you could push re lastIndex-needle length into the results array     Edit  adding link   Tim Down s answer uses the results object from RegExp exec    and all my Javascript resources gloss over its use  apart from giving you the matched string    So when he uses result index  that s some sort of unnamed Match Object   In the MDC description of exec  they actually describe this object in decent detail

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the below code will do the job for you    function indexes source  find      var result         for i 0 i lt str length    i           If you want to search case insensitive use         if  source substring i  i   find length  toLowerCase      find        if  source substring i  i   find length     find          result push i               return result     indexes  hello  how are you    ar

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I would recommend Tim s answer  However  this comment by  blazs states  quot Suppose searchStr aaa and that str aaaaaa  Then instead of finding 4 occurences your code will find only 2 because you re making skips by searchStr length in the loop  quot   which is true by looking at Tim s code  specifically this line here  startIndex   index   searchStrLen  Tim s code would not be able to find an instance of the string that s being searched that is within the length of itself  So  I ve modified Tim s answer   x000D   x000D  function getIndicesOf searchStr  str  caseSensitive        var startIndex   0  index  indices           if   caseSensitive            str   str toLowerCase            searchStr   searchStr toLowerCase              while   index   str indexOf searchStr  startIndex    gt  -1            indices push index           startIndex   index   1            return indices    var searchStr   prompt  Enter a string     var str   prompt  What do you want to search for in the string     var indices   getIndicesOf str  searchStr    document getElementById  output   innerHTML   indices       x000D   lt div id  output  gt  lt  div gt  x000D   x000D   x000D   Changing it to   1 instead of   searchStrLen will allow the index 1 to be in the indices array if I have an str of aaaaaa and a searchStr of aaa  P S  If anyone would like comments in the code to explain how the code works  please say so  and I ll be happy to respond to the request

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Use String prototype match   Here is an example from the MDN docs itself   var str    ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz   var regexp     A-E  gi  var matches array   str match regexp    console log matches array        A    B    C    D    E    a    b    c    d    e

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var str    I learned to play the Ukulele in Lebanon   var regex    le gi  result  indices       while    result   regex exec str           indices push result index       UPDATE  I failed to spot in the original question that the search string needs to be a variable  I ve written another version to deal with this case that uses indexOf  so you re back to where you started  As pointed out by Wrikken in the comments  to do this for the general case with regular expressions you would need to escape special regex characters  at which point I think the regex solution becomes more of a headache than it s worth    x000D   x000D  function getIndicesOf searchStr  str  caseSensitive    x000D      var searchStrLen   searchStr length  x000D      if  searchStrLen    0    x000D          return     x000D        x000D      var startIndex   0  index  indices       x000D      if   caseSensitive    x000D          str   str toLowerCase    x000D          searchStr   searchStr toLowerCase    x000D        x000D      while   index   str indexOf searchStr  startIndex    gt  -1    x000D          indices push index   x000D          startIndex   index   searchStrLen  x000D        x000D      return indices  x000D    x000D   x000D  var indices   getIndicesOf  le    I learned to play the Ukulele in Lebanon     x000D   x000D  document getElementById  output   innerHTML   indices       x000D   lt div id  output  gt  lt  div gt  x000D   x000D   x000D

User · Answer

I am a bit late to the party  by almost 10 years  2 months   but one way for future coders is to do it using while loop and indexOf   let haystack    quot I learned to play the Ukulele in Lebanon  quot   let needle    quot le quot   let pos   0     Position Ref let result          Final output of all index s  let hayStackLower   haystack toLowerCase        Loop to check all occurrences  while  hayStackLower indexOf needle  pos     -1      result push hayStackLower indexOf needle   pos      pos   hayStackLower indexOf needle   pos    1     console log  quot Final  quot   result      Returns all indexes or empty array if not found

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Check this solution which will able to find same character string too  let me know if something missing or not right    x000D   x000D  function indexes source  find    x000D      if   source    x000D        return     x000D        x000D      if   find    x000D          return source split     map function    i    return i      x000D        x000D      source   source toLowerCase    x000D      find   find toLowerCase    x000D      var result       x000D      var i   0  x000D      while i  lt  source length    x000D        if  source substring i  i   find length     find  x000D          result push i     x000D        else x000D          i   x000D        x000D      return result  x000D      x000D    console log indexes  aaaaaaaa    aaaaaa    x000D    console log indexes  aeeaaaaadjfhfnaaaaadjddjaa    aaaa    x000D    console log indexes  wordgoodwordgoodgoodbestword    wordgood    x000D    console log indexes  I learned to play the Ukulele in Lebanon     le    x000D   x000D   x000D

[javascript] How to find indices of all occurrences of one string in another in JavaScript?

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