Monday, October 31, 2011

Get All indexOf From An Array - Array.allIndexOf()

This array function is an extension of Array.indexOf(), if it exists, and will return all indexes of the search element. It's named allIndexOf(). This code extends javascript and does not require jQuery. It is designed to work in older versions of IE as well, albeit a tiny bit slower.

Originally if you wanted the second instance of a search element, you'd have to call "indexOf" twice. The second time with the starting index of the first result (plus one), or once with a guessed starting index. This should simplify the process for getting any or all of the indexes.

/*
Array.allIndexOf(searchElement)
  Array [Array] - the array to search within for the searchElement
  searchElement [String] - the desired element with which to find starting indexes
*/
(function(){
  Array.prototype.allIndexOf = function(searchElement) {
    if (this === null) { return [-1]; }
    var len = this.length,
    hasIndexOf = Array.prototype.indexOf, // you know, because of IE
    i = (hasIndexOf) ? this.indexOf(searchElement) : 0,
    n,
    indx = 0,
    result = [];
    if (len === 0 || i === -1) { return [-1]; }
    if (hasIndexOf) {
      // Array.indexOf does exist
      for (n = 0; n <= len; n++) {
        i = this.indexOf(searchElement, indx);
        if (i !== -1) {
          indx = i + 1;
          result.push(i);
        } else {
          return result;
        }
      }
      return result;
    } else {
    // Array.indexOf doesn't exist
      for (n = 0; n <= len; n++) {
        if (this[n] === searchElement) {
          result.push(n);
        }
      }
      return (result.length > 0) ? result : [-1];
    }
  };
})();
Use it as follows:
var s = ["red","green","blue","red","yellow","blue","green","purple","red"];
s.allIndexOf("r"); // result [ -1 ]
s.allIndexOf("red"); // result [ 0,3,8 ]
s.allIndexOf("blue"); // result [ 2,5 ]
Try out your own strings in the demo below or full screen.