jQuery find next/prev elements of a certain class but not necessarily siblings

Try this. It will mark your Element, create a set of Elements matching your selector and collect all Elements from the set following your element.

$.fn.findNext = function ( selector ) {
    var set = $( [] ), found = false;
    $( this ).attr( "findNext" , "true" );
    $( selector ).each( function( i , element ) {
        element = $( element );
        if ( found == true ) set = set.add( element )
        if ( element.attr("findNext") == "true" ) found = true;
    })
    $( this ).removeAttr( "findNext" )
    return set
}

EDIT

much simpler solution using jquerys index method. the element you call the method from needs to be selectable by the same selector though

$.fn.findNext = function( selector ){
    var set = $( selector );
    return set.eq( set.index( this, ) + 1 )
}

to free the function from this handicap, we could youse the browsers own compareDocumentposition

$.fn.findNext = function ( selector ) {
  // if the stack is empty, return the first found element
  if ( this.length < 1 ) return $( selector ).first();
  var found,
      that = this.get(0);
  $( selector )
    .each( function () {
       var pos = that.compareDocumentPosition( this );
       if ( pos === 4 || pos === 12 || pos === 20 ){
       // pos === 2 || 10 || 18 for previous elements 
         found = this; 
         return false;
       }    
    })
  // using pushStack, one can now go back to the previous elements like this
  // $("#someid").findNext("div").remove().end().attr("id")
  // will now return "someid" 
  return this.pushStack( [ found ] );
},  

EDIT 2 this is far easier using jQuery's $.grep. here's the new code

   $.fn.findNextAll = function( selector ){
      var that = this[ 0 ],
          selection = $( selector ).get();
      return this.pushStack(
         // if there are no elements in the original selection return everything
         !that && selection ||
         $.grep( selection, function( n ){
            return [4,12,20].indexOf( that.compareDocumentPosition( n ) ) > -1
         // if you are looking for previous elements it should be [2,10,18]
         })
      );
   }
   $.fn.findNext = function( selector ){
      return this.pushStack( this.findNextAll( selector ).first() );
   }

when compressing variable names this becomes a mere two liner.

Edit 3 using bitwise operations, this function may be even faster?

$.fn.findNextAll = function( selector ){
  var that = this[ 0 ],
    selection = $( selector ).get();
  return this.pushStack(
    !that && selection || $.grep( selection, function(n){
       return that.compareDocumentPosition(n) & (1<<2);
       // if you are looking for previous elements it should be & (1<<1);
    })
  );
}
$.fn.findNext = function( selector ){
  return this.pushStack( this.findNextAll( selector ).first() );
}

I was working on this problem myself today, here's what I came up with:

/**
 * Find the next element matching a certain selector. Differs from next() in
 *  that it searches outside the current element's parent.
 *  
 * @param selector The selector to search for
 * @param steps (optional) The number of steps to search, the default is 1
 * @param scope (optional) The scope to search in, the default is document wide 
 */
$.fn.findNext = function(selector, steps, scope)
{
    // Steps given? Then parse to int 
    if (steps)
    {
        steps = Math.floor(steps);
    }
    else if (steps === 0)
    {
        // Stupid case :)
        return this;
    }
    else
    {
        // Else, try the easy way
        var next = this.next(selector);
        if (next.length)
            return next;
        // Easy way failed, try the hard way :)
        steps = 1;
    }

    // Set scope to document or user-defined
    scope = (scope) ? $(scope) : $(document);

    // Find kids that match selector: used as exclusion filter
    var kids = this.find(selector);

    // Find in parent(s)
    hay = $(this);
    while(hay[0] != scope[0])
    {
        // Move up one level
        hay = hay.parent();     
        // Select all kids of parent
        //  - excluding kids of current element (next != inside),
        //  - add current element (will be added in document order)
        var rs = hay.find(selector).not(kids).add($(this));
        // Move the desired number of steps
        var id = rs.index(this) + steps;
        // Result found? then return
        if (id > -1 && id < rs.length)
            return $(rs[id]);
    }
    // Return empty result
    return $([]);
}

So in your example

<div><span id="click">hello</span></div>
<div><p class="find">world></p></div>

you could now find and manipulate the 'p' element using

$('#click').findNext('.find').html('testing 123');

I doubt it will perform well on large structures, but here it is :)

My solution would involve adjusting your markup a bit to make the jQuery much easier. If this is not possible or not an appealing answer, please ignore!

I would wrap a 'parent' wrapper around what you want to do...

<div class="find-wrapper">
    <div><span id="click">hello</span></div>
    <div><p class="find">world></p></div>
</div>

Now, to find the find:

$(function() {
    $('#click').click(function() {
        var $target = $(this).closest('.find-wrapper').find('.find');
        // do something with $target...
    });
});

This gives you the flexibility to have whatever kind of markup and hierarchy you'd like inside the wrapper I suggested, and still reliably find your target.

Good luck!