jQuery find next and prev element

I try to find a next or prev element of current element. But next() and prev() function c开发者_JAVA技巧an only work in a scope, it can not reach outside. For an example this is what I want to achieve:

<ul id="ul1">
  <li>1</li>
  <li>2</li>
  <li>3</li>
  <li>
    <ul id="ul2">
      <li>4</li>
      <li>
        <ul id="ul3">
          <li>5</li>
          <li>6</li>
        </ul>
      </li>
      <li>7</li>
      <li>8</li>
    </ul>
  </li>
  <li>9</li>
</ul>

If current element is ul1, next element is <li>1</li>, prev element is null.

If current element is <li>1</li>, next element is <li>2</li>, prev element is ul1

If current element is <li>8</li>, next element is <li>9</li>, prev element is <li>7</li>

First of all, this was a really neat problem. Thanks for posting!

I accomplished this with a chained jQuery statement. Only thing is, if you run previous() on the top ul you'll get body. Which I think technically makes sense. However, if this isn't the desired behavior, see my update below.

Usage: next("#ul3") returns <li>5</li>

Next:

function next(selector) {
    var $element = $(selector);

    return $element
        .children(":eq(0)")
        .add($element.next())
        .add($element.parents().filter(function() {
            return $(this).next().length > 0;
        }).next()).first();
}

Previous:

function previous(selector) {
    var $element = $(selector);

    return $element
        .prev().find("*:last")   
        .add($element.parent())    
        .add($element.prev())
        .last();     
}

Update If you want to limit the upper most node previous can be, you could do:

function previous(selector, root) {
    var $element = $(selector);

    return $element
        .prev().find("*:last")   
        .add($element.parent())     
        .add($element.prev())
        .last().not($(root).parent());      
}

http://jsfiddle.net/andrewwhitaker/n89dz/

Another Update: Here's a jQuery plugin for convenience:

(function($) {
    $.fn.domNext = function() {
        return this
            .children(":eq(0)")
            .add(this.next())
            .add(this.parents().filter(function() {
                return $(this).next().length > 0;
            }).next()).first();        
    };

    $.fn.domPrevious = function() {
        return this
            .prev().find("*:last")   
            .add(this.parent())     
            .add(this.prev())
            .last();         
    };
})(jQuery);

Expanded example here: http://jsfiddle.net/andrewwhitaker/KzyAY/

What an interesting problem! Sounds like you're flattening the recursive structure of HTML so that it's loop-able - I can see it coming in handy.

You can solve the problem by breaking .next() down into the following cases:

1. Element has children --> next() is the first child.
2. Element has no children, and is not the last child --> next() is the same as jQuery's .next()
3. Element has no children, is last child:
   1. Find closest parent that is not the last child --> next() is that parent's next sibling
   2. If no such parent exists --> next() is null

Code would be (calling it flatNext to preserve original jQuery .next behaviour):

$.fn.flatNext = function () {
   if (this.children().length) return this.children().eq(0);
   if (!this.is(":last-child")) return this.next();
   if (this.closest(":not(:last-child)").length)
       return this.closest(":not(:last-child)").next();
   return $([]); // Return empty jQuery object rather than null
};

And $.fn.flatPrev should follow..

Depending on your data set, you could just flatten the whole thing (presuming $.find is stable, which I believe it is), like this:

var eles = [];
$("#ul1").find("li").each(function (e) {
    eles.push(e);
});

The next element after ele[3] is ele[4] regardless of the level of indentation.

Example http://jsfiddle.net/xRpmL/