Apologies for the subject line, I'm struggling to explain the problem
in a concise manner.

I have the following document:

        <body>
                <h1 class="page-editable" widget="line">Page Title Goes 
Here</h1>

                <div widget="repeat-area">
                        <div widget="repeat">
                                <h2 widget="line">Section Heading</h2>

                <div widget="repeat-area">

                    <div widget="repeat">
                                                <h3 widget="line">Section 
Title</h3>
                                                <p 
widget="textarea">Content&hellip;</p>
                                                <img src="test.png" 
widget="image" />
                    </div>

                    <div widget="repeat">
                                                <h3 widget="line">Section 
Title</h3>
                                                <p 
widget="textarea">Content&hellip;</p>
                                                <img src="test.png" 
widget="image" />
                    </div>
                </div>

                        </div>
                </div>


        </body>

I am trying to recursively access each element with the attribute
'widget'. The first level of recursion is easy:

$('[widget]:not([widget] [widget])')

Which matches all widgets that are not descendants of other widgets,
ie the top level.

The problem is with the next recursive steps, I can't figure out how
to find just the first level of widgets that are a descendents of the
current node.

$('[widget]:not([widget] [widget])').each(function(i){

   // For example, I need a query I'd put here to access the next
level of elements.

});

There is no guarantee that the widgets would be direct children of
each other, or that they'd be on the same tree level as their
'sibling' widgets.

Basically I need a way of finding all widgets that are descendents of
the current working node, but with no other widgets in-between.

If anything needs clarifying let me know.

Regards,
Andrew Ingram

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