Cool, it will be good to see your continued work. We may be among the last
holdouts in this mailing list - but I figure since the thread started here
we can carry on.
BTW you may notice one bit of annoying repeated boilerplate in that last
version. See how the callback functions for all the .each() loops begin with
something like:
var $foobar = $(foobar);
Why don't we extend jQuery with a better version of .each() that does that
for us?
Let's call it .$each() as a reminder that it gives the callback a DOM
element that's already been wrapped with a $(). It would look like this:
jQuery.fn.$each = function( callback ) {
for( var element, i = -1; element = this[++i]; )
callback( jQuery(element), i );
};
Note that in addition to calling jQuery(element), it also reverses the
callback arguments - in most cases you only need the element inside the
callback - it's much less often that you need the index. (In your code, only
one of the four loops uses the index.)
So, using that function, the code can be written as:
function parseXml( xml ) {
var html = [];
html.push( '<div>' );
$(xml).find('sites').$each( function( $site ) {
$site.find('>element').$each( function( $parent, i ) {
html.push( '<br/>', $parent.attr('label'), i + 1, '<br/>' );
$parent.find('>element').$each( function( $child ) {
html.push( $child.attr('label'), '<br/>' );
$child.find('>element').$each( function( $inner ) {
html.push( $inner.attr('label'), ': ', $inner.text(),
'<br/>' );
});
});
});
});
html.push( '</div>' );
$('#container').append( html.join('') );
}
That's getting pretty clean!
Note that the code still uses the naming convention of $foobar for a jQuery
object - it helps make it clear which variables are jQuery objects.
-Mike
On Tue, Feb 2, 2010 at 10:27 PM, augur <312...@gmail.com> wrote:
> OK, so this is much better than a lesson in parsing XML. Showing the
> errors that I had made in my process, you have also shown sound jQuery
> which I was having a hard time picking up from the documentation. I
> tend to learn best by doing and getting feedback. Mike, thank you. I
> am going to keep this chain going with my mods as I make them.
>
