Yes in that case you would, and before deleting the level 2 append level3 to
level 2's parent.
Kind of (not at all tested) like this..
for each (var item:XML in myxml..level2) {
item.parent.append(item..level3);
delete item;
}
you may actually need 2 loops one for the delete. If the above technique
doesnt work try the copy()
item.parent.append(item..level3.copy());
sorry no time to actually test anything here.
jason
-----Message d'origine-----
De : [email protected] [mailto:[EMAIL PROTECTED] la
part de phipzkillah
Envoyé : mercredi 10 janvier 2007 21:58
À : [email protected]
Objet : [flexcoders] Re: XML manipulation
That makes sense and I guess I should have rephrased my question. I
have multiple levels of level1 and I need to preserve the hierarchy.
I would assume you would need a for loop to do this. Any ideas?
<level1>
<level2>
<level3/>
<level3/>
<level3/>
</level2>
</level1>
<level1>
<level2>
<level3/>
<level3/>
</level2>
</level1>
Transformed to...
<level1>
<level3/>
<level3/>
<level3/>
</level1>
<level1>
<level3/>
<level3/>
</level1>
--- In [email protected], "Jason Hawryluk" <[EMAIL PROTECTED]> wrote:
>
> Yes; basically
>
> Grab level3 into an xmllist var
>
> delete level 2 append level 3 to level 1.
>
> var tempxml:XMLList = myxml..level3;
> delete myxml.level2;
> myxml.append(tempxml);
>
> the above code is off the top of my head. May and may not work :)
>
> jason
>
>
> -----Message d'origine-----
> De : [email protected] [mailto:[EMAIL PROTECTED] la
> part de phipzkillah
> Envoyé : mercredi 10 janvier 2007 21:28
> À : [email protected]
> Objet : [flexcoders] Re: XML manipulation
>
>
> Ben,
>
> Thanks for your help. It works great!
>
> I have another question that I'm not sure if you will be able to
> answer or not. If I have 3 levels of hierarchy in my XML and I want
> to remove level 2 (but not its children - level 3) how would you go
> about doing that? Eliminate level 2 while promoting level 3.
>
> <level1>
> <level2>
>
> <level3/>
> <level3/>
> <level3/>
>
> </level2>
> </level1>
>
> Transformed to...
>
> <level1>
>
> <level3/>
> <level3/>
> <level3/>
>
> </level1>
>
> --- In [email protected], "ben.clinkinbeard"
> <ben.clinkinbeard@> wrote:
> >
> > Wow, I had no idea deleting a set of nodes would be this hard.
> > Apparently this is how its done:
> >
> > for each(var node:XML in myXML..remove)
> > {
> > delete(myXML..remove[node.childIndex()]);
> > }
> >
> > HTH,
> > Ben
> >
> >
> > --- In [email protected], "phipzkillah" <pkrasko@> wrote:
> > >
> > > How can I remove a selection of children from an xml tree.
> > >
> > > var myXML:XML =
> > > <root>
> > > <cluster>
> > > <remove>
> > > <garbage/>
> > > <garbage/>
> > > <garbage>
> > > <garbage1/>
> > > <garbage2/>
> > > <garbage3/>
> > > </garbage>
> > > </remove>
> > > <remove>
> > > <garbage>
> > > <garbage1/>
> > > <garbage2/>
> > > </garbage>
> > > <garbage/>
> > > </remove>
> > >
> > > <keep>
> > > <info/>
> > > <info/>
> > > <info>
> > > <goodStuff/>
> > > <goodStuff/>
> > > <goodStuff>
> > > <good1/>
> > > <good2/>
> > > </goodStuff>
> > > </info>
> > > </keep>
> > > </cluster>
> > > </root>
> > >
> > > ......................
> > >
> > > Now I want to keep all the good stuff.
> > >
> > > I want the final xml tree to look like this:
> > > <root>
> > > <cluster>
> > > <keep>
> > > <info/>
> > > <info/>
> > > <info>
> > > <goodStuff/>
> > > <goodStuff/>
> > > <goodStuff>
> > > <good1/>
> > > <good2/>
> > > </goodStuff>
> > > </info>
> > > </keep>
> > > </cluster>
> > > </root>
> > >
> > > I tried doing this with no luck:
> > >
> > > myXML = delete myXML..remove;
> > >
> > > Any ideas why that doesn't work?
> > >
> >
>
