Re: [xml] how to add namespace-less node in python

[Hannu Krosing <hannu tm ee>]

John Fleck kirjutas L, 20.12.2003 kell 19:17:
> Try this:
>
> #!/usr/bin/python -u
> import libxml2
>
> doc = libxml2.newDoc("1.0")
> root = doc.newChild(None, "doc", None)
> docns = root.newNs("http://www.krosing.net/testns","a";)
> para = root.newChild(docns, "para", None)
> title = doc.newDocNode(None, "title", None)
> para.addChild(title)
> print doc.serialize(None,1)

Thanks!

Any idea why it is done this way ?

> On Sat, 2003-12-20 at 09:50, Hannu Krosing wrote:
> > Hallo
> >
> > I need to add a node that has no namespace prefix while building xml doc
> > in python (for my home-made SOAP server/client lib).
> >
> > I know that such a tree is possible, as it is correctly generated by
> > libxml2.parseFile :
> >
> > > > > d = libxml2.parseFile('nstest.xml')
> > > > > print d.serialize(None,1)
> > <?xml version="1.0"?>
> > <doc xmlns:a="http://www.krosing.net/testns";>
> >   <a:para>
> >     <title/>
> >   </a:para>
> > </doc>
> >
> >
> > But I am unable to make similar tree programmatically:
> >  
> > > > > doc = libxml2.newDoc("1.0")
> > > > > root = doc.newChild(None, "doc", None)
> > > > > docns = root.newNs("http://www.krosing.net/testns","a";)
> > > > > para = root.newChild(docns, "para", None)
> > > > > title = para.newChild(None, "title", None)
> > > > > print doc.serialize(None,1)
> > <?xml version="1.0"?>
> > <doc xmlns:a="http://www.krosing.net/testns";>
> >   <a:para>
> >     <a:title/>
> >   </a:para>
> > </doc>
> >
> > I want to get <title/> and not <a:title/> .
> >
> > What am I doing wrong ?
> >
> > ----------------------
> > Hannu
> >
> >
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>
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