url fetching from xml depending upon selection

Tue May 18 04:08:48 EDT 2010

On May 18, 12:04 pm, Stefan Behnel <stefan... at behnel.de> wrote:
> shanti bhushan, 18.05.2010 07:18:
>
>
>
>
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> > I have a sample.XML file
> > the code is like this
>
> > <?xml version="1.0" encoding="UTF-8"?>
> > <opml version="1.0">
> > <head>
> >    <title>My Podcasts</title>
> >    <dateCreated>Sun, 07 Mar 2010 15:53:26
>
> > GMT</dateCreated>
> >    <dateModified>Sun, 07 Mar 2010 15:53:26
>
> > GMT</dateModified>
> > </head>
> > <body>
> >    <TestCase name="Sprint_001">
> >      <Input url="http://first.co.jp" />
> >      <Input url="http://www.google.com" />
> >      <Input url="http://www.epaper.times.india.com" />
> >    </TestCase>
> >    <TestCase name="Sprint_002">
> >      <Input url="http://second.co.jp" />
> >      <Input url="http://www.google.com" />
> >      <Input url="http://www.epaper.times.india.com" />
> >    </TestCase>
> >    <TestCase name="Sprint_003">
> >      <Input url="http://third.co.jp" />
> >      <Input url="http://www.google.com" />
> >      <Input url="http://www.epaper.times.india.com" />
> >    </TestCase>
> > </body>
> > </opml>
>
> > This my python code
> > from xml.etree import ElementTree
>
> > with open('our.xml', 'rt') as f:
> >      tree = ElementTree.parse(f)
>
> > for node, value in tree.findall('.//TestCase/Input'):
>
> >          url=node.attrib.get('url')
> >          print url
>
> > i want to print the url depending on name="sprint_001".
>
>      for test_case in tree.findall('.//TestCase'):
>          if test_case.get('name') == 'sprint_002':
>             for input_el in test_case:
>                 print input_el.get('url')
>
> Stefan- Hide quoted text -
>
> - Show quoted text -

Hi Stevan
i tried your logic i am not getting any out put for this