Python XML parse this doc -


i couldn't find this, , can't use beautifulsoap. i've xml url doc. wanna parse items called contenido, has atribs , don't know how them. i've tried xml.sax don't know how use attrib

<contenido> <tipo>evento</tipo> <atributos idioma="es"> <atributo nombre="id-evento">8006941</atributo> <atributo nombre="titulo">582 mapas, compañía teatral “la cola del pavo”</atributo> <atributo nombre="tipo"> /contenido/actividades/recitalespresentacionesactosliterarios </atributo> </atributos> </contenido>

this example using xml.sax

import xml.sax  class myhandler( xml.sax.contenthandler ):      def __init__(self):         self.is_atributo = false      def startelement(self, tag, attributes):        if tag == 'atributo':            self.is_atributo = true      def characters(self, content):         if self.is_atributo:             print(content)  if __name__ == '__main__':     parser = xml.sax.make_parser()     parser.setfeature(xml.sax.handler.feature_namespaces, 0)      handler = myhandler()     parser.setcontenthandler(handler)     parser.parse('your.xml')

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