用lxml嵌套命名空间解析XML属性

Question

我想解析RankByCountry中的国家代码属性。我该怎么做呢？

意思是-打印一个列表'GB'，'US'，'O‘

<aws:UrlInfoResponse xmlns:aws="http://alexa.amazonaws.com/doc/2005-10-05/"><aws:Response xmlns:aws="http://awis.amazonaws.com/doc/2005-07-11"><aws:OperationRequest><aws:RequestId>122bfdc6-ae8e-d2a2-580e-3841ab33b966</aws:RequestId></aws:OperationRequest><aws:UrlInfoResult><aws:Alexa>
 <aws:TrafficData>
  <aws:DataUrl type="canonical">androidjones.com/</aws:DataUrl>
  <aws:RankByCountry>
    <aws:Country Code="GB">
      <aws:Rank>80725</aws:Rank>
      <aws:Contribution>
        <aws:PageViews>30.6%</aws:PageViews>
        <aws:Users>41.3%</aws:Users>
      </aws:Contribution>
    </aws:Country>
    <aws:Country Code="US">
      <aws:Rank>354356</aws:Rank>
      <aws:Contribution>
        <aws:PageViews>39.1%</aws:PageViews>
        <aws:Users>28.9%</aws:Users>
      </aws:Contribution>
    </aws:Country>
    <aws:Country Code="O">
      <aws:Rank/>
      <aws:Contribution>
        <aws:PageViews>30.2%</aws:PageViews>
        <aws:Users>29.8%</aws:Users>
      </aws:Contribution>
    </aws:Country>
  </aws:RankByCountry>
 </aws:TrafficData>
</aws:Alexa></aws:UrlInfoResult><aws:ResponseStatus xmlns:aws="http://alexa.amazonaws.com/doc/2005-10-05/"><aws:StatusCode>Success</aws:StatusCode></aws:ResponseStatus></aws:Response></aws:UrlInfoResponse>

已经试过了--

namespaces = {"aws": "http://awis.amazonaws.com/doc/2005-07-11"}
RankByCountry = tree.xpath("//aws:Country/Code", namespaces=namespaces)

但没有运气。

还包括：

for country in tree.xpath('//Country'):
   for attrib in country.attrib:
      print '@' + attrib + '=' + country.attrib[attrib]

Answer 1

文档看起来很奇怪，因为它两次使用aws名称空间前缀。您需要使用更具体的命名空间，因为这会用前缀aws覆盖全局命名空间。其实你这样做是对的。

问题在于xpath表达式本身，它应该如下所示：

for country in tree.xpath('//aws:RankByCountry/aws:Country/@Code', namespaces=namespaces):
    print(country) 

注意，<aws:RankByCountry>没有Code属性，但是<aws:Country>有。