使用Python解析Alexa XML [英] Parsing Alexa XML with Python

问题描述

我有一个非常类似的问题：
python alexa结果我正在想知道如何解析第二个 DataUrl 的lxml.etree 。

I have a pretty similiar question to: python alexa result parsing with lxml.etree.

这意味着我想获取 DataUrl 变量，该变量位于 TrafficData 下，而不是下的变量ContentData 。 （获取 people.com 而不是 google.com ）

I'm wondering how to parse the second DataUrl. That means I want to get the DataUrl variable which is under TrafficData and not the one under ContentData. (get people.com and not google.com)

我还在使用lxml，它的数据与他描述的完全相同。

I'm also using lxml with the exact same data as he described.

这是代码：

<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>ccf3f263-ab76-ab63-db99-244666044e85</aws:RequestId>
    </aws:OperationRequest>
    <aws:UrlInfoResult>
      <aws:Alexa>
        <aws:ContentData>
          <aws:DataUrl type="canonical">google.com/</aws:DataUrl>
          <aws:SiteData>
        <aws:Title>Google</aws:Title>
            <aws:Description>Enables users to search the world's information, including webpages, images, and videos. Offers unique features and search technology.</aws:Description>
            <aws:OnlineSince>15-Sep-1997</aws:OnlineSince>
          </aws:SiteData>
          <aws:LinksInCount>3453627</aws:LinksInCount>
        </aws:ContentData>
        <aws:TrafficData>
          <aws:DataUrl type="canonical">people.com/</aws:DataUrl>
          <aws:Rank>1</aws:Rank>
        </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>

推荐答案

我需要做：

namespaces = {"aws": "http://awis.amazonaws.com/doc/2005-07-11"}
texts = doc.xpath("//aws:TrafficData/aws:DataUrl/text()", namespaces=namespaces)
print texts[0]

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