用lxml.etree解析python结果

Question

我正在使用aws的alexa，但是我发现很难解析结果以得到我想要的结果。

alexa返回一个对象树<type 'lxml.etree._ElementTree'>

我用这段代码打印树

from lxml import etree
root = tree.getroot()
print etree.tostring(root)

下面是xml

<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">google.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>

我使用root.find('LinksInCount').text获取元素的值，但它不起作用。

我想知道如何获得文本3453627 of aws:LinksInCount

Answer 1

你遇到了两个挑战：

• 使用命名空间的XML
• 共享相同名称空间前缀的两个命名空间

两个不同名称空间的具有可重用前缀的XML文档

您可以看到"aws:"前缀，但它用于两个不同的名称空间：

xmlns:aws="http://alexa.amazonaws.com/doc/2005-10-05/"
xmlns:aws="http://awis.amazonaws.com/doc/2005-07-11"

在XML中使用相同的命名空间前缀是完全合法的。规则是，后者是有效的。

xmlstr = """
<?xml version="1.0"?>
<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">google.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>
"""

下一个挑战是，如何搜索命名空间元素。

我更喜欢使用xpath，对于它，您可以在xpath表达式中使用您喜欢的任何名称空间，但是您必须告诉xpath调用这些前缀的含义。这是由namespaces字典完成的：

from lxml import etree
doc = etree.fromstring(xmlstr.strip())

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