使用包含命名空间的lxml解析xml

Question

我需要在lxml中的特定标签后获取一些信息.xml doc看起来像这样

<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app xmlns="http://java.sun.com/xml/ns/j2ee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/
ns/j2ee/web-app_2_4.xsd"
    version="2.4">
    <display-name>Community Bank</display-name>
    <description>WebGoat for Cigital</description>

        <context-param>
                <param-name>PropertiesPath</param-name>
<param-value>/WEB-INF/properties.txt</param-value>
                <description>This is the path to the properties file from the servlet root</description>
        </context-param>

    <servlet>
        <servlet-name>Index</servlet-name>
<servlet-class>com.cigital.boi.servlet.index</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>Index</servlet-name>
        <url-pattern>/index</url-pattern>
    </servlet-mapping>

    <servlet-mapping>
        <servlet-name>Index</servlet-name>
        <url-pattern>/index.html</url-pattern>
    </servlet-mapping>

我想阅读com.cigital.boi.servlet.index.

我已经使用这段代码来读取servlet下的所有内容

    context = etree.parse(handle)
    list = parser.xpath('//servlet')
    print list

list只包含更多信息:迭代上下文字段我找到了这些行.

<Element {http://java.sun.com/xml/ns/j2ee}servlet-name at 2ad19e6eca48>
<Element {http://java.sun.com/xml/ns/j2ee}servlet-class at 2ad19e6ecaf8>

我在想,因为我在搜索时没有包含名称空间,输出是空列表.请建议在servlet-class标签中阅读"com.cigital.boi.servlet.index"

Answer 1

试试以下:

from lxml import etree
context = etree.parse(handle)
print next(x.text for x in context.xpath('.//*[local-name()="servlet-class"]'))

替代方案:

from lxml import etree
context = etree.parse(handle)
nsmap = context.getroot().nsmap.copy()
nsmap['xmlns'] = nsmap.pop(None)
print next(x.text for x in context.xpath('.//xmlns:servlet-class', namespaces=nsmap))