Question

我有以下XML格式，我想使用python的xml.etree.ElementTree模块提取名称，区域和状态的值。

但是，到目前为止，我尝试获取此信息的尝试都没有成功。

<feed>
    <entry>
        <id>uuid:asdfadsfasdf123123</id>
        <title type="text"></title>
        <content type="application/xml">
            <NamespaceDescription xmlns="http://schemas.microsoft.com/netservices/2010/10/servicebus/connect" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
                <Name>instancename</Name>
                <Region>US</Region>
                <Status>Active</Status>
            </NamespaceDescription>
        </content>
    </entry>
    <entry>
        <id>uuid:asdfadsfasdf234234</id>
        <title type="text"></title>
        <content type="application/xml">
            <NamespaceDescription xmlns="http://schemas.microsoft.com/netservices/2010/10/servicebus/connect" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
                <Name>instancename2</Name>
                <Region>US2</Region>
                <Status>Active</Status>
            </NamespaceDescription>
        </content>
    </entry>
</feed>

我的代码尝试：

NAMESPACE = '{http://www.w3.org/2005/Atom}'
root = et.fromstring(XML_STRING)
entry_root = root.findall('{0}entry'.format(NAMESPACE))
for child in entry_root:
    content_node = child.find('{0}content'.format(NAMESPACE))
    for content in content_node:
        for desc in content.iter():
            print desc.tag
            name = desc.find('{0}Name'.format(NAMESPACE))
            print name

desc.tag给了我想要访问的节点，但名称返回None。任何想法我的代码有什么问题？

desc.tag的输出：

{http://schemas.microsoft.com/netservices/2010/10/servicebus/connect}Name
{http://schemas.microsoft.com/netservices/2010/10/servicebus/connect}Region
{http://schemas.microsoft.com/netservices/2010/10/servicebus/connect}Status

Answer 1

我不知道为什么我之前没有看到这个。但是，我能够获得价值。

root = et.fromstring(XML_STRING)
entry_root = root.findall('{0}entry'.format(NAMESPACE))
for child in entry_root:
    content_node = child.find('{0}content'.format(NAMESPACE))
    for descr in content_node:
        name_node = descr.find('{0}Name'.format(NAMESPACE))
        print name_node.text

Answer 2

您可以使用lxml.etree和默认命名空间映射来解析XML，如下所示：

content = '''
<feed>
    <entry>
        <id>uuid:asdfadsfasdf123123</id>
        <title type="text"></title>
        <content type="application/xml">
            <NamespaceDescription xmlns="http://schemas.microsoft.com/netservices/2010/10/servicebus/connect" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
                <Name>instancename</Name>
                <Region>US</Region>
                <Status>Active</Status>
            </NamespaceDescription>
        </content>
    </entry>
    <entry>
        <id>uuid:asdfadsfasdf234234</id>
        <title type="text"></title>
        <content type="application/xml">
            <NamespaceDescription xmlns="http://schemas.microsoft.com/netservices/2010/10/servicebus/connect" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
                <Name>instancename2</Name>
                <Region>US2</Region>
                <Status>Active</Status>
            </NamespaceDescription>
        </content>
    </entry>
</feed>'''

from lxml import etree

tree = etree.XML(content)
ns = {'default': 'http://schemas.microsoft.com/netservices/2010/10/servicebus/connect'}

names = tree.xpath('//default:Name/text()', namespaces=ns)
regions = tree.xpath('//default:Region/text()', namespaces=ns)
statuses = tree.xpath('//default:Status/text()', namespaces=ns)

print(names)
print(regions)
print(statuses)

<强>输出

['instancename', 'instancename2']
['US', 'US2']
['Active', 'Active']

可以调整此XPath /命名空间功能，以便以您需要的任何格式输出数据。

使用ElementTree解析嵌套XML

2 个答案: