良好的python XML解析器用于处理名称空间繁重的文档。

Question

Python elementTree似乎无法使用名称空间。我有什么选择？BeautifulSoup在名称空间方面也很垃圾。我不想把他们剥掉。

关于特定python库如何获取命名空间元素的示例，它们的集合都是+1。

编辑：，你能用你选择的库来提供处理这个真实世界用例的代码吗？

如何获得字符串‘换行’、'2.6‘和一个列表'PYTHON’、'XML‘、’XML-命名空间‘

<?xml version="1.0" encoding="UTF-8"?>
<zs:searchRetrieveResponse
    xmlns="http://unilexicon.com/vocabularies/"
    xmlns:zs="http://www.loc.gov/zing/srw/"
    xmlns:dc="http://purl.org/dc/elements/1.1/"
    xmlns:lom="http://ltsc.ieee.org/xsd/LOM">
    <zs:records>
        <zs:record>
            <zs:recordData>
                <srw_dc:dc xmlns:srw_dc="info:srw/schema/1/dc-schema">
                    <name>Line Break</name>
                    <dc:title>Processing XML namespaces using Python</dc:title>
                    <dc:description>How to get contents string from an element,
                        how to get a collection in a list...</dc:description>
                    <lom:metaMetadata>
                        <lom:identifier>
                            <lom:catalog>Python</lom:catalog>
                            <lom:entry>2.6</lom:entry>
                        </lom:identifier>
                    </lom:metaMetadata>
                    <lom:classification>
                        <lom:taxonPath>
                            <lom:taxon>
                                <lom:id>PYTHON</lom:id>
                            </lom:taxon>
                        </lom:taxonPath>
                    </lom:classification>
                    <lom:classification>
                        <lom:taxonPath>
                            <lom:taxon>
                                <lom:id>XML</lom:id>
                            </lom:taxon>
                        </lom:taxonPath>
                    </lom:classification>
                    <lom:classification>
                        <lom:taxonPath>
                            <lom:taxon>
                                <lom:id>XML-NAMESPACES</lom:id>
                            </lom:taxon>
                        </lom:taxonPath>
                    </lom:classification>
                </srw_dc:dc>
            </zs:recordData>
        </zs:record>
        <!-- ... more records ... -->
    </zs:records>
</zs:searchRetrieveResponse>

Answer 1

lxml是命名空间感知的.

>>> from lxml import etree
>>> et = etree.XML("""<root xmlns="foo" xmlns:stuff="bar"><bar><stuff:baz /></bar></root>""")
>>> etree.tostring(et, encoding=str) # encoding=str only needed in Python 3, to avoid getting bytes
'<root xmlns="foo" xmlns:stuff="bar"><bar><stuff:baz/></bar></root>'
>>> et.xpath("f:bar", namespaces={"b":"bar", "f": "foo"})
[<Element {foo}bar at ...>]

编辑:在您的示例中：

from lxml import etree

# remove the b prefix in Python 2
# needed in python 3 because
# "Unicode strings with encoding declaration are not supported."
et = etree.XML(b"""...""")

ns = {
    'lom': 'http://ltsc.ieee.org/xsd/LOM',
    'zs': 'http://www.loc.gov/zing/srw/',
    'dc': 'http://purl.org/dc/elements/1.1/',
    'voc': 'http://www.schooletc.co.uk/vocabularies/',
    'srw_dc': 'info:srw/schema/1/dc-schema'
}

# according to docs, .xpath returns always lists when querying for elements
# .find returns one element, but only supports a subset of XPath
record = et.xpath("zs:records/zs:record", namespaces=ns)[0]
# in this example, we know there's only one record
# but else, you should apply the following to all elements the above returns

name = record.xpath("//voc:name", namespaces=ns)[0].text
print("name:", name)

lom_entry = record.xpath("zs:recordData/srw_dc:dc/"
                         "lom:metaMetadata/lom:identifier/"
                         "lom:entry",
                         namespaces=ns)[0].text

print('lom_entry:', lom_entry)

lom_ids = [id.text for id in
           record.xpath("zs:recordData/srw_dc:dc/"
                        "lom:classification/lom:taxonPath/"
                        "lom:taxon/lom:id",
                        namespaces=ns)]

print("lom_ids:", lom_ids)

输出：

name: Frank Malina
lom_entry: 2.6
lom_ids: ['PYTHON', 'XML', 'XML-NAMESPACES']

Answer 2

不如：

http://docs.python.org/library/pyexpat.html

Answer 3

libxml (http://xmlsoft.org/)是xml解析的最佳、更快的库。有用于python的实现。