适合处理命名空间复杂文档的Python XML解析器
Python的elementTree在处理命名空间时似乎不好用。我还有什么其他选择呢?BeautifulSoup在处理命名空间方面也不太好。我不想把命名空间去掉。
如果有关于某个Python库如何获取带命名空间的元素及其集合的例子,那就太好了。
编辑:能否提供一些代码,展示如何用你选择的库来处理这个实际的使用案例?
你会如何获取字符串'Line Break'、'2.6'和一个列表['PYTHON', 'XML', 'XML-NAMESPACES']?
<?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>
3 个回答
0
libxml(http://xmlsoft.org/)是一个很棒的库,用来解析XML文件。它的速度很快,是处理XML的最佳选择。这个库也有适用于Python的版本。
13
lxml 是一个支持命名空间的库。
>>> 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 ...>]
补充说明:关于你的例子:
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)
输出结果:
name: Frank Malina
lom_entry: 2.6
lom_ids: ['PYTHON', 'XML', 'XML-NAMESPACES']