[英]Good python XML parser to work with namespace heavy documents
Python elementTree似乎無法使用命名空間。 我有什么選擇? BeautifulSoup也很容易使用名稱空間。 我不想把它們剝掉。
特定python庫如何獲取命名空間元素及其集合的示例均為+1。
編輯:您能否使用您選擇的庫提供代碼來處理這個真實世界的用例?
你將如何獲得字符串'Line Break','2.6'和列表['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>
lxml是名稱空間感知的。
>>> 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 ...>]
編輯: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)
輸出:
name: Frank Malina
lom_entry: 2.6
lom_ids: ['PYTHON', 'XML', 'XML-NAMESPACES']
libxml(http://xmlsoft.org/)用於xml解析的最佳,更快的lib。 有python的實現。
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