好的python XML解析器能够处理命名空间中的重文档

Question

Python elementTree似乎对命名空间不可用。我有什么选择？ BeautifulSoup也是名字空间的垃圾。 我不想将它们去掉。

一个特定的python库如何获取名称空间元素及其集合的例子都是+1。

编辑：你能提供代码来处理这个真实世界的用例使用你选择的库吗？

你会如何去得到字符串 '换行'， '2.6' 和一个列表[ 'PYTHON'， 'XML'， 'XML的命名空间']

<?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是名称空间感知的。

>>> 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：

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

Answer 2

如何：

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

Answer 3

的libxml（http://xmlsoft.org/） 最佳，XML解析更快库。 python有实现。