Python and libxml2: how to iterate in xml nodes with XPATH

问题

I have a problem with retrieving information from a XML tree.

My XML has this shape:

<?xml version="1.0"?>
<records xmlns="http://www.mysyte.com/foo">
  <record>
    <id>first</id>
    <name>john</name>
    <papers>
      <paper>john_1</paper>
      <paper>john_2</paper>
    </papers>
  </record>
  <record>
    <id>second</id>
    <name>mike</name>
    <papers>
      <paper>mike_a</paper>
      <paper>mike_b</paper>
    </papers>
  </record>
  <record>
    <id>third</id>
    <name>albert</name>
    <papers>
      <paper>paper of al</paper>
      <paper>other paper</paper>
    </papers>
  </record>
</records>

What I want to do is to extract tuples of data like the follow:

[{'code': 'first', 'name': 'john'}, 
 {'code': 'second', 'name': 'mike'}, 
 {'code': 'third', 'name': 'albert'}]

Now I wrote this python code:

try:
  doc = libxml2.parseDoc(xml)
except (libxml2.parserError, TypeError):
  print "Problems loading XML"

ctxt = doc.xpathNewContext()
ctxt.xpathRegisterNs("pre", "http://www.mysyte.com/foo")

record_nodes = ctxt.xpathEval('/pre:records/pre:record')

for record_node in record_nodes:
  id = record_node.xpathEval('id')[0].content
  name = record_node.xpathEval('name')[0].content
  ret_list.append({'code': id, 'name': name})

My problem is that I don't have any result and I have the impression that I'm doing something wrong with the XPATH when I iterate on the nodes.

I also tried with these XPATHs for the id and the name:

/id
/name
/record/id
/record/name
/pre:id
/pre:name

and so on, but with any result (BTW if I use the prefix in the sub queries I have an error).

Any idea?

回答1:

Here is a suggestion. Note the setContextNode() method:

import libxml2

xml = "test.xml"
doc = libxml2.parseFile(xml) 

ctxt = doc.xpathNewContext() 
ctxt.xpathRegisterNs("pre","http://www.mysyte.com/foo") 

ret_list = []
record_nodes = ctxt.xpathEval('/pre:records/pre:record') 

for node in record_nodes:
    ctxt.setContextNode(node)
    _id = ctxt.xpathEval('pre:id')[0].content
    name = ctxt.xpathEval('pre:name')[0].content
    ret_list.append({'code': _id, 'name': name}) 

print ret_list

回答2:

You can select all the elements you need with a single XPath expression:

/pre:records/pre:record/*[self::pre:id or self::pre:name]

Then just process the selected nodes in python.

回答3:

If it is possible to switch to lxml, here is one way it could be done:

import lxml.etree as le
root=le.XML(content)
result=[]
namespaces={'pre':'http://www.mysyte.com/foo'}
for record in root:
    id=record.xpath('pre:id',namespaces=namespaces)[0]
    name=record.xpath('pre:name',namespaces=namespaces)[0]
    result.append({'code':id.text,'name':name.text})
print(result)
# [{'code': 'first', 'name': 'john'}, {'code': 'second', 'name': 'mike'}, {'code': 'third', 'name': 'albert'}]

Building off of Dimitre Novatchev's XPath expression, you could do this:

id_name_nodes = iter(ctxt.xpathEval('/pre:records/pre:record/*[self::pre:id or self::pre:name]'))

ret_list=[]
for id,name in zip(id_name_nodes,id_name_nodes):
    ret_list.append({'code':id.content,'name':name.content})
print(ret_list)

This libxml2 code, relies on every record having an id and name. If an id or name is missing, the ret_list will pair the wrong id and name, failing silently. Under the same circumstance, the lxml code would raise an error.

回答4:

libxslt lacks such an important namespace support for some reason, but we can pre-parse the xml file, pre-read namespaces from it and then call xsltproc with those namespaces

def xpath(xml, xpathexpression):
    f=open(xml)
    fcontent = f.read()
    f.close()

    doc=libxml2.parseFile(xml)
    xp = doc.xpathNewContext()
    for nsdeclaration in re.findall('xmlns:*\w*="[^"]*"', fcontent):
        m = re.match('xmlns:(\w+)=.*', nsdeclaration)
        if m:
            ns = m.group(1)
        else:
            ns = "default"
        url = nsdeclaration[nsdeclaration.find('"')+1:nsdeclaration.rfind('"')]
        xp.xpathRegisterNs(ns, url)
    a=xp.xpathEval(xpathexpression)
    if len(a):
        return a[0].content
    return ""

来源：https://stackoverflow.com/questions/3366032/python-and-libxml2-how-to-iterate-in-xml-nodes-with-xpath