如何递归地遍历 xml 文件并访问子节点/元素并使用 Python 存储它们的数据？

Question

I have a XML file like below. Now I need to access port->name , port->wire->direction , port->wire->driver->defval . The XML file is very large in size.

How do I approach this?

<spirit:Bus> 
    <spirit:Ports>   
        <spirit:port>
            <spirit:name>ABCPORT</spirit:name>
            <spirit:description>SOME DESCRIPTION</spirit:description>
            <spirit:wire>
                <spirit:direction>INPUT</spirit:direction>
                <spirit:driver>
                    <spirit:defaultValue>0</spirit:defaultValue>
                </spirit:driver>
            </spirit:wire>
        </spirit:port>
        <spirit:port>
            <spirit:name>PQRPORT</spirit:name>
            <spirit:description>SOME DESCRIPTION</spirit:description>
            <spirit:wire>
                <spirit:direction>OUTPUT</spirit:direction>
            </spirit:wire>
        </spirit:port>        
    </spirit:ports>
</spirit:Bus>

Answer 1

I believe the best way to address this is with lxml and xpath:

from lxml import etree

#the xml below is somewhat different than the one in the question, because of a type and the declare namespaces

spirit = """<?xml version="1.0" encoding="UTF-8"?>
<doc xmlns:spirit="http://example.com">
   <spirit:Bus>
      <spirit:Ports>
         <spirit:port>
            <spirit:name>ABCPORT</spirit:name>
            <spirit:description>SOME DESCRIPTION</spirit:description>
            <spirit:wire>
               <spirit:direction>INPUT</spirit:direction>
               <spirit:driver>
                  <spirit:defaultValue>0</spirit:defaultValue>
               </spirit:driver>
            </spirit:wire>
         </spirit:port>
         <spirit:port>
            <spirit:name>PQRPORT</spirit:name>
            <spirit:description>SOME DESCRIPTION</spirit:description>
            <spirit:wire>
               <spirit:direction>OUTPUT</spirit:direction>
            </spirit:wire>
         </spirit:port>
      </spirit:Ports>
   </spirit:Bus>
</doc>
"""

doc = etree.XML(spirit.encode('utf-8'))
ports = doc.xpath('//*[local-name()="port"]')
for port in ports:
    try:
        print("Port-",port.xpath('.//*[local-name()="name"]')[0].text)
        print("Direction",port.xpath('.//*[local-name()="direction"]')[0].text)
        print("Default value",port.xpath('.//*[local-name()="defaultValue"]')[0].text)
    except:
        continue

Output:

Port- ABCPORT
Direction INPUT
Default value 0
Port- PQRPORT
Direction OUTPUT

Answer 2

To have properly formatted XML, I added the namespace to your sample:

<spirit:Bus xmlns:spirit="http://dummy.com">
    ...
</spirit:Bus>

but Bus is still the root node, as in your sample. Of course, you can change the given URL to whatever you wish.

To do your task solely in ElementTree you can use the following code:

import xml.etree.ElementTree as et

tree = et.parse('Input.xml')
root = tree.getroot()
ns = {'spirit': 'http://dummy.com'}
for nd in root.findall('spirit:Ports/spirit:port', ns):
    print(nd.tag.split('}')[1], nd.findtext('spirit:name', namespaces=ns),
        nd.findtext('spirit:wire/spirit:direction', namespaces=ns),
        nd.findtext('spirit:wire/spirit:driver/spirit:defaultValue', namespaces=ns))

Note that your XML contains a namespace specification, so you have also to specify it in the code.

My code shows also how to get the local name of a node (without the namespace).

The result, for your sample is:

port ABCPORT INPUT 0
port PQRPORT OUTPUT None