使用minidom从URL解析XML与Python

Question

I have a problem during parsing the following xml from a url.

sample XML in my url path:

<?xml version="1.0" encoding="utf-8"?> 
<Documents>
    <class>
        <mid name="yyyyyyyyyyyyy"></mid>
        <person name="yyyyyyyyyy"></person>
        <url name="yyyyyyyyy"></url>
    </class>
    <class>
        <mid name="xxxxx"></mid>
        <person name="xxxxxxxxxx"></person>
        <url name="xxxxxxxxxxx"></url>
    </class>
</Documents>

Below is my python code;

def staff_list(request):

    url = http://path.to.url/
    dom = minidom.parse(urlopen(url))
    person = dom.getElementsByTagName('person')
    for i in person:
        print i.attributes['name'].value

within forloop I want to print the person and url tag values in xml that belongs to same parent class.

I tried the following method with iteration but get the "too many values to unpack" ERROR

def staff_list(request):

    url = http://path.to.url/
    dom = minidom.parse(urlopen(url))
    person = dom.getElementsByTagName('person')
    mid = dom.getElementsByTagName('mid')
    url = dom.getElementsByTagName('url')
    for i,j,k in person,mid,url:
        print i.attributes['name'].value,j.attributes['name'].value,k.attributes['name'].value

Any suggestions ?

Answer 1

You want to use zip() to combine the elements, I think:

for i,j,k in zip(person, mid, url):

Do yourself a big favour though and use the ElementTree API instead; that API is far pythononic and easier to use than the XML DOM API.

Answer 2

If you want to stick with minidom you can change your loop to:

for cls in dom.getElementsByTagName('class'):
    person = cls.getElementsByTagName('person')[0]
    mid = cls.getElementsByTagName('mid')[0]
    url = cls.getElementsByTagName('url')[0]

    print person.attributes['name'].value
    print mid.attributes['name'].value
    print url.attributes['name'].value

As @Martijn Pieters said, have a look at ElementTree as an alternative API. For example:

import xml.etree.ElementTree as ET
documents = ET.fromstring(xmlstr)
for cls in documents.iter('class'):
    person = cls.find('person')
    mid = cls.find('mid')
    url = cls.find('url')

    print person.get('name'), mid.get('name'), url.get('name')

Answer 3

I would use xpath and lxml.html: A minimalist approach:

import lxml.html as lh
doc=lh.parse(test.xml)

In [70]: persons = doc.xpath('.//person/@name')

In [71]: urls=doc.xpath('.//person[@name]/following-sibling::url/@name')

In [72]: mids=doc.xpath('.//person[@name]/preceding-sibling::mid/@name')

In [73]: [[p,m,u]for p,m,u in zip(persons, mids, urls)]
Out[73]: 
[['yyyyyyyyyy', 'yyyyyyyyyyyyy', 'yyyyyyyyy'],
 ['xxxxxxxxxx', 'xxxxx', 'xxxxxxxxxxx']]