How to consume XML from RESTful web services using Django / Python?

Question

Should I use PyXML or what\'s in the standard library?

Answer 1

There's also BeautifulSoup, which has an API some might prefer. Here's an example on how you can extract all tweets that have been favorited from Twitter's Public Timeline:

from BeautifulSoup import BeautifulStoneSoup
import urllib

url = urllib.urlopen('http://twitter.com/statuses/public_timeline.xml').read()
favorited = []

soup = BeautifulStoneSoup(url)
statuses = soup.findAll('status')

for status in statuses:
    if status.find('favorited').contents != [u'false']:
        favorited.append(status)

Answer 2

ElementTree is provided as part of the standard Python libs. ElementTree is pure python, and cElementTree is the faster C implementation:

# Try to use the C implementation first, falling back to python
try:
    from xml.etree import cElementTree as ElementTree
except ImportError, e:
    from xml.etree import ElementTree

Here's an example usage, where I'm consuming xml from a RESTful web service:

def find(*args, **kwargs):
    """Find a book in the collection specified"""

    search_args = [('access_key', api_key),]
    if not is_valid_collection(kwargs['collection']):
        return None
    kwargs.pop('collection')
    for key in kwargs:
        # Only the first keword is honored
        if kwargs[key]:
            search_args.append(('index1', key))
            search_args.append(('value1', kwargs[key]))
            break

    url = urllib.basejoin(api_url, '%s.xml' % 'books')
    data = urllib.urlencode(search_args)
    req = urllib2.urlopen(url, data)
    rdata = []
    chunk = 'xx'
    while chunk:
        chunk = req.read()
        if chunk:
            rdata.append(chunk)
    tree = ElementTree.fromstring(''.join(rdata))
    results = []
    for i, elem in enumerate(tree.getiterator('BookData')):
        results.append(
               {'isbn': elem.get('isbn'),
                'isbn13': elem.get('isbn13'),
                'title': elem.find('Title').text,
                'author': elem.find('AuthorsText').text,
                'publisher': elem.find('PublisherText').text,}
             )
    return results

Answer 3

I always prefer to use the standard library when possible. ElementTree is well known amongst pythonistas, so you should be able to find plenty of examples. Parts of it have also been optimized in C, so it's quite fast.

http://docs.python.org/library/xml.etree.elementtree.html