Parse XML from URL into python object

Question

The goodreads website has this API for accessing a user’s ‘shelves:’ https://www.goodreads.com/review/list/20990068.xml?key=nGvCqaQ6tn9w4HNpW8kquw&v=2&shelf=toread

It returns XML. I’m trying to create a django project that shows books on a shelf from this API. I’m looking to find out how (or if there is a better way than) to write my view so I can pass an object to my template. Currently, this is what I’m doing:

import urllib2

def homepage(request):
    file = urllib2.urlopen('https://www.goodreads.com/review/list/20990068.xml?key=nGvCqaQ6tn9w4HNpW8kquw&v=2&shelf=toread')
    data = file.read()
    file.close()
    dom = parseString(data)

I’m not entirely sure how to manipulate this object if I’m doing this correctly. I’m following this tutorial.

Asked By: smilebomb

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Source

Answer 1

I’d use xmltodict to make a python dictionary out of the XML data structure and pass this dictionary to the template inside the context:

import urllib2
import xmltodict

def homepage(request):
    file = urllib2.urlopen('https://www.goodreads.com/review/list/20990068.xml?key=nGvCqaQ6tn9w4HNpW8kquw&v=2&shelf=toread')
    data = file.read()
    file.close()

    data = xmltodict.parse(data)
    return render_to_response('my_template.html', {'data': data})

Answered By: alecxe

Answer 2

xmltodict using urllib3

import traceback
import urllib3
import xmltodict

def getxml():
    url = "https://yoursite/your.xml"

    http = urllib3.PoolManager()

    response = http.request('GET', url)
    try:
        data = xmltodict.parse(response.data)
    except:
        print("Failed to parse xml from response (%s)" % traceback.format_exc())
    return data

Answered By: gies0r

Answer 3

xmltodict using requests

import requests
import xmltodict

url = "https://yoursite/your.xml"
response = requests.get(url)
data = xmltodict.parse(response.content)

Answered By: Vincent

Parse XML from URL into python object

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