The server employs User-Agent sniffing; it looks at the User-Agent header and if it doesn't like what it sees it'll return an empty response.

You can set the header yourself:

Copyimport urllib2
import shutil

headers = {'User-Agent': 'Mozilla'}
urlfile = "http://www.equibase.com/premium/eqbLateChangeXMLDownload.cfm"

request = urllib2.Request(urlfile, headers=headers)
response = urllib2.urlopen(request)
with open("c:\\test.xml", 'wb') as outfile:
    shutil.copyfileobj(response, outfile)

The 'Mozilla' User-Agent string is apparently enough to convince the server to serve the file.

I used a combination of urllib2 (an updated version of the urllib library) and shutil.copyfileobj() to handle setting additional headers, then copying the response data to a file. urllib.urlretrieve() does not support adding headers, and urllib2 has no urlretrieve() equivalent.

Here is the solution

import requests
import xml.etree.ElementTree as ET


response = requests.get('https://registers.esma.europa.eu/solr/esma_registers_firds_files/select?q=*&fq=publication_date:%5B2021-01-17T00:00:00Z+TO+2021-01-19T23:59:59Z%5D&wt=xml&indent=true&start=0&rows=100')
root = ET.fromstring(response.text)
for i in root.findall('result'):
    for j in i.findall('doc'):
        for k in j:
            link = j.find('.//str[@name="download_link"]').text
            print(link)
            req = requests.get(link)

Try to read the XML first, and then pass the result to elementtree :

.....
file_details = url_opener.open(url).read()
root = ET.fromstring(file_details)

And if somehow the tree is needed as well :

tree = ET.ElementTree(root)

instead of using urlopen I just use and retrieve.

def getXML(self,subreddit):
        url = 'http://www.reddit.com'+subreddit+'.rss'
        source = urllib.request.urlretrieve(url,'rss.xml')
        self.getXmlData()

First suggestion: do what even the official urllib docs says and don't use urllib, use requests instead.

Your problem is that you use .writelines() and it expects a list of lines, not a bytes objects (for once in Python the error message is not very helpful). Use .write() instead

import requests
resp = requests.get('URL')
with open('file.xml', 'wb') as foutput:
   foutput.write(resp.content)

To import this file as JSON data (not XML) you need the JSON library

import urllib.request
import json
from pprint import pprint

testfile = urllib.request.URLopener()
file_name = 'http://localhost:8080/otp/routers/default/plan?fromPlace=48.40915,%20-71.04996&toPlace=48.41428,%20-71.06996&date=2017/12/04&time=8:00:00&mode=TRANSIT,WALK'
testfile.retrieve(file_name, "file.json")

data = json.load(open('file.json'))
pprint(data)

• json.load reads the JSON data and convert into a Python object (https://docs.python.org/2/library/json.html?highlight=json%20load#json.load)
• pprint is for "Pretty printing" the JSON data (https://docs.python.org/2/library/pprint.html)

This should help. The url you mentioned gives a zip. You can download that and extract it to get your XML.

EX:

import requests 
import zipfile 
import StringIO

URL='http://oasis.caiso.com/oasisapi/SingleZip?queryname=PRC_FUEL&fuel_region_id=ALL&startdatetime=20130919T07:00-0000&enddatetime=20130928T07:00-0000&version=1'
r = requests.get(URL, stream=True)
z = zipfile.ZipFile(StringIO.StringIO(r.content))
z.extractall()

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})

You don't need to write line-by-line. Just write the whole thing in one go:

>>> import urllib2
>>> url = "webservice"
>>> s = urllib2.urlopen(url)
>>> contents = s.read()
>>> file = open("export.xml", 'w')
>>> file.write(contents)
>>> file.close()