We Keep Coding

Getting values from XML Url Python

I have an account in https://es.besoccer.com/ and they have an api for getting data in a xml.
I have this code in python for print the values of the xml I need:
from xml.dom import minidom
doc = minidom.parse("datos.xml")
partidos = doc.getElementsByTagName("matches")
for partido in partidos:
local = partido.getElementsByTagName("local")[0]
visitante = partido.getElementsByTagName("visitor")[0]
print("local:%s" % local.firstChild.data)
print("visitante:%s" % visitante.firstChild.data)
canales=partido.getElementsByTagName("channels")
for canal in canales:
nombre=canal.getElementsByTagName("name")[0]
print("canal:%s" % nombre.firstChild.data)
The problem is thatthe XML of this site is a url so I don´t know how to read the xml directly form the url. Other problem is that the xml contains some tags that are a link, and python throughs a error with that tags that contains a url.

Read the API docs here: https://www.besoccer.com/api/documentacion
After you understand which API call you need to use, prepare the URL and the query arguments and use a library like requests in order to read the data.
Once you have the reply (assuming it is XML based) - you can use your code and parse it.

Weird json value urllib python

I'm trying to manipulate a dynamic JSON from this site:
http://esaj.tjsc.jus.br/cposgtj/imagemCaptcha.do
It has 3 elements, imagem, a base64, labelValorCaptcha, just a message, and uuidCaptcha, a value to pass by parameter to play a sound in this link bellow:
http://esaj.tjsc.jus.br/cposgtj/somCaptcha.do?timestamp=1455996420264&uuidCaptcha=sajcaptcha_e7b072e1fce5493cbdc46c9e4738ab8a
When I enter in the first site through a browser and put in the second link the uuidCaptha after the equal ("..uuidCaptcha="), the sound plays normally. I wrote a simple code to catch this elements.
import urllib, json
url = "http://esaj.tjsc.jus.br/cposgtj/imagemCaptcha.do"
response = urllib.urlopen(url)
data = json.loads(response.read())
urlSound = "http://esaj.tjsc.jus.br/cposgtj/somCaptcha.do?timestamp=1455996420264&uuidCaptcha="
print urlSound + data['uuidCaptcha']
But I dont know what's happening, the caught value of the uuidCaptcha doesn't work. Open a error web page.
Someone knows?
Thanks!

It works for me.
$ cat a.py
#!/usr/bin/env python
# encoding: utf-8
import urllib, json
url = "http://esaj.tjsc.jus.br/cposgtj/imagemCaptcha.do"
response = urllib.urlopen(url)
data = json.loads(response.read())
urlSound = "http://esaj.tjsc.jus.br/cposgtj/somCaptcha.do?timestamp=1455996420264&uuidCaptcha="
print urlSound + data['uuidCaptcha']
$ python a.py
http://esaj.tjsc.jus.br/cposgtj/somCaptcha.do?timestamp=1455996420264&uuidCaptcha=sajcaptcha_efc8d4bc3bdb428eab8370c4e04ab42c

As I said #Charlie Harding, the best way is download the page and get the JSON values, because this JSON is dynamic and need an opened web link to exist.
More info here.

The JSON syntax vs html/xml tags

The JSON syntax definition say that
html/xml tags (like the <script>...</script> part) are not part of
valid json, see the description at http://json.org.
A number of browsers and tools ignore these things silently, but python does
not.
I'd like to insert the javascript code (google analytics) to get info about the users using this service (place, browsers, OS ...).
What do you suggest to do?
I should solve the problem on [browser output][^1] or [python script][^2]?
thanks,
Antonio
[^1]: Browser output
<script>...</script>
[{"key": "value"}]
[^2]: python script
#!/usr/bin/env python
import urllib2, urllib, json
url="http://.........."
params = {}
url = url + '?' + urllib.urlencode(params, doseq=True)
req = urllib2.Request(url)
headers = {'Accept':'application/json;text/json'}
for key, val in headers.items():
req.add_header(key, val)
data = urllib2.urlopen(req)
print json.load(data)

These sound like two different kinds of services--one is a user-oriented web view of some data, with visualizations, formatting, etc., and one is a machine-oriented data service. I would keep these separate, and maybe build the user view as an extension to the data service.

Is it possible to use python suds to read a wsdl file from the file system?

From suds documentation, I can create a Client if I have a url for the WSDL.
from suds.client import Client
url = 'http://localhost:7080/webservices/WebServiceTestBean?wsdl'
client = Client(url)
I currently have the WSDL file on my file system. Is it possible to use suds to read the WSDL file from my file system instead of hosting it on a web server?

try to use url='file:///path/to/file'

Oneliner
# Python 3
import urllib, os
url = urllib.parse.urljoin('file:', urllib.request.pathname2url(os.path.abspath("service.xml")))
This is a more complete one liner that will:
let you specify just the local path,
get you the absolute path,
and then format it as a file-url.
Based upon:
the comments in the accepted answer and
this https://stackoverflow.com/a/14298190/622276
and thanks to user Sebastian the updated Python 3 implementation since we should avoid writing legacy python at this time.
Original for reference
# Python 2 (Legacy Python)
import urlparse, urllib, os
url = urlparse.urljoin('file:', urllib.pathname2url(os.path.abspath("service.xml")))

Using pathlib:
from pathlib import Path
url = Path('resources/your_definition.wsdl').absolute().as_uri()

Given a URL to a text file, what is the simplest way to read the contents of the text file?

In Python, when given the URL for a text file, what is the simplest way to access the contents off the text file and print the contents of the file out locally line-by-line without saving a local copy of the text file?
TargetURL=http://www.myhost.com/SomeFile.txt
#read the file
#print first line
#print second line
#etc

Edit 09/2016: In Python 3 and up use urllib.request instead of urllib2
Actually the simplest way is:
import urllib2 # the lib that handles the url stuff
data = urllib2.urlopen(target_url) # it's a file like object and works just like a file
for line in data: # files are iterable
print line
You don't even need "readlines", as Will suggested. You could even shorten it to: *
import urllib2
for line in urllib2.urlopen(target_url):
print line
But remember in Python, readability matters.
However, this is the simplest way but not the safe way because most of the time with network programming, you don't know if the amount of data to expect will be respected. So you'd generally better read a fixed and reasonable amount of data, something you know to be enough for the data you expect but will prevent your script from been flooded:
import urllib2
data = urllib2.urlopen("http://www.google.com").read(20000) # read only 20 000 chars
data = data.split("\n") # then split it into lines
for line in data:
print line
* Second example in Python 3:
import urllib.request # the lib that handles the url stuff
for line in urllib.request.urlopen(target_url):
print(line.decode('utf-8')) #utf-8 or iso8859-1 or whatever the page encoding scheme is

I'm a newbie to Python and the offhand comment about Python 3 in the accepted solution was confusing. For posterity, the code to do this in Python 3 is
import urllib.request
data = urllib.request.urlopen(target_url)
for line in data:
...
or alternatively
from urllib.request import urlopen
data = urlopen(target_url)
Note that just import urllib does not work.

The requests library has a simpler interface and works with both Python 2 and 3.
import requests
response = requests.get(target_url)
data = response.text

There's really no need to read line-by-line. You can get the whole thing like this:
import urllib
txt = urllib.urlopen(target_url).read()

import urllib2
for line in urllib2.urlopen("http://www.myhost.com/SomeFile.txt"):
print line

Another way in Python 3 is to use the urllib3 package.
import urllib3
http = urllib3.PoolManager()
response = http.request('GET', target_url)
data = response.data.decode('utf-8')
This can be a better option than urllib since urllib3 boasts having
Thread safety.
Connection pooling.
Client-side SSL/TLS verification.
File uploads with multipart encoding.
Helpers for retrying requests and dealing with HTTP redirects.
Support for gzip and deflate encoding.
Proxy support for HTTP and SOCKS.
100% test coverage.

import urllib2
f = urllib2.urlopen(target_url)
for l in f.readlines():
print l

For me, none of the above responses worked straight ahead. Instead, I had to do the following (Python 3):
from urllib.request import urlopen
data = urlopen("[your url goes here]").read().decode('utf-8')
# Do what you need to do with the data.

requests package works really well for simple ui
as #Andrew Mao suggested
import requests
response = requests.get('http://lib.stat.cmu.edu/datasets/boston')
data = response.text
for i, line in enumerate(data.split('\n')):
print(f'{i} {line}')
o/p:
0 The Boston house-price data of Harrison, D. and Rubinfeld, D.L. 'Hedonic
1 prices and the demand for clean air', J. Environ. Economics & Management,
2 vol.5, 81-102, 1978. Used in Belsley, Kuh & Welsch, 'Regression diagnostics
3 ...', Wiley, 1980. N.B. Various transformations are used in the table on
4 pages 244-261 of the latter.
5
6 Variables in order:
Checkout kaggle notebook on how to extract dataset/dataframe from URL

I do think requests is the best option. Also note the possibility of setting encoding manually.
import requests
response = requests.get("http://www.gutenberg.org/files/10/10-0.txt")
# response.encoding = "utf-8"
hehe = response.text

Just updating here the solution suggested by #ken-kinder for Python 2 to work with Python 3:
import urllib
urllib.request.urlopen(target_url).read()

You can use this, as well for simple methodology:
import requests
url_res = requests.get(url= "http://www.myhost.com/SomeFile.txt")
with open(filename + ".txt", "wb") as file:
file.write(url_res.content)