I am trying to fetch a page using urllib2.urlopen (actually, I am using mechanize, but this is the method that mechanize calls) When I fetch the page, I am getting incomplete responses; the page gets truncated. However, if I access the non-HTTPS version of the page, I get the complete page.
I am on Arch Linux (3.5.4-1-ARCH x86_64). I am running openssl 1.0.1c. This problem occurs on another Arch Linux machine I own, but not when using Python 3 (3.3.0).
This problem seems to be related to urllib2 not retrieving entire HTTP response.
I tested it on the only online Python interpreter that would let me use urllib2 (Py I/O) and it worked as expected. Here is the code:
import urllib2
u = urllib2.urlopen('https://wa151.avayalive.com/WAAdminPanel/login.aspx?ReturnUrl=%2fWAAdminPanel%2fprivate%2fHome.aspx')
print u.read()[-100:]
The last lines should contain the usual </body></html>.
When I try urllib.urlretrieve on my machines, I get:
ContentTooShortError: retrieval incomplete: got only 11365 out of 13805 bytes
I cannot test urlretrieve on the online interpreter because it will not let users write to temporary files. Later in the evening, I will try fetching the URL from my machine, but from a different location.
I'm getting the same error, using Python 2.7, on a different Linux system:
>>> urllib.urlretrieve('https://wa151.avayalive.com/WAAdminPanel/login.aspx?ReturnUrl=%2fWAAdminPanel%2fprivate%2fHome.aspx')
---------------------------------------------------------------------------
ContentTooShortError Traceback (most recent call last)
...
ContentTooShortError: retrieval incomplete: got only 11365 out of 13805 bytes
However, the same operation can be done (and actually works for me) using requests:
>>> import requests
>>> r = requests.get('https://wa151.avayalive.com/WAAdminPanel/login.aspx?ReturnUrl=%2fWAAdminPanel%2fprivate%2fHome.aspx')
>>> with open(somefilepath, 'w') as f:
... f.write(r.text)
Is that working for you?
Related
I'm writing a python script to parse jenkins job results. I'm using urllib2 to fetch consoleText, but the file that I receive isn't full. The code to fetch the file is:
data = urllib2.urlopen('http://<server>/job/<jobname>/<buildid>/consoleText')
lines = data.readlines()
And the number of lines I get is 2306, while the actual number of lines in the console log is 37521. I can check that buy fetching the file via wget:
$ wget 'http://<server>/job/<jobname>/<buildid>/consoleText'
$ wc -l consoleText
37521
Why does urlopen not give me the full result?
UPDATE:
Using requests (as suggested by #svrist) instead of urllib2 doesn't have such a problem, so I'm switching to it. My new code is:
data = requests.get('http://<server>/job/<jobname>/<buildid>/consoleText')
lines = [l for l in data.iter_lines()]
But I still have no idea why urllib2.urlopen doesn't work properly.
The Jenkins build log is returned using a chunked encoding response.
Transfer-Encoding: chunked
Based on a couple of other questions, it seems like urllib2 does not handle the entire response and as you've observed, only returns the first chunk.
I also recommend using the requests package.
Using urllib2 in Python 2.7.4, I can readily download an Excel file:
output_file = 'excel.xls'
url = 'http://www.nbmg.unr.edu/geothermal/GEOTHERM-30Jun11.xls'
file(output_file, 'wb').write(urllib2.urlopen(url).read())
This results in the expected file that I can use as I wish.
However, trying to download just an HTML file gives me an empty file:
output_file = 'webpage.html'
url = 'http://www.nbmg.unr.edu/geothermal/mapfiles/nvgeowel.html'
file(output_file, 'wb').write(urllib2.urlopen(url).read())
I had the same results using urllib. There must be something simple I'm missing or don't understand. How do I download an HTML file from a URL? Why doesn't my code work?
If you want to download files or simply save a webpage you can use urlretrieve(from urllib library)instead of use read and write.
import urllib
urllib.urlretrieve("http://www.nbmg.unr.edu/geothermal/mapfiles/nvgeowel.html","doc.html")
#urllib.urlretrieve("url","save as..")
If you need to set a timeout you have to put it at the start of your file:
import socket
socket.setdefaulttimeout(25)
#seconds
It also Python 2.7.4 in my OS X 10.9, and the codes work well on it.
So I think there maybe other problems prevent its working. Can you open "http://www.nbmg.unr.edu/geothermal/GEOTHERM-30Jun11.xls" in your browser?
This may not directly answer the question, but if you're working with HTTP and have sufficient privileges to install python packages, I'd really recommend doing this with 'requests'. There's a related answered here - https://stackoverflow.com/a/13137873/45698
I have a program where I need to open many webpages and download information in them. The information, however, is in the middle of the page, and it takes a long time to get to it. Is there a way to have urllib only retrieve x lines? Or, if nothing else, don't load the information afterwards?
I'm using Python 2.7.1 on Mac OS 10.8.2.
The returned object is a file-like object, and you can use .readline() to only read a partial response:
resp = urllib.urlopen(url)
for i in range(10):
line = resp.readline()
would read only 10 lines, for example. Note that this won't guarantee a faster response.
OK, this is driving me nuts.
I am trying to read from the Crunchbase API using Python's Urllib2 library. Relevant code:
api_url="http://api.crunchbase.com/v/1/financial-organization/venrock.js"
len(urllib2.urlopen(api_url).read())
The result is either 73493 or 69397. The actual length of the document is much longer. When I try this on a different computer, the length is either 44821 or 40725. I've tried changing the user-agent, using Urllib, increasing the time-out to a very large number, and reading small chunks at a time. Always the same result.
I assumed it was a server problem, but my browser reads the whole thing.
Python 2.7.2, OS X 10.6.8 for the ~40k lengths. Python 2.7.1 running as iPython for the ~70k lengths, OS X 10.7.3. Thoughts?
There is something kooky with that server. It might work if you, like your browser, request the file with gzip encoding. Here is some code that should do the trick:
import urllib2, gzip
api_url='http://api.crunchbase.com/v/1/financial-organization/venrock.js'
req = urllib2.Request(api_url)
req.add_header('Accept-encoding', 'gzip')
resp = urllib2.urlopen(req)
data = resp.read()
>>> print len(data)
26610
The problem then is to decompress the data.
from StringIO import StringIO
if resp.info().get('Content-Encoding') == 'gzip':
g = gzip.GzipFile(fileobj=StringIO(data))
data = g.read()
>>> print len(data)
183159
I'm not sure if this is a valid answer, since it's a different module entirely but using the requests module, I get a ~183k response:
import requests
url = r'http://api.crunchbase.com/v/1/financial-organization/venrock.js'
r = requests.get(url)
print len(r.text)
>>>183159
So if it's not too late into the project, check it out here: http://docs.python-requests.org/en/latest/index.html
edit: Using the code you provided, I also get a len of ~36k
Did a quick search and found this: urllib2 not retrieving entire HTTP response
I'm currently trying to initiate a file upload with urllib2 and the urllib2_file library. Here's my code:
import sys
import urllib2_file
import urllib2
URL='http://aquate.us/upload.php'
d = [('uploaded', open(sys.argv[1:]))]
req = urllib2.Request(URL, d)
u = urllib2.urlopen(req)
print u.read()
I've placed this .py file in my My Documents directory and placed a shortcut to it in my Send To folder (the shortcut URL is ).
When I right click a file, choose Send To, and select Aquate (my python), it opens a command prompt for a split second and then closes it. Nothing gets uploaded.
I knew there was probably an error going on so I typed the code into CL python, line by line.
When I ran the u=urllib2.urlopen(req) line, I didn't get an error;
alt text http://www.aquate.us/u/55245858877937182052.jpg
instead, the cursor simply started blinking on a new line beneath that line. I waited a couple of minutes to see if something would happen but it just stayed like that. To get it to stop, I had to press ctrl+break.
What's up with this script?
Thanks in advance!
[Edit]
Forgot to mention -- when I ran the script without the request data (the file) it ran like a charm. Is it a problem with urllib2_file?
[edit 2]:
import MultipartPostHandler, urllib2, cookielib,sys
import win32clipboard as w
cookies = cookielib.CookieJar()
opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cookies),MultipartPostHandler.MultipartPostHandler)
params = {"uploaded" : open("c:/cfoot.js") }
a=opener.open("http://www.aquate.us/upload.php", params)
text = a.read()
w.OpenClipboard()
w.EmptyClipboard()
w.SetClipboardText(text)
w.CloseClipboard()
That code works like a charm if you run it through the command line.
If you're using Python 2.5 or newer, urllib2_file is both unnecessary and unsupported, so check which version you're using (and perhaps upgrade).
If you're using Python 2.3 or 2.4 (the only versions supported by urllib2_file), try running the sample code and see if you have the same problem. If so, there is likely something wrong either with your Python or urllib2_file installation.
EDIT:
Also, you don't seem to be using either of urllib2_file's two supported formats for POST data. Try using one of the following two lines instead:
d = ['uploaded', open(sys.argv[1:])]
## --OR-- ##
d = {'uploaded': open(sys.argv[1:])}
First, there's a third way to run Python programs.
From cmd.exe, type python myprogram.py. You get a nice log. You don't have to type stuff one line at a time.
Second, check the urrlib2 documentation. You'll need to look at urllib, also.
A Request requires a URL and a urlencoded encoded buffer of data.
data should be a buffer in the
standard
application/x-www-form-urlencoded
format. The urllib.urlencode()
function takes a mapping or sequence
of 2-tuples and returns a string in
this format.
You need to encode your data.
If you're still on Python2.5, what worked for me was to download the code here:
http://peerit.blogspot.com/2007/07/multipartposthandler-doesnt-work-for.html
and save it as MultipartPostHandler.py
then use:
import urllib2, MultipartPostHandler
opener = urllib2.build_opener(MultipartPostHandler.MultipartPostHandler())
opener.open(url, {"file":open(...)})
or if you need cookies:
import urllib2, MultipartPostHandler, cookielib
cj = cookielib.CookieJar()
opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cj), MultipartPostHandler.MultipartPostHandler())
opener.open(url, {"file":open(...)})