python urllib2 urlopen response:
<addinfourl at 1081306700 whose fp = <socket._fileobject object at 0x4073192c>>
expected:
{"token":"mYWmzpunvasAT795niiR"}
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3 Answers
You need to bind the resultant file-like object to a variable, otherwise the interpreter just dumps it via repr:
>>> import urllib2
>>> urllib2.urlopen('http://www.google.com')
<addinfourl at 18362520 whose fp = <socket._fileobject object at 0x106b250>>
>>>
>>> f = urllib2.urlopen('http://www.google.com')
>>> f
<addinfourl at 18635448 whose fp = <socket._fileobject object at 0x106b950>>
To get the actual data you need to perform a read().
>>> data = f.read()
>>> data[:50]
'<!doctype html><html itemscope="itemscope" itemtyp'
To see the headers returned:
>>> print f.headers
Date: Thu, 23 Aug 2012 00:46:22 GMT
Expires: -1
Cache-Control: private, max-age=0
... etc ...
2 Comments
Sidharth Samant
I have one question here. If I don't store the contents of
f to data, and simply perform f.read(), I get the contents only once. If I do f.read() again, it prints an empty string. Why is that?
mhawke
@SidharthSamant: because you have consumed all of the data from the stream - it is not stored internally by
urllib2.Add the following after your call to urlopen
print feed.read()
Comments
Perhaps you will find using the requests library more intuitive to use than urllib2.
