slashdottir
slashdottir

Reputation: 8546

How to create a 'pure' virtualenv?

I'm trying to create a virtualenv and carefully track all dependencies. I created the env this way:

virtualenv --no-site-packages purenv

cd purenv
source bin/activate

Then run this python script:

url = "http://localhost:6543/foo/",

hdrz = {
    "Accept" : "text/html",
    "account-code":"foo1234",
    'Content-Type': 'application/json'
}
request = urllib2.Request(url, headers=hdrz)

I get this error:

  File "foo.py", line 10, in <module>
    request = urllib2.Request(url, headers=hdrz)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 229, in __init__
    self.__original = unwrap(url)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib.py", line 1075, in unwrap
    url = url.strip()
AttributeError: 'tuple' object has no attribute 'strip'

which python
/Users/foo.bar/workspace/purenv/bin/python

So, I'm using the python installed to the virtualenv, but the error is coming from outside the virtualenv (/System/Library/Frameworks/... and not purenv/lib/python2.7/site-packages... or the like).

How can I create a virtualenv where no external files are called?

Upvotes: 1

Views: 437

Answers (1)

Gianmar
Gianmar

Reputation: 515

Change:

url = "http://localhost:6543/foo/",

To:

url = "http://localhost:6543/foo/"

You are sending a tuple, not a str. Sorry for the bad english.

Upvotes: 1

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