我正在尝试创建一个虚拟环境并仔细跟踪所有依赖项。我是这样创建环境的:
virtualenv --no-site-packages purenv
cd purenv
source bin/activate
然后运行python脚本:
url = "http://localhost:6543/foo/",
hdrz = {
"Accept" : "text/html",
"account-code":"foo1234",
'Content-Type': 'application/json'
}
request = urllib2.Request(url, headers=hdrz)
我得到这个错误:
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
所以,我使用的是安装到virtualenv的python,但错误来自virtualenv外部(/System/Library/Frameworks/...
而不是purenv/lib/python2.7/site-packages...
或类似的)。
如何创建一个不调用外部文件的虚拟环境?
变化:
url = "http://localhost:6543/foo/",
:
url = "http://localhost:6543/foo/"
你正在发送一个元组,而不是一个str。对不起,我的英语不好。