restapi以.json格式输出

Question

我正在创建一个脚本，它将在python中调用restapi并以JSON格式输出结果。我在我的代码中发现了一些追溯错误。我该怎么解决这个问题呢。在

'import sitecustomize' failed; use -v for traceback
Traceback (most recent call last):
  File "/home/Desktop/Sync.py", line 12, in <module>
    url = urllib2.Request(request)
  File "/usr/lib/python2.7/urllib2.py", line 202, in __init__
    self.__original = unwrap(url)
  File "/usr/lib/python2.7/urllib.py", line 1057, in unwrap
    url = url.strip()
  File "/usr/lib/python2.7/urllib2.py", line 229, in __getattr__
    raise AttributeError, attr
AttributeError: strip

代码如下：

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Answer 1

来自Request类的documentation：

url should be a string containing a valid URL.

您当前正在将另一个Request对象传递给它的构造函数，所以这就是您看到的错误的原因。正确的方法是：

request=urllib2.Request( "http.google.com")
request.add_header("Authorization","Basic xxxxxxxxxxxxxxxxxxxxxxxx=")
socket = urllib2.urlopen(request)

Answer 2

你似乎有个问题：

request=urllib2.Request( "http.google.com")
request.add_header("Authorization","Basic xxxxxxxxxxxxxxxxxxxxxxxx=")
url = urllib2.Request(request)
socket = urllib2.urlopen(url)

您正试图通过将请求对象传递到构造函数中来创建名为“url”的请求对象。在

见http://docs.python.org/2/library/urllib2.html#urllib2.Request

试试这个：

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