Question

我正在尝试访问REST API，需要使用一行XML来调用它以获取过滤条件。我很抱歉提供其他人无法访问的代码。当我执行此代码时，我收到下面列出的错误消息。

import urllib2
import urllib
import hashlib
import hmac
import time
import random
import base64

def MakeRequest():
  url = 'https://api01.marketingstudio.com/API/Gateway/9'
  publickey = ''
  privatekey = ''
  method = 'Query'
  nonce = random.randrange(123400, 9999999)
  age = int(time.time())
  final = str(age) + '&' + str(nonce) + '&' + method.lower() + '&' + url.lower()  

  converted = hmac.new(privatekey, final, hashlib.sha1).digest()
  authorization = 'AMS ' + publickey + ':' + base64.b64encode(converted)

  xml_string = "<list><FilterItems><FilterItem attribute='pageNumber' value='1'/></FilterItems></list>"
  form = {'XML':xml_string}
  data = urllib.urlencode(form)
  headers = {'Content-Type': 'application/xml'}
  req = urllib2.Request(url,data,headers)
  req.add_header('ams-method', method)
  req.add_header('ams-nonce', nonce)
  req.add_header('ams-age', age)
  req.add_header('Authorization', authorization)  
  r = urllib2.urlopen(req)

  print r.read()

MakeRequest();

以下是错误消息。

Data at the root level is invalid. Line 1, position 1.
   at Aprimo.REST.Core.RESTService.GetRequest(String URI, HttpRequest req)
   at Aprimo.REST.RESTHandler.GetRequest(String apiUrl, HttpContext context)
   at Aprimo.REST.RESTHandler.ProcessRequest(HttpContext context)

我认为这有正确的逻辑和过滤条件，我应该怎样看才能使其工作。谢谢。

Per @ Mark的建议我删除了XML字符串的urlencode并获得了以下TraceBack：

Traceback (most recent call last):
  File "file.py", line 36, in <module>
    MakeRequest(); 
  File "file.py", line 32, in MakeRequest
    r = urllib2.urlopen(req)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 126, in urlopen
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 392, in open
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 410, in _open
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 370, in _call_chain
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1194, in https_open
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1155, in do_open
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 941, in request
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 975, in _send_request
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 937, in endheaders
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 801, in _send_output
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 773, in send
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/ssl.py", line 207, in sendall
TypeError: unhashable type

Answer 1

所以问题在于表单变量的格式化和我试图做的编码。修改以下行可以调用工作。我不需要指定标题。

 xml_string = "<list><FilterItems><FilterItem attribute='pageNumber' value='1'/></FilterItems></list>"
 data = (xml_string)
 req = urllib2.Request(url,data)

使用Python中的urllib将XML传递给HTTP Post api的问题

1 个答案: