mod_wsgi和zeromq錯誤（python）

Question

我正在嘗試制作一個在網絡上運行的搜索引擎。 因此，我使用mod_wsgi通過python從網頁獲取查詢。 然后，我使用zeromq將查詢發送到C ++搜索程序。

但是問題是，似乎zeromq和mod_wsgi不能一起工作。 python可以導入zmq（我已經對其進行了測試），這是絕對正確的，但是當它在網絡上運行時，它會顯示錯誤消息（實際上是來自Apache的錯誤日志）

File "D:/wsgi_app/wsgi_app.py", line 2, in <module>, referer: http://localhost/
         import zmq, referer: http://localhost/

File "D:\\util\\Python27\\lib\\site-packages\\zmq\\__init__.py", line 35, in <module>, referer: http://localhost/
         from zmq.utils import initthreads # initialize threads, referer: http://localhost/

ImportError: DLL load failed: \xc1\xf6\xc1\xa4\xb5\xc8 \xb8\xf0\xb5\xe2\xc0\xbb \xc3\xa3\xc0\xbb \xbc\xf6 \xbe\xf8\xbd\xc0\xb4\xcf\xb4\xd9., referer: http://localhost/

我不知道為什么wsgi無法導入zmq。 順便說一下，下面的源代碼是完整的python代碼

from cgi import parse_qs, escape
import zmq
def application( # It accepts two arguments:
      # environ points to a dictionary containing CGI like environment variables
      # which is filled by the server for each received request from the client
      environ,
      # start_response is a callback function supplied by the server
      # which will be used to send the HTTP status and headers to the server
      start_response):

    # get a query from the webpage :)
   data = parse_qs(environ['QUERY_STRING'])

   query = data.get('query', [''])[0]
   query = escape(query) #prevent script injection

   context = zmq.Context()
   socket = context.socket(zmq.REQ)
   socket.connect ("tcp://localhost:5555") #connect to C++ search server database
   socket.send (query)

   # build the response body possibly using the environ dictionary
   response_body = 'The request method was %s' % environ['REQUEST_METHOD']

   # HTTP response code and message
   status = '200 OK'

   # These are HTTP headers expected by the client.
   # They must be wrapped as a list of tupled pairs:
   # [(Header name, Header value)].
   response_headers = [('Content-Type', 'text/plain'),
                       ('Content-Length', str(len(response_body)))]

   # Send them to the server using the supplied function
   start_response(status, response_headers)

   # Return the response body.
   # Notice it is wrapped in a list although it could be any iterable.
   return [response_body]

Answer 1

在wsgi腳本中，顯式添加依賴項的位置：

import site
site.addsitedir(path_to_zeromq)