python jsonrpc錯誤錯誤的網關

Question

我想了解python中的JSON rpc。我的http服務器如下所示。

#!/usr/bin/env python 
# coding: utf-8 

    import pyjsonrpc 

    def add(a, b): 
     """Test function""" 
     return a + b 

    class RequestHandler(pyjsonrpc.HttpRequestHandler): 

     # Register public JSON-RPC methods 
     methods = { 
      "add": add 
     } 

    # Threading HTTP-Server 
    http_server = pyjsonrpc.ThreadingHttpServer(
     server_address = ('localhost', 8080), 
     RequestHandlerClass = RequestHandler 
) 
    print "Starting HTTP server ..." 
    print "URL: http://localhost:8080" 
    http_server.serve_forever() 

我的客戶端看起來如下。

#!/usr/bin/env python 
# coding: utf-8 

    import pyjsonrpc 

    http_client = pyjsonrpc.HttpClient(
     url = "http://example.com/jsonrpc", 
     username = "Username", 
     password = "Password" 
) 
print http_client.call("add", 1, 2) 
# Result: 3 

# It is also possible to use the *method* name as *attribute* name. 
print http_client.add(1, 2) 
15 # Result: 3 

我開始HTTP服務器如下

'$蟒蛇http_server.py'

Starting HTTP server ... 
URL: http://localhost:8080 

然後我開始客戶端如下

$蟒蛇http_client.py

我收到以下錯誤。

Traceback (most recent call last): 
    File "http_client.py", line 11, in <module> 
    print http_client.call("add", 1, 2) 
    File "/usr/local/lib/python2.7/dist-packages/pyjsonrpc/http.py", line 98, in call 
    password = self.password 
    File "/usr/local/lib/python2.7/dist-packages/pyjsonrpc/http.py", line 33, in http_request 
    response = urllib2.urlopen(request) 
    File "/usr/lib/python2.7/urllib2.py", line 126, in urlopen 
    return _opener.open(url, data, timeout) 
    File "/usr/lib/python2.7/urllib2.py", line 406, in open 
    response = meth(req, response) 
    File "/usr/lib/python2.7/urllib2.py", line 519, in http_response 
    'http', request, response, code, msg, hdrs) 
    File "/usr/lib/python2.7/urllib2.py", line 444, in error 
    return self._call_chain(*args) 
    File "/usr/lib/python2.7/urllib2.py", line 378, in _call_chain 
    result = func(*args) 
    File "/usr/lib/python2.7/urllib2.py", line 527, in http_error_default 
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) 
urllib2.HTTPError: HTTP Error 502: Bad Gateway 

Answer 1

在客戶端腳本中，將您的URL更改爲localhost：8080以匹配服務器設置。你也不需要這個簡單實現的用戶名或密碼。我在下面包含了客戶端的工作版本。

#!/usr/bin/env python 
# coding: utf-8 

import pyjsonrpc 

http_client = pyjsonrpc.HttpClient(
    url = "http://localhost:8080" 
) 

print http_client.call("add", 1, 2) 
# Result: 3 

# It is also possible to use the *method* name as *attribute* name. 
print http_client.add(1, 2) 
# Result: 3