【问题标题】:Send file from client to server using XMLRPC?使用 XMLRPC 将文件从客户端发送到服务器?
【发布时间】:2012-02-24 07:56:35
【问题描述】:

我想编写 Python 代码将文件从客户端发送到服务器。服务器需要保存客户端发送的文件。但是我的代码有一些我无法修复的错误。以下是我的服务器代码:

# server.py
from SimpleXMLRPCServer import SimpleXMLRPCServer
import os

server = SimpleXMLRPCServer(('localhost', 9000))

def save_data(data):
    handle = open("x123.dat", "wb")
    handle.write(data)
    handle.close()

server.register_function(save_data, 'save_data')
server.serve_forever()

以及客户端代码:

# client.py
import sys, xmlrpclib

proxy = xmlrpclib.Server('http://localhost:9000')
handle = open(sys.argv[1], "rb")
proxy.save_data(handle.read())
handle.close()

然后我运行我的代码,客户端返回以下错误(这是在 Windows 上):

Traceback (most recent call last):
File "client.py", line 6, in <module> proxy.save_data(handle.read())
File "c:\python27\lib\xmlrpclib.py", line 1224, in __call__
  return self.__send(self.__name, args)
File "c:\python27\lib\xmlrpclib.py", line 1575, in __request
  verbose=self.__verbose
File "c:\python27\lib\xmlrpclib.py", line 1264, in request
  return self.single_request(host, handler, request_body, verbose)
File "c:\python27\lib\xmlrpclib.py", line 1297, in single_request
  return self.parse_response(response)
File "c:\python27\lib\xmlrpclib.py", line 1473, in parse_response
  return u.close()
File "c:\python27\lib\xmlrpclib.py", line 793, in close
  raise Fault(**self._stack[0])
xmlrpclib.Fault: <Fault 1: "<class 'xml.parsers.expat.ExpatError'>:not well-formed (invalid token): line 7, column 1">

我有一些问题:

  1. 如何修复上述错误?

  2. 我的代码有时需要传输一些大文件。由于我的方法非常简单,我怀疑它对于移动大数据是否有效。有人可以建议一种更好的方法来移动大文件吗? (当然在 Python 上使用 XMLRPC 更好)

【问题讨论】:

    标签: python xml-rpc sendfile


    【解决方案1】:

    服务器端:

    def server_receive_file(self,arg):
            with open("path/to/save/filename", "wb") as handle:
                handle.write(arg.data)
                return True
    

    客户端:

    with open("path/to/filename", "rb") as handle:
        binary_data = xmlrpclib.Binary(handle.read())
    client.server_receive_file(binary_data)
    

    这对我有用。

    【讨论】:

      【解决方案2】:

      您想查看 xmlrpclib Binary object。使用此类,您可以对 base64 字符串进行编码和解码。

      【讨论】:

        【解决方案3】:

        这是你的做法:

        #!/usr/bin/env python3.7
        
        # rpc_server.py
        
        # Fix missing module issue: ModuleNotFoundError: No module named 'SimpleXMLRPCServer'
        #from SimpleXMLRPCServer import SimpleXMLRPCServer
        from xmlrpc.server import SimpleXMLRPCServer
        
        import os
        
        # Put in your server IP here
        IP='10.198.16.73'
        PORT=64001
        
        server = SimpleXMLRPCServer((IP, PORT))
        
        def server_receive_file(arg, filename):
            curDir = os.path.dirname(os.path.realpath(__file__))
            output_file_path = curDir + '/' + filename
            print('output_file_path -> ({})'.format(output_file_path))
            with open(output_file_path, "wb") as handle:
                handle.write(arg.data)
                print('Output file: {}'.format(output_file_path))
                return True
        
        server.register_function(server_receive_file, 'server_receive_file')
        print('Control-c to quit')
        server.serve_forever()
        
        ### rpc_client.py
        #!/usr/bin/env python3.7
        
        import os
        
        
        # client.py
        import sys
        
        # The answer is that the module xmlrpc is part of python3
        
        import xmlrpc.client
        
        #Put your server IP here
        IP='10.198.16.73'
        PORT=64001
        
        
        url = 'http://{}:{}'.format(IP, PORT)
        ###server_proxy = xmlrpclib.Server(url)
        client_server_proxy = xmlrpc.client.ServerProxy(url)
        
        curDir = os.path.dirname(os.path.realpath(__file__))
        filename = sys.argv[1]
        fpn = curDir + '/' + filename
        print(' filename -> ({})'.format(filename))
        print(' fpn -> ({})'.format(fpn))
        if not os.path.exists(fpn):
            print('Missing file -> ({})'.format(fpn))
            sys.exit(1)
        
        with open(fpn, "rb") as handle:
            binary_data = xmlrpc.client.Binary(handle.read())
            client_server_proxy.server_receive_file(binary_data, filename)
        

        【讨论】:

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