Send file from client to server using XMLRPC?
문제
I want to write Python code to send a file from client to server. server needs to save the file sent from the client. But my code have some bugs which I cannot fix. Below is my server code:
# 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()
And the client code:
# client.py
import sys, xmlrpclib
proxy = xmlrpclib.Server('http://localhost:9000')
handle = open(sys.argv[1], "rb")
proxy.save_data(handle.read())
handle.close()
But then I run my code, the client returns the following error (this is on Windows):
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">
I have some questions:
How to fix the above bug?
My code needs to transfer some big files sometimes. Since my method is so simple, I doubt that it is efficient for moving big data. Could anybody please suggest a better method to move big files? (Of course it is better to use XMLRPC on Python)
해결책
Server side:
def server_receive_file(self,arg):
with open("path/to/save/filename", "wb") as handle:
handle.write(arg.data)
return True
Client side:
with open("path/to/filename", "rb") as handle:
binary_data = xmlrpclib.Binary(handle.read())
client.server_receive_file(binary_data)
This worked for me.
다른 팁
You want to look into the xmlrpclib Binary object. With this class you can encode and decode to/from a base64 string.
Here is how you do it:
#!/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)