客户端如何获取服务器上传的文件名和扩展名的原始文件

Question

请帮助我，我只是python的初学者，我想学习这个。我不知道如何从服务器部分获取原始文件名和扩展名。

我尝试了很多方法和研究，但仍然无法工作。我见过许多类型的例子，这些示例只能在客户端部分上传带有with open('received_file','.txt','wb') as f:的文本文件，而不能上传文件的多种类型扩展名。我知道因为'.txt'所以只为文本文件工作。我不知道如何声明以获取多个扩展名和原始文件名。这是我的原始代码。

client

import socket
import os
TCP_IP = 'localhost'
TCP_PORT = 9001
BUFFER_SIZE = 8192
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect((TCP_IP, TCP_PORT))
#data = s.recv(BUFFER_SIZE)

with open('received_file','.txt','wb') as f:
    print ('file opened')
    while True:
        print('receiving data...')
        data = s.recv(BUFFER_SIZE)
        print('data=%s', (data))
        if not data:
            f.close()
            print ('file close()')
            break
        # write data to a file
        f.write(data)
print('Successfully get the file')
s.close()
print('connection closed')

块引用

server

import socket
from threading import Thread
from socketserver import ThreadingMixIn
import tkinter
import tkinter.filedialog
TCP_IP = 'localhost'
TCP_PORT = 9001
BUFFER_SIZE = 8192
tkinter.Tk().withdraw() 
in_path = tkinter.filedialog.askopenfilename( )
class ClientThread(Thread):
    def __init__(self,ip,port,sock):
        Thread.__init__(self)
        self.ip = ip
        self.port = port
        self.sock = sock
        print (" New thread started for "+ip+":"+str(port))
    def run(self):
        filename= in_path
        f = open(filename,'rb')
        while True:
            l = f.read(BUFFER_SIZE)
            while (l):
                self.sock.send(l)
                l = f.read(BUFFER_SIZE)
            if not l:
                f.close()
                self.sock.close()
                break
tcpsock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
tcpsock.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
tcpsock.bind((TCP_IP, TCP_PORT))
threads = []
while True:
    tcpsock.listen(5)
    print ("Waiting for incoming connections...")
    (conn, (ip,port)) = tcpsock.accept()
    print ('Got connection from ', (ip,port))
    newthread = ClientThread(ip,port,conn)
    newthread.start()
    threads.append(newthread)
for t in threads:
    t.join()

名称的输出文件received_file不带扩展名。

Answer 1

您需要定义一个协议来传输文件名，然后传输数据。 下面是一个示例，它将input.txt的内容作为文件名传输到客户端output.txt。 客户端读取文件名，然后将数据写入该文件名。 我这样做只是因为客户端和服务器在同一台计算机上运行，并且在同一目录中读/写文件。

server.py

import socketserver
class MyTCPHandler(socketserver.BaseRequestHandler):
    def handle(self):
        filename = 'output.txt'
        self.request.sendall(filename.encode() + b'rn')
        with open('input.txt','rb') as f:
            self.request.sendall(f.read())
if __name__ == "__main__":
    HOST, PORT = "localhost", 9001
    with socketserver.ThreadingTCPServer((HOST, PORT), MyTCPHandler) as server:
        server.serve_forever()

client.py

import socket
import os
SERVER = 'localhost',9001
s = socket.socket()
s.connect(SERVER)
# autoclose f and s when with block is exited.
# makefile treats the socket as a file stream.
# Open in binary mode so the bytes of the file are received as is.
with s,s.makefile('rb') as f:
    # The first line is the UTF-8-encoded filename.  Strip the line delimiters.
    filename = f.readline().rstrip(b'rn').decode()
    with open(filename,'wb') as out:
        out.write(f.read())