python - 使用 Python 发送文件和文件名

Question

我想制作一个简单的服务器，它接受来自客户端的不同类型的文件。问题是，我无法获得从客户端发送过来的文件名，所以所有文件都保存为 1 个文件名，无论文件叫什么。

这是我的client.py:

import socket
with socket.socket() as s:
    s.connect(('192.168.1.2',10000))
    with open('Image.jpg','rb') as f:
        s.sendall(f.read())

这是我的 server.py:

import socket
print("Server started")
while True:
        with socket.socket() as s:
            s.bind(('192.168.1.2',10000))
            s.listen(1)
            with s.accept()[0] as c:
                chunks = []
                while True:
                    chunk = c.recv(4096)
                    if not chunk: break
                    chunks.append(chunk)
                    print("Receiving file")
            with open('Image.jpg','wb') as f:
                f.write(b''.join(chunks))
                print("File Received")

哪个工作正常。问题开始了，如果我发送另一个文件，例如“data.csv”，它仍然会在服务器上保存为“Image.jpg”。

有没有办法将文件名与文件一起保存？

Answer 1

我通常做的是在数据之前发送文件名:

import socket
print("Server started")
while True:
        with socket.socket() as s:
            s.bind(('192.168.1.2',10000))
            s.listen(1)
            with s.accept()[0] as c:
                chunks = []
                while True:
                    filename = c.recv(1600)
                    filename = filename.decode()
                    chunk = c.recv(4096)
                    if not chunk: break
                    chunks.append(chunk)
                    print("Receiving file")
            with open(filename,'wb') as f:
                f.write(b''.join(chunks))
                print("File Received")

客户:

import socket
import time
with socket.socket() as s:
    s.connect(('192.168.1.2',10000))
    with open('Image.jpg','rb') as f:
        s.sendall('Image.jpg'.encode())
        time.sleep(3)
        s.sendall(f.read())