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Send the file and the filename using Python

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3

1

I want to make a dead simple server that accepts different types of files from clients. The problem is, that I cannot get the filename sent over from the client, so all the files are saved as 1 filename, whatever the file is called.

This is my 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())

And this is my 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")

Which works fine. The problem starts, that if I send another file, for example "data.csv" it will still be saved as "Image.jpg" on the server.

Is there any way to also save the filename along with the file?

python sockets

share|improve this question

edited Mar 8 at 0:07

orde

4,76262533

asked Mar 8 at 0:02

Ioannis Tzon TsolakidisIoannis Tzon Tsolakidis

161

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Maybe you could include it as a header, or as a message before the picture?
  
  – Reedinationer
  Mar 8 at 0:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sending files in clear text isn't a great idea nowadays, have you considered using scp ?
  
  – Benoît Pilatte
  Mar 8 at 0:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You'll need to create a wire protocol that lets you include the file name (or use one that already can do so, like e.g. ftp...)
  
  – thebjorn
  Mar 8 at 0:22
  

  
  
  
  

add a comment | 

3

1

I want to make a dead simple server that accepts different types of files from clients. The problem is, that I cannot get the filename sent over from the client, so all the files are saved as 1 filename, whatever the file is called.

This is my 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())

And this is my 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")

Which works fine. The problem starts, that if I send another file, for example "data.csv" it will still be saved as "Image.jpg" on the server.

Is there any way to also save the filename along with the file?

python sockets

share|improve this question

edited Mar 8 at 0:07

orde

4,76262533

asked Mar 8 at 0:02

Ioannis Tzon TsolakidisIoannis Tzon Tsolakidis

161

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Maybe you could include it as a header, or as a message before the picture?
  
  – Reedinationer
  Mar 8 at 0:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sending files in clear text isn't a great idea nowadays, have you considered using scp ?
  
  – Benoît Pilatte
  Mar 8 at 0:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You'll need to create a wire protocol that lets you include the file name (or use one that already can do so, like e.g. ftp...)
  
  – thebjorn
  Mar 8 at 0:22
  

  
  
  
  

add a comment | 

I want to make a dead simple server that accepts different types of files from clients. The problem is, that I cannot get the filename sent over from the client, so all the files are saved as 1 filename, whatever the file is called.

This is my 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())

And this is my 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")

Which works fine. The problem starts, that if I send another file, for example "data.csv" it will still be saved as "Image.jpg" on the server.

Is there any way to also save the filename along with the file?

python sockets

share|improve this question

edited Mar 8 at 0:07

orde

4,76262533

asked Mar 8 at 0:02

Ioannis Tzon TsolakidisIoannis Tzon Tsolakidis

161

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())

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")

share|improve this question

edited Mar 8 at 0:07

orde

4,76262533

asked Mar 8 at 0:02

Ioannis Tzon TsolakidisIoannis Tzon Tsolakidis

161

share|improve this question

edited Mar 8 at 0:07

orde

4,76262533

asked Mar 8 at 0:02

Ioannis Tzon TsolakidisIoannis Tzon Tsolakidis

161

edited Mar 8 at 0:07

orde

4,76262533

edited Mar 8 at 0:07

orde

4,76262533

asked Mar 8 at 0:02

Ioannis Tzon TsolakidisIoannis Tzon Tsolakidis

161

asked Mar 8 at 0:02

Ioannis Tzon TsolakidisIoannis Tzon Tsolakidis

161