Connecting via SMB in Python

Question

I'm trying to send a file to an SMB server, using some code:

from smb.SMBConnection import SMBConnection

file_obj = open('VA.jpg', 'rb')

connection = SMBConnection(username = 'Admin',
                           password = 'Pa$$w0rd',
                           my_name = '',
                           remote_name = 'ARM4',
                           domain = 'WORKGROUP',
                           use_ntlm_v2=True)

connection.connect(ip='192.168.34.105')

connection.storeFile(service_name='Share',
                     path='\\192.168.34.105\Share',
                     file_obj=file_obj)
connection.close()

But when I run it I get an error:

Exception has occurred: OperationFailure
Failed to store 192.168.34.105\Share on Share: Unable to open file
==================== SMB Message 0 ====================
SMB Header:
-----------
Command: 0x03 (SMB2_COM_TREE_CONNECT) 
Status: 0x00000000 
Flags: 0x00 
PID: 5628 
MID: 3 
TID: 0 
Data: 52 bytes 
b'0900000048002c005c005c00330034004f00540044002d00410052004d0034002d004c00450042005c0053006800610072006500' 
SMB Data Packet (hex):
----------------------
b'fe534d4240000000000000000300000000000000000000000300000000000000fc150000000000001500000400a00000000000000000000000000000000

and so on, SMB Message 1 ... SMB Message 3

there is an error in the code in the line:

line 14, in <module>
    connection.storeFile(service_name='Share',
smb.smb_structs.OperationFailure: Failed to store 192.168.34.105\Share on Share: Unable to open file

Could you tell me please what I'm doing wrong? How to correctly specify parameters?

Answer 1

figured it out.

it is necessary to specify the name of the file that will be on the server in the path parameter where connection.storeFile, i.e. like this:

connection.storeFile(service_name='Share',
                     path='VA.jpg',
                     file_obj=file_obj)