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Using WMI-Client-Wrapper to execute an exe and get output logs

Objective:
I am using Ubuntu 16.04 and am using WMI-CLient-Wrapper module to connect to a remote Windows Machine and send an executable to it(eg. Process Explorer) and further execute it and collect the logs it creates and fetch them back to my Linux Machine for further processing. Using WMI CLient Wrapper is the only option available as WMI Module doesn't work with Linux.
Problem:
I am able to send the file to the remote Windows machine, by establishing a connection using WMI-Client-Wrapper and SMB File Transfer Mechanism. After that when I try to create a Process for the same and try to execute that process it gives me an error stating that some of the attributes that WMI actually has, are not supported by WMI client Wrapper.
What I tried
Python Code:
import os
import wmi_client_wrapper as wmic
from socket import *
import time
wmic = wmic.WmiClientWrapper(
host ="192.168.115.128",
username = "LegalWrongDoer",
password = "sasuke14"
)
SW_SHOWNORMAL = 1
str = "smbclient //192.168.115.128/C$ -U LegalWrongDoer%sasuke14 -c \'put \"procexp64.exe\"\'"
os.system(str)
print("Folder sent")
process_startup = wmic.Win32_ProcessStartup.new()
process_startup.ShowWindow = SW_SHOWNORMAL
process_id, result = wmic.Win32_Process.Create(CommandLine="C:/procexp64.exe", ProcessStartupInformation=process_startup)
process_startup.ShowWindow = SW_SHOWNORMAL
if result == 0:
print("Process started successfully")
else:
print("Sorry, but can't execute Process!")
When I run this python file, it gives me the output to the initial query I make. But the Process_StartUp fails.
Further Traceback Calls:
Traceback (most recent call last):
File "WMIClient.py", line 22, in <module>
process_startup = wmic.Win32_ProcessStartup.new()
AttributeError: 'WmiClientWrapper' object has no attribute 'Win32_ProcessStartup'
I'd be extremely grateful if anyone of you can be able to help me through this. Thanks in advance :)

Well I finally managed to get a work-around for this whole scenario, and it might look a little messy but it sure does work for me.
Firstly I use smbclient to transfer the executable to the end-point where I want to execute it. Inside my code I use os.system() calls to make this happen.
import os
str1 = "smbclient //'<HostMachineIP>'/admin$ -U '<domain>\\<username>%<password>' -c \'lcd /usr/local/acpl/bin/endPoint/; put \"EndPointForeignsics.exe\"\'"
os.system(str1)
This helps me put the executable in desired shared folder that the user has access(Admin in my case) to and then use WMI-query through a tool called Winexe to get access to the console/command prompt of the end-point. I use another os.system() call to execute this again.
str2 = r'/usr/local/bin/winexe -U "<domain>\\<username>%<password>" //<HostMachineIP> "cmd /c c:\windows\EndPointForeignsics.exe '
os.system(str2)
P.S:-- Winexe is a tool that you'll have to download off the internet and compile it. It may take some time and effort to do that, but is quite achievable. You'll get a lot of help on the same from StackOverflow and Documentation of the tool.

How do I write remote machine Path in askopenfilename

I am writing a program in python on Ubuntu, to remove a file from remote machine(raspberrypi) by accessing it, connected with network.
For file selection I am using a command called askopenfilename.
But I am struggling in specifying the Path of RaspberryPi correctly.
Can you please guide me on how do I specify Raspberrypi path correctly?
IP = "192.168.2.34"
Username = "pi"
Password ="raspberry"
Path="/home/pi/python"
Below is my code
from tkFileDialog import askopenfilename
import paramiko
client = paramiko.SSHClient()
client.set_missing_host_key_policy(paramiko.AutoAddPolicy())
client.connect('192.168.2.34', username='pi', password='raspberry')
checkdir = "/home/pi/python"
name1= askopenfilename(title = "Select File For Removal", initialdir = checkdir)
stdin, stdout, stderr = client.exec_command('ls -l')
for line in stdout:
print '... ' + line.strip('\n')
client.close()

I am fairly sure the tk file dialog is only able to browse the Ubuntu machine (which it seems to run on) filesystem - not the RPi filesystem over SSH, so you will never see RPi directories.
if You can read the RPi directory listing there, so you could create your own file browser component or try to find an existing one that works over SSH.
Then again it looks like you are 'inventing the wheel again' (which of course is ok for learning purposes) - gFTP, FileZilla, many of the Linux desktop file browsers etc. (or WinSCP if you were using a Windows box) are ready made tools for this.

How to get linux screen title from command line

How can I fetch the title of a screen session from the command line?

I came up with a very small and simple python script with pexpect to do it.
It is handy in multiuser environments where some host is reserved and status is written to screen title by user.
It works for me, feel free to make it better.
In order to fetch specific session title, you need to modify the script and call for correct session.
If you run this through remote connection as local script (through SSH for example), remember to set export TERM=xterm before execution.
try:
import pexpect
import sys
child=pexpect.spawn('screen -x')
child.sendcontrol('a');
child.send('A');
i = child.expect('Set window.*')
child.sendcontrol('c');
child.sendcontrol('a');
child.send('d');
TITLE=str(child.after)
TITLE_P=TITLE.split('7m')
if str(TITLE_P[-1]) == '':
print 'Title not found'
else:
print str(TITLE_P[-1])
except:
print 'Could not check screen Title'

all python windows service can not start{error 1053}

all python code service can install but cannot start
Error 1053: The service did not respond to the start or control request in a timely fashion".
since my service can install and start in my server.
i think my code has no problem.
but i still wonder is there a solution that i can solve this error in code
my service:
import win32serviceutil
import win32service
import win32event
import time
import traceback
import os
import ConfigParser
import time
import traceback
import os
import utils_func
from memcache_synchronizer import *
class MyService(win32serviceutil.ServiceFramework):
"""Windows Service."""
os.chdir(os.path.dirname(__file__))
conf_file_name = "memcache_sync_service.ini"
conf_parser = ConfigParser.SafeConfigParser()
conf_parser.read(conf_file_name)
_svc_name_, _svc_display_name_, _svc_description_ = utils_func.get_win_service(conf_parser)
def __init__(self, args):
if os.path.dirname(__file__):
os.chdir(os.path.dirname(__file__))
win32serviceutil.ServiceFramework.__init__(self, args)
# create an event that SvcDoRun can wait on and SvcStop can set.
self.stop_event = win32event.CreateEvent(None, 0, 0, None)
def SvcDoRun(self):
self.Run()
win32event.WaitForSingleObject(self.stop_event, win32event.INFINITE)
def SvcStop(self):
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
win32event.SetEvent(self.stop_event)
LoggerInstance.log("memcache_sync service is stopped")
self.ReportServiceStatus(win32service.SERVICE_STOPPED)
sys.exit()
def Run(self):
try:
LoggerInstance.log("\n******\n\memcache_sync_service is running, configuration: %s\n******" % (self.conf_file_name,))
if ((not self.conf_parser.has_section('Memcache')) or
(not self.conf_parser.has_option('Memcache', 'check_interval'))):
LoggerInstance.log('memcache_sync_service : no Memcache service parameters')
self.SvcStop()
# set configuration parameters from ini configuration
self.check_interval = self.conf_parser.getint('Memcache', 'check_interval')
ms = MemcacheSynchronizer()
while 1:
ms.Sync()
time.sleep(self.check_interval)
except:
LoggerInstance.log("Unhandled Exception \n\t%s" % (traceback.format_exc(),))
if __name__ == '__main__':
win32serviceutil.HandleCommandLine(MyService)
execute result of "sc query [name]" cmd:
SERVICE_NAME: NewsMonitoringMemcacheSynchronizer
TYPE : 10 WIN32_OWN_PROCESS
STATE : 1 STOPPED
(NOT_STOPPABLE,NOT_PAUSABLE,IGNORES_SHUTDOWN)
WIN32_EXIT_CODE : 0 (0x0)
SERVICE_EXIT_CODE : 0 (0x0)
CHECKPOINT : 0x0
WAIT_HINT : 0x0
update:
i can run this service with debug mode, cmd:
memcache_syn_service.py debug

Had the same problem using pypiwin32 (version: 220) and python (version: 3.6). I had to copy :
"\Python36-32\Lib\site-packages\pypiwin32_system32\pywintypes36.dll"
to
"\Python36-32\Lib\site-packages\win32"
for the service to start (was working in debug mode)

If:
python your_service.py debug works, whilst
python your_service.py install + start it as a service fails with error 1053,
this command may help python C:\Python27\Scripts\pywin32_postinstall.py.

all my python coded windows service cannot run on my computer.
but all of them can start at our dev-server which means my code is correct.
but i found a alternative solution, run in debug mode:
any_service.py debug

Make sure you run the application with a different user than the default Local System user. Replace it with the user you successfully be able to run the debug command with.
To replace the user go to the windows services (start > services.msc)
Right click on the service you created > properties > Log On
Uncheck the Local System Account and enter your own.

All of the known fixes have failed me, and this one worked:
In services window:
right-click your installed service;
Go to Log On tab;
Select "This Account" and enter your user ID and pass;
Restart PC.
Has to do with Windows Permissions I was explained...
Method won't work if there's no password set for Windows User.

In my case the problem was from python37.dll not being at C:\Python37-x64\Lib\site-packages\win32.
Just copy it there and it will solve the problem

I had similar problem with a python service and found out that it was missing DLLs since the 'System Path' (not the user path) was not complete. Check the path in your dev-server and whether it matches the one at your computer (System path if service is installed as a LocalSystem service). For me I was missing python dlls' path c:\python27 (windows).

I had this issue and solved it two times in the same way, simply adding the Environment Variables.
I opened Environment Variables, and in system variable PATH added
C:\Users\MyUser\AppData\Local\Programs\Python\PythonXXX
C:\Users\MyUser\AppData\Local\Programs\Python\PythonXXX\Scripts
(Obviously change User name and XXX with Python version)

Launch default image viewer from pygtk program

I'm writing a PyGTK GUI application in Ubuntu to browse some images, and I'd like to open an image in the default image viewer application when it is double-clicked (like when it is opened in Nautilus).
How can I do it?

I don't know specifically using PyGTK but: xdg-open opens the default app for a file so running something like this should work:
import os
os.system('xdg-open ./img.jpg')
EDIT: I'd suggest using the subprocess module as in the comments. I'm not sure exactly how to use it yet so I just used os.system in the example to show xdg-open.

In GNU/Linux use xdg-open, in Mac use open, in Windows use start. Also, use subprocess, if not you risk to block your application when you call the external app.
This is my implementation, hope it helps: http://goo.gl/xebnV
import sys
import subprocess
import webbrowser
def default_open(something_to_open):
"""
Open given file with default user program.
"""
# Check if URL
if something_to_open.startswith('http') or something_to_open.endswith('.html'):
webbrowser.open(something_to_open)
return 0
ret_code = 0
if sys.platform.startswith('linux'):
ret_code = subprocess.call(['xdg-open', something_to_open])
elif sys.platform.startswith('darwin'):
ret_code = subprocess.call(['open', something_to_open])
elif sys.platform.startswith('win'):
ret_code = subprocess.call(['start', something_to_open], shell=True)
return ret_code

GTK (>= 2.14) has gtk_show_uri:
gtk.show_uri(screen, uri, timestamp)
Example usage:
gtk.show_uri(None, "file:///etc/passwd", gtk.gdk.CURRENT_TIME)
Related
How to open a file with the standard application?