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How can I launch an Android app on a device through Python?

I have consulted several topics on the subject, but I didn't see any related to launching an app on a device directly using a ppadb command.
I managed to do this code:
import ppadb
import subprocess
from ppadb.client import Client as AdbClient
# Create the connect functiun
def connect():
client = AdbClient(host='localhost', port=5037)
devices = client.devices()
for device in devices:
print (device.serial)
if len(devices) == 0:
print('no device connected')
quit()
phone = devices[0]
print (f'connected to {phone.serial}')
return phone, client
if __name__ == '__main__':
phone, client = connect()
import time
time.sleep(5)
# How to print each app on the emulator
list = phone.list_packages()
for truc in list:
print(truc)
# Launch the desired app through phone.shell using the package name
phone.shell(????????????????)
From there, I have access to each app package (com.package.name). I would like to launch it through a phone.shell() command but I can't access the correct syntax.
I can execute a tap or a keyevent and it's perfectly working, but I want to be sure my code won't be disturbed by any change in position.

From How to start an application using Android ADB tools, the shell command to launch an app is
am start -n com.package.name/com.package.name.ActivityName
Hence you would call
phone.shell("am start -n com.package.name/com.package.name.ActivityName")
A given package may have multiple activities. To find out what they are, you can use dumpsys package as follows:
def parse_activities(package, connection, retval):
out = ""
while True:
data = connection.read(1024)
if not data: break
out += data.decode('utf-8')
retval.clear()
retval += [l.split()[-1] for l in out.splitlines() if package in l and "Activity" in l]
connection.close()
activities = []
phone.shell("dumpsys package", handler=lambda c: parse_activities("com.package.name", c, activities))
print(activities)

Here is the correct and easiest answer:
phone.shell('monkey -p com.package.name 1')
This method will launch the app without needing to have acces to the ActivityName

Using AndroidViewClient/cluebra, you can launch the MAIN Activity of a package as follows:
#! /usr/bin/env python3
# -*- coding: utf-8 -*-
from com.dtmilano.android.viewclient import ViewClient
ViewClient.connectToDeviceOrExit()[0].startActivity(package='com.example.package')
This connects to the device (waiting if necessary) and then invokes startActivity() just using the package name.
startActivity() can also receive a component which is used when you know the package and the activity.

how to reboot remote desktop using python script?

I have a python script that reboot the local machine using the win32api package. but when i try to reboot a remote desktop the system crash and display the below error :
win32api.InitiateSystemShutdown(user, message, timeout, bForce, bReboot)
pywintypes.error: (5, 'InitiateSystemShutdown', 'Access is denied.')
what is the problem here and why i can't get the required privilege's yet we are on the same network?
code:
import win32security
import win32api
import sys
import time
from ntsecuritycon import *
def AdjustPrivilege(priv, enable=1):
# Get the process token
flags = TOKEN_ADJUST_PRIVILEGES | TOKEN_QUERY
htoken = win32security.OpenProcessToken(win32api.GetCurrentProcess(), flags)
# Get the ID for the system shutdown privilege.
idd = win32security.LookupPrivilegeValue(None, priv)
# Now obtain the privilege for this process.
# Create a list of the privileges to be added.
if enable:
newPrivileges = [(idd, SE_PRIVILEGE_ENABLED)]
else:
newPrivileges = [(idd, 0)]
# and make the adjustment
win32security.AdjustTokenPrivileges(htoken, 0, newPrivileges)
def RebootServer(user=None,message='Rebooting', timeout=30, bForce=0, bReboot=1):
AdjustPrivilege(SE_SHUTDOWN_NAME)
try:
win32api.InitiateSystemShutdown(user, message, timeout, bForce, bReboot)
finally:
# Now we remove the privilege we just added.
AdjustPrivilege(SE_SHUTDOWN_NAME, 0)
def AbortReboot():
AdjustPrivilege(SE_SHUTDOWN_NAME)
try:
win32api.AbortSystemShutdown(None)
finally:
AdjustPrivilege(SE_SHUTDOWN_NAME, 0)
#test the functions:
# Reboot computer
RebootServer("XXX.XXX.XX.XX")
# Wait 10 seconds
time.sleep(10)
print ('Aborting shutdown')
# Abort shutdown before its execution
AbortReboot()

Have you tried turning it off and on again? (Sorry)
You could try setting the owner of the script to an Administrator by right-clicking the file and selecting Preferences. You can also try running the python executable as an administrator.

Python-Application .desktop-shortcut causing malfunction

I wanted to create a desktop launcher for my Python application. The application executes various ssh operations over pexpect with publickey-authentication. The problem is however, when I start my app with the .desktop launcher it doesn't work properly. The ssh connections ask for a password and don't use the publickeys. But it works fine via commandline execution.
The .desktop File looks like this:
[Desktop Entry]
Version=1.0
Name=SSH-Manager
Comment=XYZ
Exec=python /home/userx/SSH-Manager/startup.py
Icon=/home/userx/SSH-Manager/resources/icon.png
Path=/home/userx/repos/SSH-Manager
Terminal=true
Type=Application
Categories=Utility;Application;
StartupNotify=false
The desktop environment is KDE and the desktop user is the same as the commandline user.
Can someone explain why I get such strange behavior with the launcher?
Edit: Example function
def run(self):
self.a_signal.emit("Retrieving Data")
try:
session = pxssh()
session.force_password = False
hostname = self.client
username = "root"
session.login(hostname, username)
session.sendline("ls -a")
session.prompt()
session.logout()
except ExceptionPxssh as e:
print ("pxssh failed: ")
self.error_signal.emit("failed", str(e))
print e
return
self.process_output()
self.finish_signal.emit("done")

As Mirosław Zalewski suspected in the comments, the problem was the ssh-agent was not running for the desktop-environment because ssh-add was initially used in the /etc/sources. Executing ssh-add in the X-users ~./profile therefore solves the problem.

Python 3 Windows Service starts only in debug mode

I first posted an answer in this post, but it didn't conform to the forum standards. I hope this time te answer fits the forum standards. This code should be more clear and easy to read.
In Python 3+ I have the following class that I use to build a Windows Service (it does nothing, just writes a log file):
#MyWindowsService.py
import win32serviceutil
import servicemanager
import win32service
import win32event
import sys
import logging
import win32api
class MyWindowsService(win32serviceutil.ServiceFramework):
_svc_name_ = 'ServiceName'
_svc_display_name_ = 'Service Display Name'
_svc_description_ = 'Service Full Description'
logging.basicConfig(
filename = 'c:\\Temp\\{}.log'.format(_svc_name_),
level = logging.DEBUG,
format = '%(levelname)-7.7s # %(asctime)s: %(message)s'
)
def __init__(self, *args):
self.log('Initializing service {}'.format(self._svc_name_))
win32serviceutil.ServiceFramework.__init__(self, *args)
self.stop_event = win32event.CreateEvent(None, 0, 0, None)
def SvcDoRun(self):
self.ReportServiceStatus(win32service.SERVICE_START_PENDING)
try:
self.log('START: Service start')
self.ReportServiceStatus(win32service.SERVICE_RUNNING)
self.start()
win32event.WaitForSingleObject(self.stop_event, win32event.INFINITE)
except Exception as e:
self.log('Exception: {}'.format(e))
self.SvcStop()
def SvcStop(self):
self.log('STOP: Service stopping...')
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
self.stop()
win32event.SetEvent(self.stop_event)
self.ReportServiceStatus(win32service.SERVICE_STOPPED)
def log(self, msg):
servicemanager.LogInfoMsg(str(msg)) #system log
logging.info(str(msg)) #text log
def start(self):
self.runflag = True
while self.runflag:
win32api.Sleep((2*1000), True)
self.log('Service alive')
def stop(self):
self.runflag = False
self.log('Stop received')
if __name__ == '__main__':
win32serviceutil.HandleCommandLine(MyWindowsService)
In the script I use a log file to check if it's working properly. I'm running python3.6 (also tried with python3.4) on Windows 7 and I'm experiencing the following problem. When I run python MyWindowsService.py install the prompt says that the service has been installed (but nothing is written in the log file). If I try to start the service, I get Service Error: 1 - More info NET HELPMSG 3547 which doesn't say much about the error. If I run python MyWindowsService.py debug, the program runs just fine (the log file is written), but still I don't have any control over the service: if I open another prompt and try to stop/start the service I still got the same results as stated above.
I also tryed to insert some debug code inside the init function, and when I run python MyWindowsService.py install it seems it doesn't get called. Is it possible?
I've checked for multiple solution and workarounds around the net, but I didn't find anything suitable. What am I missing?

As pointed out by eriksun in the comment to the first post, the problem came from the location of the python script, that was in a drive mapped with an UNC path - I'm working with a virtual machine. Moving the python script in the same drive as the python installation did the job.
To sum it up for future uses, if the service fails to start and you're pretty sure about your code, these are helpful actions to try and solve your issues:
use sc start ServiceName, sc query ServiceName and sc stop ServiceName to get info about the service.
check if your file is in a physical drive or in a UNC-mapped drive. If the latter try to run the script using the UNC path (for example python \\Server\share\python\your-folder\script.py) or move your script in the same drive as the python installation
make sure that "python36.dll", "vcruntime140.dll", and "pywintypes36.dll" are either symlink'd to the directory that has PythonService.exe; or symlink'd to the System32 directory; or that the directories with these DLLs are in the system (not user) Path
Check the system register with command reg query HKLM\System\CurrentControlSet\Services\your_service_name /s to get more information about the script
PLease, feel free to complete, change, modify the last so that it can be usefull for anyone that like me encounder this issue.
EDIT: One more thing... My project was thought to work actually with network folders (and UNC-mapped drive) and it failed when I tried to make it run as service. One very useful (day-saving) resource that I used to make it work is the SysinternalsSuite by Mark Russinovich that I found in this post. Hope this helps.

Executable out of script containing serial_for_url

I have developed a python script for making a serial communication to a digital pump. I now need to make an executable out of it. However even though it works perfectly well when running it with python and py2exe does produce the .exe properly when I run the executable the following error occurs:
File: pump_model.pyc in line 96 in connect_new
File: serial\__init__.pyc in line 71 in serial_for_url
ValueError: invalid URL protocol 'loop' not known
The relevant piece of my code is the following:
# New serial connection
def connect_new(self, port_name):
"""Function for configuring a new serial connection."""
try:
self.ser = serial.Serial(port = port_name,\
baudrate = 9600,\
parity = 'N',\
stopbits = 1,\
bytesize = 8,\
timeout = self.timeout_time)
except serial.SerialException:
self.ser = serial.serial_for_url('loop://',\
timeout = self.timeout_time) # This line BLOWS!
except:
print sys.exc_info()[0]
finally:
self.initialize_pump()
I should note that the application was written in OSX and was tested on Windows with the Canopy Python Distribution.

I had the exact same problem with "socket://" rather than "loop://"
I wasn't able to get the accepted answer to work however the following seems to succeed:
1) Add an explicit import of the offending urlhandler.* module
import serial
# explicit import for py2exe - to fix "socket://" url issue
import serial.urlhandler.protocol_socket
# explicit import for py2exe - to fix "loop://" url issue (OP's particular prob)
import serial.urlhandler.protocol_loop
# use serial_for_url in normal manner
self._serial = serial.serial_for_url('socket://192.168.1.99:12000')
2) Generate a setup script for py2exe (see https://pypi.python.org/pypi/py2exe/) -- I've installed py2exe to a virtualenv:
path\to\env\Scripts\python.exe -m py2exe myscript.py -W mysetup.py
3) edit mysetup.py to include option
zipfile="library.zip" # default generated value is None
(see also http://www.py2exe.org/index.cgi/ListOfOptions)
3) build it:
path\to\env\Scripts\python.exe mysetup.py py2exe
4) run it
dist\myscript.exe

Found it!
It seems that for some reason the 'loop://' arguement can't be recognised after the .exe production.
I figured out by studying the pyserial/init.py script that when issuing the command serial.serial_for_url(‘loop://') you essentially call:
sys.modules['serial.urlhandler.protocol_loop’].Serial(“loop://“)
So you have to first import the serial.urlhandler.protocol_loop
and then issue that command in place of the one malfunctioning.
So you can now type:
__import__('serial.urlhandler.protocol_loop')
sys.modules[‘serial.urlhandler.protocol_loop’].Serial("loop://")
After this minor workaround it worked fine.