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.
Related
I am trying to read event logs from Applications and Services log using python. However the output are not as expected. (Actual 10 vs output 838)
I am using the following code. Was wondering if there is a mistake with the parameters.
import win32evtlog
server = 'localhost'
logtype = "Microsoft-Windows-Storage-Storport/Operational"
hand = win32evtlog.OpenEventLog(server, logtype)
flags = win32evtlog.EVENTLOG_FORWARDS_READ | win32evtlog.EVENTLOG_SEQUENTIAL_READ
while True:
events = win32evtlog.ReadEventLog(hand, flags,0)
if events:
for event in events:
print ('Source Name:', event.SourceName)
print ('Event ID:', event.EventID)
print ('Time Generated:', event.TimeGenerated)
Found a method to get the information through the use of powershell using python.
import subprocess
getinfo = subprocess.check_output(
['powershell.exe', 'get-Winevent Microsoft-Windows-xxx/Operational'])
where xxx is a variable
I want to make hotstrings in python that converts one word when typed into another after some processing, since AHK is very limiting when it comes to determining which word to type. Right now, I am using a hotstring in ahk that runs code on the command line that runs a python script with the word that I typed as arguments. Then I use pyautogui to type the word. However, this is very slow and does not work when typing at speed. I'm looking for a way to do this all with python and without ahk, but I have not found a way to do hotstrings in python. For example, every time I type the word "test" it replaces it with "testing." Thanks for your help. I'm running the latest version of Python and Windows 10 if that is useful to anyone by the way.
(if you want to process it as each letter is typed(t,te,tes,test), you should edit your question)
I call my SymPy functions using ahk hotkeys. I register the python script as a COM server and load it using ahk.
I do not notice any latency.
you'll need pywin32, but don't download using pip install pywin32
download from https://github.com/mhammond/pywin32/releases
OR ELSE IT WON'T WORK for AutoHotkeyU64.exe, it will only work for AutoHotkeyU32.exe.
make sure to download amd64, (I downloaded pywin32-300.win-amd64-py3.8.exe)
here's why: how to register a 64bit python COM server
toUppercase COM server.py
class BasicServer:
# list of all method names exposed to COM
_public_methods_ = ["toUppercase"]
#staticmethod
def toUppercase(string):
return string.upper()
if __name__ == "__main__":
import sys
if len(sys.argv) < 2:
print("Error: need to supply arg (""--register"" or ""--unregister"")")
sys.exit(1)
else:
import win32com.server.register
import win32com.server.exception
# this server's CLSID
# NEVER copy the following ID
# Use "print(pythoncom.CreateGuid())" to make a new one.
myClsid="{C70F3BF7-2947-4F87-B31E-9F5B8B13D24F}"
# this server's (user-friendly) program ID
myProgID="Python.stringUppercaser"
import ctypes
def make_sure_is_admin():
try:
if ctypes.windll.shell32.IsUserAnAdmin():
return
except:
pass
exit("YOU MUST RUN THIS AS ADMIN")
if sys.argv[1] == "--register":
make_sure_is_admin()
import pythoncom
import os.path
realPath = os.path.realpath(__file__)
dirName = os.path.dirname(realPath)
nameOfThisFile = os.path.basename(realPath)
nameNoExt = os.path.splitext(nameOfThisFile)[0]
# stuff will be written here
# HKEY_LOCAL_MACHINE\SOFTWARE\Classes\CLSID\${myClsid}
# HKEY_LOCAL_MACHINE\SOFTWARE\Classes\CLSID\{C70F3BF7-2947-4F87-B31E-9F5B8B13D24F}
# and here
# HKEY_LOCAL_MACHINE\SOFTWARE\Classes\${myProgID}
# HKEY_LOCAL_MACHINE\SOFTWARE\Classes\Python.stringUppercaser
win32com.server.register.RegisterServer(
clsid=myClsid,
# I guess this is {fileNameNoExt}.{className}
pythonInstString=nameNoExt + ".BasicServer", #toUppercase COM server.BasicServer
progID=myProgID,
# optional description
desc="return uppercased string",
#we only want the registry key LocalServer32
#we DO NOT WANT InProcServer32: pythoncom39.dll, NO NO NO
clsctx=pythoncom.CLSCTX_LOCAL_SERVER,
#this is needed if this file isn't in PYTHONPATH: it tells regedit which directory this file is located
#this will write HKEY_LOCAL_MACHINE\SOFTWARE\Classes\CLSID\{C70F3BF7-2947-4F87-B31E-9F5B8B13D24F}\PythonCOMPath : dirName
addnPath=dirName,
)
print("Registered COM server.")
# don't use UseCommandLine(), as it will write InProcServer32: pythoncom39.dll
# win32com.server.register.UseCommandLine(BasicServer)
elif sys.argv[1] == "--unregister":
make_sure_is_admin()
print("Starting to unregister...")
win32com.server.register.UnregisterServer(myClsid, myProgID)
print("Unregistered COM server.")
else:
print("Error: arg not recognized")
you first need to register the python COM server:
first, get your own CLSID: just use a python shell.
import pythoncom
print(pythoncom.CreateGuid())
then, set myClsid to that output
to register:
python "toUppercase COM server.py" --register
to unregister:
python "toUppercase COM server.py" --unregister
hotstring python toUppercase.ahk
#NoEnv ; Recommended for performance and compatibility with future AutoHotkey releases.
#SingleInstance, force
SendMode Input ; Recommended for new scripts due to its superior speed and reliability.
SetWorkingDir %A_ScriptDir% ; Ensures a consistent starting directory.
SetBatchLines, -1
#KeyHistory 0
ListLines Off
#Persistent
#MaxThreadsPerHotkey 4
pythonComServer:=ComObjCreate("Python.stringUppercaser")
; OR
; pythonComServer:=ComObjCreate("{C70F3BF7-2947-4F87-B31E-9F5B8B13D24F}") ;use your own CLSID
; * do not wait for string to end
; C case sensitive
:*:hello world::
savedHotstring:=A_ThisHotkey
;theActualHotstring=savedHotstring[second colon:end of string]
theActualHotstring:=SubStr(savedHotstring, InStr(savedHotstring, ":",, 2) + 1)
send, % pythonComServer.toUppercase(theActualHotstring)
return
f3::Exitapp
you can test the speed of hotstring hello world, it's very fast for me.
Edit def toUppercase(string): to your liking
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.
I'm studying vCenter 6.5 and community samples help a lot, but in this particular situation I can't figure out, what's going on. The script:
from __future__ import with_statement
import atexit
from tools import cli
from pyVim import connect
from pyVmomi import vim, vmodl
def get_args():
*Boring args parsing works*
return args
def main():
args = get_args()
try:
service_instance = connect.SmartConnectNoSSL(host=args.host,
user=args.user,
pwd=args.password,
port=int(args.port))
atexit.register(connect.Disconnect, service_instance)
content = service_instance.RetrieveContent()
vm = content.searchIndex.FindByUuid(None, args.vm_uuid, True)
creds = vim.vm.guest.NamePasswordAuthentication(
username=args.vm_user, password=args.vm_pwd
)
try:
pm = content.guestOperationsManager.processManager
ps = vim.vm.guest.ProcessManager.ProgramSpec(
programPath=args.path_to_program,
arguments=args.program_arguments
)
res = pm.StartProgramInGuest(vm, creds, ps)
if res > 0:
print "Program executed, PID is %d" % res
except IOError, e:
print e
except vmodl.MethodFault as error:
print "Caught vmodl fault : " + error.msg
return -1
return 0
# Start program
if __name__ == "__main__":
main()
When I execute it in console, it successfully connects to the target virtual machine and prints
Program executed, PID is 2036
In task manager I see process with mentioned PID, it was created by the correct user, but there is no GUI of the process (calc.exe). RMB click does not allow to "Expand" the process.
I suppose, that this process was created with special parameters, maybe in different session.
In addition, I tried to run batch file to check if it actually executes, but the answer is no, batch file does not execute.
Any help, advices, clues would be awesome.
P.S. I tried other scripts and successfully transferred a file to the VM.
P.P.S. Sorry for my English.
Update: All such processes start in session 0.
Have you tried interactiveSession ?
https://github.com/vmware/pyvmomi/blob/master/docs/vim/vm/guest/GuestAuthentication.rst
This boolean argument passed to NamePasswordAuthentication and means the following:
This is set to true if the client wants an interactive session in the guest.
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)