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How do I check if Python is installed using a python script?

I am trying to create a python setup script that will be converted to an executable file in windows using pyinstaller. Currently, I am trying to figure out how this script would check if python is installed on the system, if not, then it will go ahead & download the installation exe file from the python website.
My current code is as follows:
import subprocess, requests, platform
architecture = platform.uname()[4]
if architecture == 'AMD64':
executable_file = "https://www.python.org/ftp/python/3.11.1/python-3.11.1-amd64.exe"
elif architecture == 'ARM64':
executable_file = "https://www.python.org/ftp/python/3.11.1/python-3.11.1-arm64.exe"
r = requests.get(executable_file)
with open("python-3.11.1-amd64.exe",'wb') as f:
f.write(r.content)
cmd = "python-3.11.1-amd64.exe"
returned_value = subprocess.call(cmd, shell=True)
if returned_value == 1602:
print("Installation Cancelled. Exitting.")
quit()
print('returned value:', returned_value)
I currently have no idea how to check if python is installed so the exe will not be downloaded & ran by the script. So I would like some ideas on how I could go with this.

Directly execute the python command to see the output, or use the winreg module to obtain the installed applications of the local machine

Open file in IDLE through python script

How would I open a specific file in IDLE through a python script?
I understand that an app could be opened through subprocess:
import subprocess
subprocess.call('C:\\program.exe')
But I can't figure out how to make it open a file.
If it helps, this:
import os.path
import sys
# Enable running IDLE with idlelib in a non-standard location.
# This was once used to run development versions of IDLE.
# Because PEP 434 declared idle.py a public interface,
# removal should require deprecation.
idlelib_dir = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))
if idlelib_dir not in sys.path:
sys.path.insert(0, idlelib_dir)
from idlelib.pyshell import main
main()
also opens IDLE. I checked, and main() does not take any parameters such as files to open.
I am using Windows 10 with Python 3.6.4.
Any help is greatly appreciated.

Here are 2 ways to open any python file through IDLE
import subprocess
p = subprocess.Popen(["idle.exe", path_to_file])
# ... do other things while idle is running
returncode = p.wait() # wait for notepad to exit
OR:
import subprocess
import os
subprocess.call([path_to_idle, path_to_file])
You can also use these methods to open any file with any installed app (How can I open files in external programs in Python?)

One can run IDLE from a command line on any platform with <python> -m idlelib <IDLE args>, where <python> could be 'python', 'python3', or something line 'py -3.8', depending on the platform. <IDLE args> are defined in the "Command line usage" subsection of the IDLE doc, also available within IDLE as Help => IDLE Help.
A possible 'IDLE arg' is the path of a file to be opened in an editor window. Relative paths are relative to the current working directory, which can be changed with the 'cd' command. A working command line can used quoted or turned into a list for a subprocess.run or subprocess.Popen call. In Python code, the working directory is changed with os.chdir('newdir').

How do I open exe file with python that is dependent on AVI files?

I'm trying to open Adv5KTCP.exe (path shown in code) but this exe actually opens three more avi.files as shown in image below. This causes the error. I've tried os, subprocess & pywinauto to call but to no avail. I've also tried adding the files path to environmental variable but I think that makes no sense. Opening the exe from command prompt does not work too. However, the exe can be opened manually by double clicking the exe file (like the usual). I just need it to be automated.
AVI Files:
The Error:
Double clicking the exe would open this window:
Here is my code:
import os
import sys, logging
import subprocess
import ctypes
from pywinauto import Application
def is_admin():
try:
return ctypes.windll.shell32.IsUserAnAdmin()
except:
return False
if is_admin():
exe = "Adv5KTCP.exe" #path set in environment variable: "C:\Program Files (x86)\Advantech\ADAM-5000TCP-6000 Utility\Program"
os.startfile(exe)
# subprocess.Popen([exe])
# application = Application(backend="uia").start(exe)
else:
ctypes.windll.shell32.ShellExecuteW(None, "runas", sys.executable, __file__, None, 1)
Does anybody have any idea on how I could go about this? Thank you in advance for your response.
New Findings:
I tried to cut the three AVI files to desktop & set desktop path to system variable & the exe gives the same error. However when i cut the exe file to desktop with the avi files as shown below, it works! Even when the other required files are not in desktop but path's set to system variable. Which means the exe is somehow registering the paths of the avi upon click, location/position or some sort which I'm not sure of.

interchangeable shebang line in Python script for dual OSes?

A script is developed both on OS X and Windows using a virtualenv. The so-called developer has already installed all required packages using a requirements.txt file, but one problem remains:
If the script is running on OS X, the beginning of each Python file must start like this:
#!/Users/os-x-username/.virtualenvs/some_env/bin/python
#!C:\Users\windows-username\Envs\some_env\Scripts\python.exe
But if developing on Windows, the line order must be switched:
#!C:\Users\windows-username\Envs\some_env\Scripts\python.exe
#!/Users/os-x-username/.virtualenvs/some_env/bin/python
How can the so-called developer avoid this tediousness?

If you don't mind adding extra steps, ou can create a launcher script launcher.py like:
#!/usr/bin/env python
import subprocess
import sys
if __name__ != "__main__":
print("This is a launcher. Please run only as a main script.")
exit(-1)
INTERPRETERS = {
"win": r"C:\Users\windows-username\Envs\some_env\Scripts\python.exe", # Windows
"darwin": "/Users/os-x-username/.virtualenvs/some_env/bin/python", # OSX
"linux": "/home/linux-user/.virtualenvs/some_env/bin/python" # Linux
# etc.
}
TARGET_SCRIPT = "original_script_name.py"
interpreter = None
for i in INTERPRETERS: # let's find a suitable interpreter for the current platform
if sys.platform.startswith(i):
interpreter = i
break
if not interpreter:
print("No suitable interpreter found for platform:", sys.platform)
exit(-1)
main_proc = subprocess.Popen([interpreter, TARGET_SCRIPT] + sys.argv[1:]) # call virtualenv
main_proc.communicate() # wait for it to finish
exit(main_proc.poll()) # redirect the return code
Since this script is there only to run the original_script_name.py in the desired interpreter for the current platform, it doesn't matter what its shebang is - as long as it picks any Python interpreter it will be fine.
It would act as a drop-in replacement for your original script (original_script_name.py) so just call launcher.py instead and it will even redirect the CLI arguments if needed.

How to launch a Window's shortcut using Python

I want to launch a shortcut named blender.ink located at "D://games//blender.ink". I have tryed using:-
os.startfile ("D://games//blender.ink")
But it failed, it only launches exe files.

The Python os.startfile function should work fine, but you need to specify a .lnk extension to be a valid Windows shortcut file:
import os
os.startfile (r"D:\games\blender.lnk")
If you need to wait for the application to complete before continuing, then a different approach would be needed as follows:
import win32com.shell.shell as shell
import win32event
se_ret = shell.ShellExecuteEx(fMask=0x140, lpFile=r"D:\games\blender.lnk", nShow=1)
win32event.WaitForSingleObject(se_ret['hProcess'], -1)