How do I open an application using Python [closed] - python

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How do I open an app like Minecraft using os or subprocess? I went through a few videos and none of them worked. Do I just use the name of the app or its location?

For this example, you could use:
subprocess.run(r'"C:\path_to_minecraft\minecraft"',shell=True)
Or if you have Windows shortcuts you add the .lnk extension in the call:
subprocess.run(r'"C:\path_to_minecraft\minecraft.lnk"',shell=True)

this works.
import os
os.system("your app.exe")

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how can I observe a change in a python var [closed]

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Hi I'm trying to catch the change in the paperclip of my computer with the library pyperclip something same as
url=pyperclip.paste()
I want to create a thread which can append all the urls into a list my script, but I don't know how to could I do it
Thanks for your time and help c:
Did you check the documentation or the source code at all? That's all I did. pyperclip supports waitForPaste and waitForNewPaste methods that can do this.

Is it possible to create a url link shortcut using Python, under Ubuntu? [closed]

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I'm trying to generate some shortcuts to specific websites using Python. Is it possible for me to do so under Ubuntu? Say If I'd like to create a shortcut to Stack Overflow in Home folder. How can I achieve that using Python?
You could do this :
with open('StackOverflow.desktop', 'w+') as f:
f.write("""[Desktop Entry]
Encoding=UTF-8
Name=Link to Stack Overflow
Type=Link
URL=http://stackoverflow.com/
Icon=text-html
Name[en_US]=StackOverflow
""")
It will create a shortcut on Ubuntu to StackOverflow

Can you use python to load a quick ruby script? [closed]

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I was curious if you could use python to load a quick ruby script? then continue with python.
The simplest way is to use call from subprocess module:
from subprocess import call
call(['ruby', 'quick_script.rb', 'some_parameter'])
Ruby should be available from the path variable.

wxPython. How to open a folder with wxWidget (python) hyperlink? [closed]

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I want to create a hyperlink in my wx widget which opens linked folder on clicking. Please help me how to build that.
Thanks.
os.system("start C:")
will launch a file window (on windows at least) ... it has nothing to do with wx
unless you are talking about opening a FilePickerDialog?
I have also been informed that
os.system("explorer C:")
would also work

How to run a file from the body of a program? [closed]

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What is the python command to run a py-file.?
For example, in my index.py file how would i launch start.py.
In index.py, simply use subprocess:
import subprocess
subprocess.check_output(["python", "start.py"])

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