open a terminal from python

Question

I'm developing a program that has a button. When pressed, I want to open a terminal that runs:

sudo apt-get update

I'm using:

os.system("gnome-terminal -e 'sudo apt-get update'")

This works fine. The only problem is that when the update is finished, the terminal closes. What can I do to leave the terminal open?

• rodion: See this post: stackoverflow.com/questions/3512055/…

Answer 1

You could do this:

os.system("gnome-terminal -e 'bash -c \"sudo apt-get update; exec bash\"'")

• mika29: And probably is the same with konsole (if I use KDE)?
• mastash3ff: os.system("gnome-terminal -e 'sudo apt-get update'"). Not sure why you need the extra 'bash' when this works just fine.
• jterrace: @mastash3ff - read the question. "when the update is finished, the terminal closes". The extra bash call makes it stay open.
• Andres Mejia: is it possible to do this with your termial of choice? For example, I am using alacritty, but making the substitution "gnome-terminal"="alacritty" does not work.
• Tanveer: that' worked for me. but how to close that terminal?

Answer 2

There are a few choices:

• add ; read -p "Hit ENTER to exit" to the end of the command line.
• add ; sleep 10 to the end of the command line to wait a bit, then exit.

• Configure gnome terminal:

  Go to the "Edit" menu and click "Current Profile". Click on the "Title and Command" tab. In there, there is a setting called "When command exits". Change it to "hold the terminal open". You could also create a new profile.

Answer 3

You can remove the -e:

os.system("gnome-terminal 'sudo apt-get update'")

• mastash3ff: Could not get this to work. Had to use the -e option as suggested above.

Answer 4

import os

os.system("Your command") You can also pass custom command as custom variable For example:

cmd_to_run = "ls -lat"

os.system(cmd_to_run)