使用 python 打开其他应用程序

Question

I am using Ubuntu 12.04 on a 64bit laptop. I am trying to open an application using python code.

import os
os.system("open /home/utsav/ab.txt")

It gives the following error:

"Couldn't get a file descriptor referring to the console 256"

What command do I want to use?

Answer 1

The command you're using - open is actually another command referred to in the man pages as openvt , which opens a virtual terminal.

I don't think that's what you want to do, so you would want to use another command (such as gnome-open, xdg-open, geany, gedit, vim, or nano).

Answer 2

I tried this one. it works for me.

import os
os.system("command of your preferred app")

the command for app can be found from Desktop Entry tab of Properties of the preferred app.

Answer 3

open works on OS-X, under ubuntu I end up using gnome-open (I don't know what the corresponding command is if you're using the k-desktop).

EDIT

based on the comment by Niklas B., you could probably also try xdg-open .

Answer 4

For what it's worth, the Python 2.7.3 documentation says that the os module is deprecated and the subprocess module should be used instead. To execute a command this way you can use subprocess.call(args, ...) ( "Using the subprocess module" ).

Based on the earlier answers you can use open for Mac OS X and gnome-open for a Linux distro running the Gnome desktop environment. (I checked gnome-open and xdg-open and both work on Fedora 16.) Windows is a bit tricky.

For Windows you need to use start , but if there are spaces in the path to the file or the filename, then it doesn't work right. Quoting the filename doesn't quite fix it, either, since start expects an unmarked argument with quotes around it to be a title (eg for a new cmd interface window). This is a problem since neither the title nor the file to open is a marked argument, so to get the call to work right you have to do something like start "DummyTitle" "Filename with spaces.ext" .

So what we've got is:

• Mac OS X: subprocess.call(['open','/path/to/file'])

• Linux running Gnome DE: subprocess.call(['gnome-open','/path/to/file'])

• Windows: subprocess.call(['start','"DummyTitle"','"C:\\\\Path\\\\to\\\\File.ext"'])

Or something like that.

---

For versions of Python before 2.7.3, you could us the os module with these args as strings as you suggest in your question (just altering the command portion of the system call). Note as well that there's os.startfile('/path/to/file') for Windows which will open the file with whatever the associated default is for that filetype, if you are going the deprecated os route.

---

Note that we still don't have a confirmed way of opening files on OSes using other desktop environments. Please suggest improvements to this answer!

Answer 5

Welcome to 2022!

You can simply make use of PYPI module AppOpener .

# pip install AppOpener
from AppOpener import run
run("whatsapp") #Opens whatsapp if installed
run("whatsapp, telegram") # Opens whatsapp & telegram

Visit AppOpener's documentation here .