使用subprocess.call打开程序

Question

I'm writing an automation program and I'm using subprocess.call to open Google Chrome. The code for this is

from subprocess import call
...
call('google-chrome')

When I run this program through the terminal Chrome launches but it remains tied to the terminal. If I try call('google-chrome &') I get No such file or directory in my traceback. ( $ google-chrome & when entered in terminal will launch Chrome but it won't tie it to the terminal).

I've also tried call(['google-chrome', '&']) but this opens Chrome still tied to the terminal but it thinks the & is an argument for a website I want to open.

Should I be using subprocess.call for this?

Answer 1

Don't use call() ; it'll wait for the process to complete.

Use the subprocess.Popen() class directly, without waiting:

from subprocess import Popen

Popen(['google-chrome'])

You can redirect its output to the bit bin, if desired:

from subprocess import Popen, STDOUT
import os

Popen(['google-chrome'], stdout=os.open(os.devnull, os.O_RDWR), stderr=STDOUT)

If all you want do do is spawn a browser with a given URL, take a look at the webbrowser module ; it uses the same call under the hood to spawn a browser process in the background (including the redirection to /dev/null ). Although Google Chrome is not listed on the documentation page, the 2.7.5 source code of the module does list google-chrome and chrome as recognized browser strings.