Checking whether a script can open a display?

[Walter Dnes]

On Tue, Mar 01, 2005 at 05:03:04PM -0500, Lennart Sorensen wrote

> if [ -z "$DISPLAY" ]; then
> 	echo "No display";
> else
> 	echo "Do something";
> fi

  That may not work.  $DISPLAY is an environmental variable.  On my
system it gets set in .bashrc and should remain set even if X doesn't
start.  Here's what I would suggest.  Execute...

ps -e

when X is running.  Here are some things that will always show up on
*MY* system when X is running...

 8739 tty10    00:00:00 startx
 8750 tty10    00:00:00 xinit
 8751 ?        00:17:07 X
 8769 tty10    00:00:22 blackbox
 8770 tty10    00:01:26 bbkeys
 8771 tty10    00:00:00 xterm
 8773 tty10    00:05:08 fbpanel

  So I would do something like...

#!/bin/bash
ps_output=`ps -e | grep " xinit"`
if [ ${#ps_output} -gt 2 ]; then
  echo "X is running.  On with the show."
else
  echo "X is not running.  Now what?"
fi

  Replace the "echo" commands with real stuff.  Your system will likely
have a different set of "guaranteed" programs that will be running
whenever X runs.  So be prepared to adjust as necessary.

-- 
Walter Dnes <waltdnes-SLHPyeZ9y/tg9hUCZPvPmw at public.gmane.org>
An infinite number of monkeys pounding away on keyboards will
eventually produce a report showing that Windows is more secure,
and has a lower TCO, than linux.
--
The Toronto Linux Users Group.      Meetings: http://tlug.ss.org
TLUG requests: Linux topics, No HTML, wrap text below 80 columns
How to UNSUBSCRIBE: http://tlug.ss.org/subscribe.shtml