Want to bypass login, go to app menu (Python)


4 posts
by Digital Larry » Sun Dec 02, 2012 3:19 am
Hi all,

I have my Raspberry Pi running a video game play timer for my kids. When that is not active I run a swell screen saver program I wrote that draws pretty patterns using trigonometric functions. That's about all. I don't use the GUI because I just found it too slow.

I'd like this thing to act like a real embedded product in that I want it to simply boot up and bypass the login and automatically start the timer program (or a menu/launcher). My stuff is written in Python.

Ideas?

Thanks,

DL
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by Digital Larry » Tue Dec 18, 2012 5:19 am
Hmm I guess the answer is just too obvious!

I have experimented with adding a script in /etc/init.d and then running update-rc.d, however I need to fix up the script itself to properly handle the start/stop/restart cases.

Strictly speaking, my program isn't a service - it's an interactive application, just so happens that for this particular device I want to dedicate it to that function most of the time and also make it easy for regular people to use.

I will also try putting the call to the program in /etc/rc.local and see if that's what I'm really looking for.
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by ShiftPlusOne » Tue Dec 18, 2012 5:24 am
I'd log in automatically and then automatically launch whatever you need.

http://elinux.org/RPi_Debian_Auto_Login
That link and the .bashrc file should be enough to do what you want.
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by Digital Larry » Tue Dec 25, 2012 12:54 am
That worked great, thanks very much.

Now, is there a way to bypass the /root/.bashrc when I log in via SSH? ;)

DL
Posts: 62
Joined: Tue Jul 24, 2012 9:10 pm
Location: Silicon Valley, CA