Rezin8r

hello,

I just started bash scripting a couple of weeks ago, so i am rather noobish. I have previously written a backup script to do batch backups of various types of firewalls. After discovering 'dialog' i decided to make an interactive mode available for single backups. I have been having some issue with returning dialog menu input to the calling function. Here is the code:

function main_menu {
dialog --clear --backtitle "Firewall Backup and Restore" \
--title "Main menu" \
--menu "Select Action" 10 25 6 \
1 "Backup" \
2 "Restore" 2> .tempfile
choice=`cat .tempfile`
rm -f .tempfile
#echo "$choice"
return $choice
}

for some reason, this will not return anything and just exits unexpectedly. Here is the wierd part, if i uncomment that '#echo "$choice"' line, then it will magically work! WTF?! Is there some kind of buffer or something i need to clear before passing the variable? or am i just being a noob?

EDIT
i forgot to mention that if i dont delete the .tempfile in the script, and manually inspect it after the program quits, i see the expected results of the menu input. ARGH!

mackenga

That is /bizarre/. Sorry, I have no idea. But at least I've incremented my post count.

Whew. I'm still boggling at it, but I can't see a problem. Sorry. The closest thing I have to a suggestion is something to do with the backticks or a line terminator but I really can't see how either of those is at fault - the line with the backticks is obviously working with the echo commented out, and the backticks will not have left a line terminator on your variable, whic shouldn't cause problems even if they did.

It really boggles the mind. All I can suggest is taking the quotes off the echo "$choice" line; it doesn't make sense, but when all else fails, rephrase the code pointlessly.