UNIX for Dummies Questions & Answers

Hi you,

This the code I have:

function show_menu {

echo " lalalalal"
echo " 1. ...."
echo " 2. ...."
echo " 3. exit"
read choice

case $choice in
1) ... ;;
2) ...;;
*) stop=1;;
esac

}

##main
while [ ! $stop ]; do
show_menu | tee -a event.log
done

before I added the tee command to generate a log file, this worked perfectly. But with tee, the loop will never end, because stop will never change to 1.
Could you explain me why?
How can I solve this?
Even when I add an 'exit 1' instead of stop=1 it won't work!

Thanks for any help.

Ben Sky

Ben,

I'm not sure but why do you use a tee. If I were you is would redirect the output to the logfile.

...
show_menu >> event.log
...

Greetings,

Chris

plz, forget my previous post. I think I was still sleeping...

Chris

I got a solution for my problem, I put the while loop into the show_menu function.
But it is still not clear for me why my old approach does not work.

The command:
show_menu
works the way you think it works. You invoked a function and the current shell can run it. When stop is changed inside the function, it affects the main shell's copy of "stop".

But:
show_menu | tee -a event.log
is a pipeline. Now a subshell must be spawned to run show_menu. And the function is changing the private version of "stop" in the subshell.

ksh is the only shell that I know that can run the last command of a pipeline in the current shell. This allows stuff like
echo hello | read variable
to work. But I don't know of any shell that can run the rest of a pipeline in the current shell.

See this post for another approach.

OK, it is true, you can find in this forum an answer for each UNIX question. Thanks a lot!