BASH Need help figuring out how to loop SELECT statement

mcdrr · 05-01-2007, 08:56 PM

Hello everyone,

I was hoping that someone could shed some light/point me in the right direction with a script that I am working on. For starters I am new to scripting so the code may not the best way to accomplish what I am trying to do but like I mentioned I am in the process of learning how to script.

here is a copy of the script. What I am trying to do is if there is a way to loop/reprint the SELECT menu after an option is selected for example if I select option 3 I get the results which works fine, but then I want the script to automatically reprint the option list over and over until the QUIT option is selected. For now the way that I am doing it is by having it echo in the script manually.

If I do "elif [ $answer = "Menu" ]
then
echo $CHOICE

I get the choices to print out but in this format
System Secure Cron Menu Quit

not like this.
1) System
2) Secure
3) Cron
4) Menu
5) Quit
Please select one of the options above for Menu press 4:

Is there anyway to accomplish this? Thanks for all the help in advance. Please let me know if you guys have any questions.

COPY OF SCRIPT
-----------------------------------------------------------------------------------
#!/bin/bash
# Date:
# Author:
# Purpose:

PS3='Please select one of the options above for Menu press 4:'
CHOICE="System Secure Cron Menu Quit"
select answer in $CHOICE
do
if [ $answer = "System" ]
then
tail /var/log/messages
elif [ $answer = "Secure" ]
then
tail /var/log/secure
elif [ $answer = "Cron" ]
then
tail /var/log/cron
elif [ $answer = "Menu" ]
then
echo "1) System"
echo "2) Secure"
echo "3) Cront"
echo "4) Menu"
echo "5) Quit"
elif [ $answer = "Quit" ]
then
exit
fi
done
----------------------------------------------------------------------------------

Oupa · 05-07-2007, 06:50 AM

Include the following lines with your script:

loop=0
while [ $loop = 0 ]
do
YOUR SCRIPT CODING COMES HERE
.......................
...................
done
exit 0
----------------------------

Modify your quit option to break the loop:
elif [ $answer = "Quit" ]
then
loop=1
break
fi
-----------------------------
More readable would be to use the case statement.
The coding would be something as follows:

#!/bin/bash
showmenu()
{
clear
cat << !menu.txt

*** Menu of Options ***

1 System
2 Secure
. ...
5 Quit

!menu.txt
}
loop=1
while [ $loop -eq 1 ]
do
showmenu
printf "\n Enter choice: "
read opt
case $opt in
1) tail /var/log/messages
;;
2) tail /var/log/secure
;;
.) ...
;;
5) loop=0
break
;;
*) printf " Invalid Choice\n"
;;
esac
printf "\n Press Enter to return to Menu"
read junk
done
exit 0
----------------------------
Brief explanation:
The showmenu funtion is read but not executed.
The while loop executes showmenu
showmenu uses the cat command to display the menu section.
Control returns to the loop at the:
printf "\n Enter choice: "
The choice is read and depending on what number was choosen the 'case' statement executes the relevant coding. The case statement ends with esac.
The While loop continues to execute until choice 5 is made.
---------------------------------------------

bigearsbilly · 05-08-2007, 04:47 AM

yes, use the case statement in the select, something like:

Code:

case $choice in

       this) do_this
                
                ;;
       that) do_that
                ;;

        *wildcard* ) etc
                  ;;
        *)  # default
            do_thing
            ;;
esac

you can use break to get out of loops, and even qualify the nesting
depth.

I personally just use CTRL-C to get out of select cos I am lazy.