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Full Discussion: Shell script menu problem
Top Forums Shell Programming and Scripting Shell script menu problem Post 302598868 by goddevil on Wednesday 15th of February 2012 03:19:17 PM
Old 02-15-2012
Quote:
Originally Posted by Corona688
I think we crossposted. Have a look at my general-purpose bash menu system there Smilie
Thanks a mil...That looks very interesting.. I will try it as soon as i have access to a terminal.

The only thing is, the number is options in my menus are usually 10 or more and the option is uaually something like "Show logs that belong to the XX service" and such. I have never used select before as well so i am a bit apprehensive. I am excited to try this out at the same time

---------- Post updated at 08:19 PM ---------- Previous update was at 08:10 PM ----------

Quote:
Originally Posted by in2nix4life
Here's a long-winded way that may help point you in the right direction by putting the menus in functions:

Code:
#!/bin/bash
 
 
function main_menu()
{
    while :
    do
        echo
        echo "MAIN MENU"
        echo
        echo "1. Main Menu Option 1"
        echo "2. Main Menu Option 2"
        echo "3. Main Menu Option 3 To Sub-Menu"
        echo "4. Main Menu Option 4 To Quit"
        echo
        read -p "Choice: " m_choice
        echo
 
        case "$m_choice" in 
            1)  echo "Main Menu Option 1"
                echo
                ;;
            2)  echo "Main Menu Option 2"
                echo
                ;;
            3)  sub_menu
                ;;
            4)  exit 0
                ;;
            *)  echo "Bad Option"
                echo
                ;;
        esac
    done
}
 
function sub_menu()
{
    while :
    do
        echo
        echo "SUB MENU"
        echo
        echo "1. Sub-Menu Option 1"
        echo "2. Sub-Menu Option 2"
        echo "3. Sub-Menu Option 3 Back To Main Menu"
        read -p "Choice: " s_choice
        echo
 
        case "$s_choice" in
            1)  echo "Sub-Menu Option 1"
                echo
                ;;
            2)  echo "Sub-Menu Option 2"
                echo
                ;;
            3)  main_menu
                ;;
            *)  echo "Bad Option"
                echo
                ;;
        esac
    done
}
 
main_menu
 
exit 0

Hope this helps.

That is exactly how i have it and it will work fine as long as you do not have another submenu "sub_menu_2" inside inside the function "sub_menu"... In this case, when you call the function "sub_menu_2" and then return back to "sub_menu", any choice you make will take you to "sub_menu_2"... This is because, "sub_menu_2" calls "sub_menu" and when you choose an option , it simply returns back after the call and still in the while loop in "sub_menu_2"
 

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