multiple looping with case and funtion showing error, Please help


 
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# 1  
Old 09-19-2011
MySQL multiple looping with case and funtion showing error, Please help

Hello All,

I have code as follows :-
Code:
while true do
  [calling a function]
   {opening a case1 statement} 
    1) 
      [Calling another function]
        {opening another case2 statement} 
        {closing case 2}
     2)
  Showing error for "2)" as Syntax error at line 59 : `)' is not expected.
     *) 
    {closing case 1} 
 done

This is the structure of my script. I am not able to figure out it. Could anyone crack this ?

Thanks
Renjesh
# 2  
Old 09-19-2011
If you expect people help you to troubbleshoot your script, then people need you to provide the code (and not only the structure of it).
# 3  
Old 09-19-2011
Here is the code :-
Code:
choice=""
tgtser=""
function menu
{
    clear
    echo "-------------------------------------------------------"
    echo " * * * * * * * Subversion Main Menu * * * * * * * * * * "
    echo "-------------------------------------------------------"
    echo "[1] check-out to test server"
    echo "[2] check-in"
    echo "[3] check-out to local machine"
    echo "[4] Exit/stop"
    echo "----------------------------------------------"
    echo "Enter your menu choice [1-4]: \n"
    read choice
}
function chkout1
{
     clear
     echo "Please select a target server .."
     echo "[1] AIX - st95esra"
     echo "[2] AIX - st95esrb"
     echo "[3] AIX - st95esre"
     echo "[4] HP-UX - st95esr0"
     echo "[5] HP-UX - st95esr1"
     echo "[6] HP-UX - st95esr2"
     echo "[7] Linux - st95esrn"
     echo "[8] Linux - st95esro"
     echo "[9] Linux - st95esrp"
     echo "[10] Linux - st95esrq"
     echo "[11] Linux - st95esrr"
     echo "Enter your menu choice [1-11]: \n"
     read tgtser
}
echo "================ MAIN SESSION =================="
while true do
     menu
     case $choice in
     1)
            chkout1
            case $tgtser in
                 1|2|3|4|5|6|7|8|9|10|11)
                         echo "Checked out to test server ..."
                         ;;
                 *)      echo "Opps !!! please select a valid option"
                         sleep 2
                         chkout1
                         ;;
            esac
     2)            echo "Checked in ..." ;;
            ;;
     3)
            echo "Checked out to local machine ..."
            ;;
     4)
            echo "Exiting ... "
            ;;
     *)
            echo "Opps !!! please select a valid option (1/2/3/4)" ;;
            sleep 2
            menu
            ;;
     esac
 done

Error showing in RED in the code :- ./test1[46]: Syntax error at line 60 : `)' is not expected.

Last edited by Scott; 09-19-2011 at 07:32 AM.. Reason: Code tags
# 4  
Old 09-19-2011
You miss ";;" after the "esac" which is just above the "2)"

Last edited by ctsgnb; 09-19-2011 at 09:42 AM..
# 5  
Old 09-19-2011
what happened ?
# 6  
Old 09-19-2011
I didn't try it so maybe it would require some fix, but in the main loop maybe it could be better with something like :
Code:
while true 
do
    menu
    case $choice in
    1)            
        while :
        do
            chkout1
            case $tgtser in
            1|2|3|4|5|6|7|8|9|10|11)
                echo "Checked out to test server ..."
                break
            ;;
            *)      echo "Opps !!! please select a valid option"
                sleep 2
            ;;
            esac
        done
    ;;
    2)
        echo "Checked in ..."
    ;;
    3)
        echo "Checked out to local machine ..."
    ;;
    4)
        echo "Exiting ... "
    ;;
    *)
        echo "Opps !!! please select a valid option (1/2/3/4)" ;;
        sleep 2
    ;;
esac
done

You must not make duplicate calls of 'chkout1' or 'menu' you must use the while loop instead.

Last edited by ctsgnb; 09-19-2011 at 09:39 AM..
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