Programming Forums - Simple menu script

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Simple menu script

Hi! Just started the old bash programming...I have got a bit stuck.

I wanted to create a menu where by a user can type the name of a file they wish to view.

The problem I have is that the file doesn't seem to either get saved to the variable, isn't retrieved from the variable, or there is a problem with the file's location. The file I would want users to open is located in the same directory as the script itself.

The code is as follows:-

#!/bin/bash

#Filename: quit_example



quit=n

while [ "$quit" = "n" ]

do

   clear

   echo

   echo "1. Show date"

   echo "2. Show hostname"

   echo "3. Display text file"

   echo "Q. Quit"

   echo

   echo "Enter choice"

   read choice

case $choice in

   1) date

      echo "Hit Enter to carry on"

      read junk;;

   2) hostname

      echo "Hit Enter to carry on"

      read junk;;

   3) echo "Enter file name to be displayed"

      read $filename

           if [ -r "$filename" -a -f "$filename" ]

             then

               clear

             cat $filename

            else

      echo "Cannot display $filename"

           fi

     echo "Hit Enter to carry on"

     read junk;;

   Q|q)quit=y;;

esac

done

Would anyone be able to show me where I am going wrong?

Cheers,

Trufla

Quote:

Originally Posted by trufla

  #!/bin/bash

  #Filename: quit_example

  

  quit=n

  while [ "$quit" = "n" ]

  do

     clear

     echo

     echo "1. Show date"

     echo "2. Show hostname"

     echo "3. Display text file"

     echo "Q. Quit"

     echo

     echo "Enter choice"

     read choice

  case $choice in

     1) date

        echo "Hit Enter to carry on"

        read junk;;

     2) hostname

        echo "Hit Enter to carry on"

        read junk;;

     3) echo "Enter file name to be displayed"

        read $filename

             if [ -r "$filename" -a -f "$filename" ]

               then

                 clear

               cat $filename

              else

        echo "Cannot display $filename"

             fi

       echo "Hit Enter to carry on"

       read junk;;

     Q|q)quit=y;;

  esac

  done

   #!/bin/bash

   #Filename: quit_example

   

   quit=n

   while [ "$quit" = "n" ]

   do

      clear

      echo

      echo "1. Show date"

      echo "2. Show hostname"

      echo "3. Display text file"

      echo "Q. Quit"

      echo

      echo "Enter choice"

      read choice

   case $choice in

      1) date

         echo "Hit Enter to carry on"

         read junk;;

      2) hostname

         echo "Hit Enter to carry on"

         read junk;;

      3) echo "Enter file name to be displayed"

 read filename // <<<==== the problem was that you had read $filename and you dont need the $ in the variable name when reading into it :)

              if [ -r "$filename" -a -f "$filename" ]

                then

                  clear

                cat $filename

               else

         echo "Cannot display $filename"

              fi

        echo "Hit Enter to carry on"

        read junk;;

      Q|q)quit=y;;

   esac

   done

look for the //<<<<<===== in my code to show you what i did and it works fine now :)