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How to read in a part of an echo line?

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# 8  
Old 10-31-2012
I think what we were trying to say was : first, you should focus on how to use a "case" statement, which is more appropriate to what you are trying to do and easy to read and maintain.

Then, if you want to investigate further, you should also train to use "getopts" as well as "select". (see the links previously provided)
This User Gave Thanks to ctsgnb For This Post:
Grueben
# 9  
Old 10-31-2012
Thanks Pamu but it doesn't work for what I want as I can't pass the $1 until I've been shown the options.
# 10  
Old 10-31-2012
Quote:
Originally Posted by Grueben
Thanks Pamu but it doesn't work for what I want as I can't pass the $1 until I've been shown the options.
So who is stopping you from doing this..?

You can do something like this.

Code:
echo "Options are"
echo "1 for --"
echo "2 for --"
read opt
#You can check is $opt is present or not
case "$opt" in

1)  echo "Sending 1 signal"
    ;;
2)  echo  "Sending 2 signal"
    ;;
3)  echo  "Sending 3 signal"
    ;;
*) echo "Signal number $opt is not processed"
   ;;
esac

I hope this helpsSmilie

pamu
This User Gave Thanks to pamu For This Post:
Grueben
# 11  
Old 10-31-2012
Are you struggling with the logic for the menu driven cd Grueben? Maybe this code snipplet can help (tested in bash, but it should work with ksh too):
Code:
#!/bin/bash

dir=""
backmsg="Back to menu"
PS3="Select directory to cd into: "

while [ "$dir" != "$backmsg" ]
do
   alldirs=$(ls -ap |grep /)
   select dir in $alldirs "$backmsg"
   do
      if [ "$dir" != "$backmsg" ]
      then
         cd $dir
         pwd
      fi
      break
   done
done
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