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Getopts in the subshell of ksh

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# 1  
Old 09-05-2013
Getopts in the subshell of ksh
Hi,

the getopts doesnt seem to be working in the subshell of the ksh. when I echo $@ and $* from the subshell it shows nothing. even when I am capturing the parameters from the outer shell and passing while invoking the file then I am still not getting it properly.

the below code is in the inner script
Code:
while getopts ":f:v:s" opt; do
#        case $opt in
#           "-f") shift;
#                sqltype ="$opt"
#                 echo "OPTION IS $opt"
#       ((sqltype != "c"|"h")) && { printf "%b\n" "${USAGE}";  exit 1; }
#       ;;
#   "-v") vname="$2"
#       ;;
#   "-s") dbstring="$2"
#        ;;
#   *) echo printf "%b\n" "${USAGE}"
#       exit 1
#       ;;
#
#       esac
echo $opt
shift
done

invoking from the outer script like this
Code:
$o=$@
. ./innerscript $o

Invoking the outer script like
Code:
./outerscript -f parameter

I am using ksh here. can some one please help me with this? in the above code its not getting in the while loop itself.

Many thanks!!
Last edited by Don Cragun; 09-05-2013 at 10:54 PM.. Reason: CODE tags; not QUOTE tags!
# 2  
Old 09-05-2013
Remove the # at the start of all of the lines in your while loop. The # and everything following it on a line is treated as a comment; not code to be executed.

PS You also need to change:
Code:
sqltype ="$opt"

to:
Code:
sqltype="$opt"

(no space before the equals sign).

PPS You'll also want to change the first line of outerscript from:
Code:
$o=$@

to:
Code:
o="$@"

(no dollar sign in front of o and double quotes around $@ even though the latter might not make any different with the arguments you're passing in this example).
Last edited by Don Cragun; 09-05-2013 at 10:58 PM.. Reason: Add PPS.
# 3  
Old 09-05-2013
Quote:
Originally Posted by Don Cragun
Remove the # at the start of all of the lines in your while loop. The # and everything following it on a line is treated as a comment; not code to be executed.
I understand that. but my point is its not even getting into the while loop and not displaying the echo statement.
# 4  
Old 09-05-2013
Quote:
Originally Posted by hitmansilentass
I understand that. but my point is its not even getting into the while loop and not displaying the echo statement.
Please see the PPS in my earlier note...
# 5  
Old 09-05-2013
Also, why are the arguments (:) before the options (f, v, s)?
# 6  
Old 09-05-2013
Thanks guy's. I just figured the issue. the code was within a function hence it was not capturing the values. anyway, I am getting another issue now that its not capturing second flag values. do you know why?

Its probably to do with the arguments but its been very long since I have worked on shell scripts so I am still recalling a lot of stuff.

Code:
while getopts "f:v:s" opt; do
echo $opt
        case $opt in
           "f") shift;
                 typeset -l sqltype=$OPTARG
       [[ $sqltype != "c" ]] &&  [[ $sqltype != "h" ]] && { printf "%b\n" "${USAGE}";  exit 1; }
       ;;
   "v") vname="$OPTARG"
       ;;
   "s") dbstring="$OPTARG"
        ;;
   *) echo printf "%b\n" "${USAGE}"
       exit 1
       ;;

        esac
done

Last edited by Scott; 09-05-2013 at 11:35 PM.. Reason: Again, CODE tags not QUOTE tags, please...
# 7  
Old 09-05-2013
You can remove the shift command, it's not required.
Code:
$ cat myScript
while getopts f:v:s opt; do
  case "$opt" in
    f) echo f has $OPTARG;;
    v) echo v has $OPTARG;;
    s) echo s has nothing;;
  esac
done

$ ./myScript -f f_opt -s -v v_opt
f has f_opt
s has nothing
v has v_opt
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