passing variables to a bash script

Question

I have a bash script "test.sh" and one parameter I want to use is --no-email.

when I run test.sh --no-email, everything works as expected and I do not receive an email status report.

However what I really want to run is "test.sh test.cnf" where the --no-email parameter is stored in the test.cnf file along with a load of other parameters. I cant for the life of me get this to work. Perhaps I am being completely stupid and not understanding?

Many thanks

echo $*|grep -se '--no-email'&>/dev/null
SEND_MAIL=`echo $?`

echo -e "DEBUG: \$*=$*"

if [ ! "$SEND_MAIL" == "0" ]; then
  echo 'Mail would have been sent!'
else
  echo 'NO MAIL WOULD HAVE BEEN SENT!'
fi

A.H. · Accepted Answer · 2012-11-17 13:21:00Z

If you cannot modify the script test.sh to support this you still can use this syntax to fetch parameters from a config file:

test.sh $(<test.cnf)

If this assumption is not true, i.e. you want to modify test.sh itself to support this then you have to be more specific about what happens inside test.sh.

Edit: Now the content of test.sh has been added to the question. Starting from there the most simple thing to do would be like this:

grep -sqe '--no-email' "$*"
SEND_MAIL=$?

But you wrote that you have a bunch of other paramaters. Doing a grep for each one might be inconvenient. In this case you can loop over the word of a cnf file like this:

#!/bin/bash

while read line; do 
    for word in $line; do
        echo "examing $word"
        case "$word" in 
        --no-email) 
            SEND_MAIL=0
        ;;
        --no-foo)
            NO_FOO=0
        ;;
        *)
            echo 1>&2 "WARNING: Unknown parameter: $word"
        ;;
    esac
    done
done < "$1"

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