Passing arguments to a script

Shell Programming and Scripting
1

I've written a script (bgrep) for a more advanced grep command (& attached a cut down version below). I'm trying allow all grep options to be used, or in any combination.

The script works fine if I type say

bgrep -i -files product

it will return a non-case sensitive list of matches for product

however if I type

bgrep -in -files product

I get the same list & no line numbers.

I know why this is happening, it's because the -n option is a valid option for the echo command I am using. I not sure how to get round this. Is there some other command I could use.

I'd appreciate any advice, feel free to criticise.

#!/bin/sh
>grep_args
files=false
forms=false
text=""

while [ $# -gt 0 ]
do
        case "$1" in
                -?)     echo $1 > bgrep.tmp
                        paste -d" " bgrep.tmp grep_args > grep_args.tmp
                        mv grep_args.tmp grep_args;;

                -files) files=true;;

                -forms) forms=true;;

                -*)     echo "usage :$0 [-grepoption] [-files] [-forms] text"
                        exit 1;;

                *)      text=$@
                        break;;
        esac
        shift
done

if [ $files = true ]
then
        echo files:
        find . -name "*.f" -print|xargs grep `cat grep_args` "$text"|pg
fi

if [ $forms = true ]
then
        echo forms:
        find . -name "*.s" -print|xargs grep `cat grep_args` "$text"|pg
fi
2

Sorry about the indenting, it seems to have been lost in the cut & paste from my system

3

I edited your post to include code tags. This fixed the indenting.

I think that
echo "" $1
will fix your problem.

4

Yeah, it works now - should have thought of that really.

Thanks.