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I'm trying to write a script where I want to check if any of the parameters passed to a bash script match a string. The way I have it setup right now is

if [ "$3" != "-disCopperBld" -a "$4" != "-disCopperBld" -a "$5" != "-disCopperBld" -a "$6" != "-disCopperBld"]

but there might be a large number of parameters, so I was wondering if there is a better way to do this?

Thanks

EDIT: I tried this chunk of code out, and called the script with the option, -disableVenusBld, but it still prints out "Starting build". Am I doing something wrong? Thanks in advance!

while [ $# -ne 0 ]
do
    arg="$1"
    case "$arg" in
        -disableVenusBld)
            disableVenusBld=true
            ;;
        -disableCopperBld)
            disableCopperBld=true
            ;;
        -disableTest)
            disableTest=true
            ;;
        -disableUpdate)
            disableUpdate=true
            ;;
        *)
            nothing="true"
            ;;
    esac
    shift
done

if [ "$disableVenusBld" != true ]; then
    echo "Starting build"
fi
share|improve this question
    
Thanks for the replies guys, I appreciate ya'll taking time out to help me. I tried a chunk of code, but I can't seem to figure out what's going wrong. Any ideas? (I've pasted the code in an edit of my original post) –  iman453 Sep 9 '10 at 15:29
    
Hmm: it works for me. I added #! /bin/sh - to the top of what you've included there, made the script executable, then ./t.sh prints "Starting build", but ./t.sh -disableVenusBld prints nothing. –  Norman Gray Sep 9 '10 at 21:47

3 Answers 3

up vote 11 down vote accepted

It looks like you're doing option handling in a shell script. Here's the idiom for that:

#! /bin/sh -

# idiomatic parameter and option handling in sh
while test $# -gt 0
do
    case "$1" in
        --opt1) echo "option 1"
            ;;
        --opt2) echo "option 2"
            ;;
        --*) echo "bad option $1"
            ;;
        *) echo "argument $1"
            ;;
    esac
    shift
done

exit 0

(There are a couple of conventions for indenting the ;;, and some shells allow you to give the options as (--opt1), to help with brace matching, but this is the basic idea)

share|improve this answer
disCopperBld=
for x; do
  if [ "$x" = "-disCopperBld" ]; then disCopperBld=1; break; fi
done
if [ -n "$disCopperBld" ]; then
  ...
fi

If you need to test only the parameters starting at $3, do the search in a function:

## Usage: search_trailing_parameters NEEDLE NUM "$@"
## Search NEEDLE amongst the parameters, skipping $1 through ${$NUM}.
search_trailing_parameters () {
  needle=$1
  shift $(($2 + 2))
  for x; do
    if [ "$x" = "$needle" ]; then return 0; fi
  done
  return 1
}
if search_trailing_parameters -disCopperBld 2 "$@"; then
  ...
fi

But I wonder why you're trying to do this in the first place, it's not a common need. Usually, you'd process options in order, as in Dennis's answer to your previous question.

share|improve this answer

This worked for me. It does exactly what you asked nothing more (no option processing). Whether that's good or bad is an exercise for the poster :)

if [[ "$*" == *YOURSTRING* ]]
then
    echo "YES"
else
    echo "NO"
fi

Takes advantage of special handling of $* and bash super-test [[ ]] brackets.

share|improve this answer

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