Check if any of the parameters to a bash script match a string

Question

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

Answer 1

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)

Answer 2

This worked for me. It does exactly what you asked and 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

This takes advantage of special handling of $* and bash super-test [[…]] brackets.

Answer 3

How about searching (with wildcards) the whole parameter space:

if [[ $@ == *'-disableVenusBld'* ]]
then

Edit: Ok, ok, so that wasn't a popular answer. How about this one, it's perfect!:

if [[ "${@#-disableVenusBld}" = "$@" ]]
then
    echo "Did not find disableVenusBld"
else
    echo "Found disableVenusBld"
fi

Edit2: Ok, ok, maybe this isn't perfect... Think it works only if -param is at the start of the list and will also match -paramXZY or -paramABC. I still think the original problem can be solved very nicely with bash string manipulation, but I haven't quite cracked it here... -Can you??

Answer 4

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.