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

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)

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.

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??