Bash script error [: !=: unary operator expected

Question

In my script I am trying to error check if the first and only argument is equal to -v but it is an optional argument. I use an if statement but I keep getting the unary oper

Answer 1

Or for what seems like rampant overkill, but is actually simplistic ... Pretty much covers all of your cases, and no empty string or unary concerns.

In the case the first arg is '-v', then do your conditional ps -ef, else in all other cases throw the usage.

#!/bin/sh
case $1 in
  '-v') if [ "$1" = -v ]; then
         echo "`ps -ef | grep -v '\['`"
        else
         echo "`ps -ef | grep '\[' | grep root`"
        fi;;
     *) echo "usage: $0 [-v]"
        exit 1;; #It is good practice to throw a code, hence allowing $? check
esac

If one cares not where the '-v' arg is, then simply drop the case inside a loop. The would allow walking all the args and finding '-v' anywhere (provided it exists). This means command line argument order is not important. Be forewarned, as presented, the variable arg_match is set, thus it is merely a flag. It allows for multiple occurrences of the '-v' arg. One could ignore all other occurrences of '-v' easy enough.

#!/bin/sh

usage ()
 {
  echo "usage: $0 [-v]"
  exit 1
 }

unset arg_match

for arg in $*
 do
  case $arg in
    '-v') if [ "$arg" = -v ]; then
           echo "`ps -ef | grep -v '\['`"
          else
           echo "`ps -ef | grep '\[' | grep root`"
          fi
          arg_match=1;; # this is set, but could increment.
       *) ;;
  esac
done

if [ ! $arg_match ]
 then
  usage
fi

But, allow multiple occurrences of an argument is convenient to use in situations such as:

$ adduser -u:sam -s -f -u:bob -trace -verbose

We care not about the order of the arguments, and even allow multiple -u arguments. Yes, it is a simple matter to also allow:

$ adduser -u sam -s -f -u bob -trace -verbose

Answer 2

Quotes!

if [ "$1" != -v ]; then

Otherwise, when $1 is completely empty, your test becomes:

[ != -v ]

instead of

[ "" != -v ]

...and != is not a unary operator (that is, one capable of taking only a single argument).