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 operator expected error.

This is the code:

if [ $1 != -v ]; then
   echo "usage: $0 [-v]"
   exit
fi

To be more specific:

This part of the script above is checking an optional argument and then after, if the argument is not entered, it should run the rest of the program.

#!/bin/bash

if [ "$#" -gt "1" ]; then
   echo "usage: $0 [-v]"
   exit
fi

if [ "$1" != -v ]; then
   echo "usage: $0 [-v]"
   exit
fi

if [ "$1" = -v ]; then
   echo "`ps -ef | grep -v '['`"
else
   echo "`ps -ef | grep '[' | grep root`"
fi

Answer 1

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).