Question

I can seem to see why this doesn’t work:

#!/bin/bash

if [ $# -ne 1 ] || [ $# -ne 2 ]; then 
# Should run if there are either 1 or 2 options specified
  echo "usage: ${0##*/} <username>"
  exit
fi

When testing to see if it works:

root@ubuntu:~# testing.sh optionone optiontwo
...Correct output...
root@ubuntu:~# testing.sh optionone
usage: testing.sh <username>

Answer 1

Note that you are executing 2 commands in:

[ $# -ne 1 ] || [ $# -ne 2 ]

[ $# -ne 1 ] is a 1st command, and the [ $# -ne 2 ] command is executed only if the previous has a non-zero error code as of the || shell operator.

In your case, it is not important, but in the case bellow, it is:

[ $? -eq 0 ] || [ $? -eq 1 ]

The 2nd command will always be true, as the 2nd $? is the return code of [ $? -eq 0 ]. You can test it with the lines bellow that will print true twice:

function f() { return $1; }
f 1
{ [ $? -eq 0 ] || [ $? -eq 1 ]; } && echo "true"
f 2
{ [ $? -eq 0 ] || [ $? -eq 1 ]; } && echo "true"

The correct way to execute a or in a single command is:

[ $? -eq 0 -o $? -eq 1 ]

This way, those bellow only print true once:

function f() { return $1; }
f 1
{ [ $? -eq 0 -o $? -eq 1 ]; } && echo "true"
f 2
{ [ $? -eq 0 -o $? -eq 1 ]; } && echo "true"

And concerning your original question, kev has already point out that there was a logic error in your test. The negative of [ $# -eq 1 ] || [ $# -eq 2 ] is NOT [ $# -eq 1 ] && NOT [ $# -eq 2 ] and this becomes [ $# -ne 1 ] && [ $# -ne 2 ] or in a single command:

[ $# -ne 1 -a $# -ne 2 ]