If-statements in Bash Syntax issues produce right answers, while right code shows wrong answers [duplicate] - bash

Question

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Why should there be spaces around '[' and ']' in Bash?
(5 answers)
Closed 5 years ago.
I am fairly new to bash scripting and am struggling with some if-statement syntax.
I have currently written up the following loop:
for (( i = 2; i < $# - 1; i++)); do
if [ $i -ne 0]; then
if [ $i -ne 1]; then
echo "$i was not 1 or 0. Please correct this then try again."
exit 1;
fi
fi
done
This code is supposed to test whether any arguments after the first are either a 1 or a 0.
While the following errors are printed:
./blink.sh: line 36: [: missing `]'
./blink.sh: line 36: [: missing `]'
...the code actually runs fine afterwards (so the errors don't kill the program).
My understanding, however, is that in bash, you put spaces before and after the expression inside the if statement. So this:
if [ $i -ne 0]; then
Becomes:
if [ $i -ne 0 ]; then
However, running this code produces the following:
2 was not 1 or 0. Please correct this then try again.
The main issue I am having with this stems from not understanding how to indirectly reference the positional arguments provided by the execution command. As such, I am confused as to what syntax must be altered to call the objects the arguments point to (in this case, hopefully either a 1 or a 0) rather than the position of the arguments themselves (argument 1, 2, 3...).
Thanks!
EDIT: Altering the question to better fit the advice #randomir provided and clear up what the actual question entails

Answer 1

Based on:
This code is supposed to test whether any arguments after the first are either a 1 or a 0.
I'm assuming you're trying to access positional arguments $2, $3, etc. To make your for loop solution work, you would have to use an indirect reference: ${!i} (see shell parameter expansion). For example, this should work:
#!/bin/bash
for (( i = 2; i <= $#; i++ )); do
if [[ ${!i} -ne 0 ]]; then
if [[ ${!i} -ne 1 ]]; then
echo "$i was not 1 or 0. Please correct this then try again."
exit 1;
fi
fi
done
Note the i running from 2 to number of arguments $#. Also, note the use of recommended and less error-prone [[ .. ]] instead of [ .. ] (otherwise you would have to write [ "${!i}" -ne 0 ], etc).
A simpler solution which avoids the unnecessary indirect referencing looks like this:
#!/bin/bash
while [[ $2 ]]; do
if (( $2 != 0 && $2 != 1 )); then
echo "$2 is neither 0, nor 1"
exit 1
fi
shift
done
We start checking the second argument ($2), use the arithmetic expression (( expr )) testing of value of the second argument, and shift positional arguments to the left by 1 at each iteration (now $3 becomes $2, etc).