BASH: read in while loop - bash

Question

while [ $done = 0 ]
do
echo -n "Would you like to create one? [y/n]: "
read answer
if [ "$(answer)" == "y" ] || [ "$(answer)" == "Y" ]; then
mkdir ./fsm_$newVersion/trace
echo "Created trace folder in build $newVersion"
$done=1
elif [ "$(answer)" == "n" ] || [ "$(answer)" == "N" ]; then
$done=2
else
echo "Not a valid answer"
fi
done
Ok so I have this simple bashscript above that simply just tries to get input from a user and validate it. However I keep getting this error
./test.sh: line 1: answer: command not found
./test.sh: line 1: answer: command not found
./test.sh: line 1: answer: command not found
./test.sh: line 1: answer: command not found
Which I have no idea why because "answer" is nowhere near line 1. So I ran into this article
Which makes sense since it's referring to line 1 and can't find answer. So it seems to be starting a new subshell. However I didn't really understand the solution and can't see how I would apply it to my case. I just wanna get this to work.

Answer 1

$(answer) doesn't substitute the value of the variable answer. It executes answer as a command, and substitutes the output of that command. You want ${answer} everywhere you have $(answer). In this case you can get away with bare $answer too, but overuse of ${...} is good paranoia.
(Are you perhaps used to writing Makefiles? $(...) and ${...} are the same in Makefiles, but the shell is different.)
By the way, you have some other bugs:
In shell, you do not put a dollar sign on the variable name on the left hand side of an assignment. You need to change $done=1 to just done=1 and similarly for $done=2.
You are not being paranoid enough about your variable substitutions. Unless you know for a fact that it does the wrong thing in some specific case, you should always wrap all variable substitutions in double quotes. This affects both the mkdir command and the condition on the while loop.
You are not being paranoid enough about arguments to test (aka [). You need to prefix both sides of an equality test with x so that they cannot be misinterpreted as switches.
== is not portable shell, use = instead (there is no difference in bash, but many non-bash shells do not support == at all).
Put it all together and this is what your script should look like:
while [ "x${done}" = x0 ]; do
echo -n "Would you like to create one? [y/n]: "
read answer
if [ "x${answer}" = xy ] || [ "x${answer}" = xY ]; then
mkdir "./fsm_${newVersion}/trace"
echo "Created trace folder in build $newVersion"
done=1
elif [ "x${answer}" = xn ] || [ "x${answer}" = xN ]; then
done=2
else
echo "Not a valid answer"
fi
done

Answer 2

Which I have no idea why because
"answer" is nowhere near line 1. So I
ran into this article
That's not your problem here.
I ran the script and did not get the error you got. I did receive the error:
./test.sh: line 1: [: -eq: unary operator expected
when I tried to compile though. Defining done fixed this. The following script should work...
#!/bin/bash
done=0
while [ $done -eq 0 ]
do
echo -n "Would you like to create one? [y/n]: "
read answer
if [[ "$(answer)" == "y" || "$(answer)" == "Y" ]]; then
mkdir ./fsm_${newVersion}/trace
echo "Created trace folder in build $newVersion"
$done=1
elif [[ "$(answer)" == "n" || "$(answer)" == "N" ]]; then
$done=2
else
echo "Not a valid answer"
fi
done
...note you were doing string comparisons on your done variable, which you apparently intended to be numeric. It's generally bad form to do string comparison on a numeric type variable, though it will work. Use -eq (arithmetic comparison operator) instead. (Also note that if you kept that test, your string equality would be inconsistent... you had "=" in one spot and "==" in another spot... nitpicking here, but it's helpful to be consistent).
Also, I suggest double brackets for your compound conditionals as they will be more readable if you have longer ones. e.g.
if [[($var1 -eq 0 && $var2 -eq 1) || ($var1 -eq 1 && $var2 -eq 0)]]; then
Just a matter of preference as you only have two conditions, but could be useful in the future.
Also you were missing braces '{' '}' around your newVersion variable.
Finally, I'd suggest putting the line #!/bin/bash on the top of your script. Otherwise its up to your environment to determine what to do with your script, which is a bad idea.