how to select multiple exit

Question

I'm new to bash scripting and I don't understand what's wrong with this specific part of my script. im trying to select all the exit that have in my script. and trying to put in conditon that if all those condition is not met then its not valid and else its valid.

if [[ `echo $?` -ne {0..4} ]]
then
         echo "Its a Valid Date."
 else
         echo " Not valid date."
fi

When I run it, I get:

kpatel138@matrix:~/Lab2> ./cal2 1994 5 4
./cal2: line 40: unexpected argument `(' to conditional binary operator
./cal2: line 40: syntax error near `(0'
./cal2: line 40: `if [[ `echo $?` -ne (0..4) ]]'

Answer 1

`echo $?` is a convoluted way of writing $?. Don't do things the complicated way when there's an obvious simpler way.

I think you want to test whether $? is one of the values 0, 1, 2, 3 or 4. The syntax you're using doesn't make sense. You can test all five equality cases:

if [[ $? -eq 0 || $? -eq 1 || $? -eq 2 || $? -eq 3 || $? -eq 4 ]]; then …

But that's overly verbose. Since you want to allow the range 0–4, test that $? is more than the minimum value and less than the maximum value.

if [[ $? -ge 0 && $? -le 4 ]]; then …

$? is never negative, so the test $? -ge 0 is superfluous. Thus:

if [[ $? -le 4 ]]; then …

This can also be written using an arithmetic expression. Note that the operator is different: conditional constructs in bracketts use -eq, -ne, -le, -ge, -lt and -gt to compare integers, while arithmetic expressions have a C-like syntax.

if (($? <= 4)); then …