Bash loop until a certain command stops failing

Question

I would like to write a loop in bash which executes until a certain command stops failing (returning non-zero exit code), like so:

while ! my_command; do
    # do something
done

But inside this loop I need to check which exit code my_command returned, so I tried this:

while ! my_command; do
    if [ $? -eq 5 ]; then
        echo "Error was 5"
    else
        echo "Error was not 5"
    fi
    # potentially, other code follows...
done

But then the special variable ? becomes 0 inside the loop body. The obvious solution is:

while true; do
    my_command
    EC=$?
    if [ $EC -eq 0 ]; then
        break
    fi
    some_code_dependent_on_exit_code $EC
done

How can I check the exit code of my_command (called in loop header) inside loop body without rewriting this example using a while true loop with a break condition as shown above?

Answer 1

In addition to the well-known while loop, POSIX provides an until loop that eliminates the need to negate the exit status of my_command.

# To demonstrate
my_command () { read number; return $number; }

until my_command; do
    if [ $? -eq 5 ]; then
        echo "Error was 5"
    else
        echo "Error was not 5"
    fi
    # potentially, other code follows...
done

Answer 2

If true command hurt your sensibility, you could write:

while my_command ; ret=$? ; [ $ret -ne 0 ];do
    echo do something with $ret
  done

This could be simplified:

while my_command ; ((ret=$?)) ;do
    echo do something with $ret
  done

But if you don't need ResultCode, you could simply:

while my_command ; [ $? -ne 0 ];do
    echo Loop on my_command
  done

or

while my_command ; (($?)) ;do
    echo Loop on my_command
  done

And maybe, why not?

while ! my_command ;do
    echo Loop on my_command
  done

But from there you could better use until as chepner suggest