How can I compare a string to multiple correct values in Bash?

Solution 1:

If the main intent is to check whether the supplied value is not found in a list, maybe you can use the extended regular expression matching built in BASH via the "equal tilde" operator (see also this answer):

if ! [[ "$cms" =~ ^(wordpress|meganto|typo3)$ ]]; then get_cms ; fi

Have a nice day

Solution 2:

Instead of saying:

if [ "$cms" != "wordpress" && "$cms" != "meganto" && "$cms" != "typo3" ]; then

say:

if [[ "$cms" != "wordpress" && "$cms" != "meganto" && "$cms" != "typo3" ]]; then

You might also want to refer to Conditional Constructs.

Solution 3:

Maybe you should better use a case for such lists:

case "$cms" in
  wordpress|meganto|typo3)
    do_your_else_case
    ;;
  *)
    do_your_then_case
    ;;
esac

I think for long such lists this is better readable.

If you still prefer the if you can do it with single brackets in two ways:

if [ "$cms" != wordpress -a "$cms" != meganto -a "$cms" != typo3 ]; then

or

if [ "$cms" != wordpress ] && [ "$cms" != meganto ] && [ "$cms" != typo3 ]; then

Solution 4:

As @Renich suggests (but with an important typo that has not been fixed unfortunately), you can also use extended globbing for pattern matching. So you can use the same patterns you use to match files in command arguments (e.g. ls *.pdf) inside of bash comparisons.

For your particular case you can do the following.

if [[ "${cms}" != @(wordpress|magento|typo3) ]]

The @ means "Matches one of the given patterns". So this is basically saying cms is not equal to 'wordpress' OR 'magento' OR 'typo3'. In normal regular expression syntax @ is similar to just ^(wordpress|magento|typo3)$.

Mitch Frazier has two good articles in the Linux Journal on this Pattern Matching In Bash and Bash Extended Globbing.

For more background on extended globbing see Pattern Matching (Bash Reference Manual).