How do I fall through in a bash case statement in OSX?

Question

I have some bash scripts that I would like to run in OSX, but some of them contain the fallthrough case terminator ';&' which doesn't seem to work on Mac. Here's sample code:

#!/bin/bash
case $1 in
    test)
        echo "go" ;&    
    test)
        echo "go2" ;;
esac

In Cygwin, .\test.sh test produces

go
go2

but on OSX,

./test.sh: line 4: syntax error near unexpected token `&'
./test.sh: line 4: `            echo "go" ;&   

Answer 1

Your code works as is on my ubuntu system, bash 4.2.24.

I think you are using an earlier version of bash which doesn't support ;& fall-through which was added since bash 4.

From wikipedia:

Fall through is done using ;& (new since Bash 4) whereas ;; acts as a case break.

Answer 2

Perhaps you could try using /bin/zsh instead of /bin/bash? Your sample code works with “zsh 4.3.9” but not “bash 3.2.48” on my MacOS X 10.6.8.

Otherwise, the script has to be rewritten in some way… but it's hard to say exactly how without knowing what it does. Simply duplicating the code as needed is probably the simplest option and should work in every case.