Case statements in bash

[Don]

I'm trying to write a bash script using a case statement. (I've already written it with "Russian-doll" nested if-then statements, and it runs, but I have more than two choices, so the if-then programming is kludgy.)

I think that I'm getting hung up on how to express a test as a case alternative. First I'm giving the case statement its data file:

case $starting_data in

I want first to test whether the user included an argument on the command line invoking the script. With an if-then, that's easy:

if [ $# -eq 0 ]
then echo "Error"
else [process a command list]

But if, after the case line above, my next line reads

[ $# -eq 0 ]) {Because this is the test for whether there were any arguments on the command line, right?}
echo "Error"
etc.

then I get

hw9case.sh: 7: hw9case.sh: Syntax error: "(" unexpected (expecting ")")

and I'd be lying through my teeth if I pretended I could decipher that. I have a feeling I'm missing something really fundamental, but I've not been able to find anything online that has enlightened me (admittedly a hard job).

Can someone set me straight? Thanks a lot.

[jlittle]

In bash case statements do a string pattern match. They're not a truth table like some languages, or an integer switch statement like others. You could

case $# in
   0)  echo Error whatever
        etc... ;;
esac

if you really wanted to. Old Bourne shell scripting lacked another fast way to test for a pattern match so you'd see case statements that were really if statements, but IMO it was a kludge, and shouldn't be used in modern bash scripting.