How do I test for an empty string in a Bash case statement?

Here's how I do it (to each their own):

#!/bin/sh

echo -en "Enter string: "
read string
> finder.txt
echo "--" >> finder.txt

for file in `find . -name '*cgi'`

do

x=`grep -i -e "$string" $file`

case $x in
"" )
     echo "Skipping $file";
;;
*)
     echo "$file: " >> finder.txt
     echo "$x" >> finder.txt
     echo "--" >> finder.txt
;;
esac

done

more finder.txt

If I am searching for a subroutine that exists in one or two files in a filesystem containing dozens of cgi files I enter the search term, e.g. 'ssn_format'. bash gives me back the results in a text file (finder.txt) that looks like this:

-- ./registry/master_person_index.cgi: SQLClinic::Security::ssn_format($user,$script_name,$local,$Local,$ssn) if $ssn ne "";

The case statement uses globs, not regexes, and insists on exact matches.

So the empty string is written, as usual, as "" or '':

case "$command" in
  "")        do_empty ;;
  something) do_something ;;
  prefix*)   do_prefix ;;
  *)         do_other ;;
esac

I use a simple fall through. no parameters passed ($1="") will be caught by the second case statement, yet the following * will catch any unknown parameter. Flipping the "") and *) will not work as *) will catch everything every time in that case, even blanks.

#!/usr/local/bin/bash
# testcase.sh
case "$1" in
  abc)
    echo "this $1 word was seen."
    ;;
  "") 
    echo "no $1 word at all was seen."
    ;;
  *)
    echo "any $1 word was seen."
    ;;
esac