How to handle missing args in shell script

Question

What's the "right" way to handle missing arguments in a shell script? Is there a pre-canned way to check for this and then throw an exception? I'm an absolute beginner.

Answer 1

Typical shell scripts begin by parsing the options and arguments passed on the command line. The number of arguments is stored in the # parameter, i.e., you get it with $#. For example, if your scripts requires exactly three arguments, you can do something like this:

if [ $# -lt 3 ]; then   echo 1>&2 "$0: not enough arguments"   exit 2 elif [ $# -gt 3 ]; then   echo 1>&2 "$0: too many arguments"   exit 2 fi # The three arguments are available as "$1", "$2", "$3" 

The built-in command exit terminates the script execution. The integer argument is the return value of the script: 0 to indicate success and a small positive integer to indicate failure (a common convention is that 1 means “not found” (think grep) and 2 means “unexpected error” (unrecognized option, invalid input file name, ...)).

If your script takes options (like -x), use getopts.

Answer 2

Example script (myscript.sh):

#!/bin/bash  file1=${1?param missing - from file.} file2=${2?param missing - to file.}   [...] 

Example:

$ ./myscript.sh file1  ./myscript.sh: line 4: 2: param missing - to file.