scripting - How to pass arguments from wrapper shell script to Java application?

Question

I want to run Java programs I am creating at on the command line (linux and mac). I don't want to type "java" and arguments all the time, so I am thinking about creating wrapper scripts. What's the best way to do this so that they work everywhere? I want to be able to pass arguments, too. I was thinking of using "shift" to do this (removing first argument).

Is there a better way to do this without using scripts at all? Perhaps make an executable that doesn't require invocation through the "java" command?

Answer 1

Assuming that you are using a shell that is compatible with the Bourne shell; e.g. sh, bash, ksh, etc, the following wrapper will pass all command line arguments to the java command:

#!/bin/sh
OPTS=... 
java $OPTS com.example.YourApp "$@"

The $@ expands to the remaining arguments for the shell script, and putting quotes around it causes the arguments to be individually quoted, so that the following will pass a single argument to Java:

$ wrapper "/home/person/Stupid Directory Name/foo.txt" 

Without the double quotes around "$@" in the wrapper script, Java would receive three arguments for the above.

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Note that this does not work with "$*". According to the bash manual entry:

"$*" is equivalent to "$1c$2c...", where c is the first character of the value of the IFS variable.

In other words, all shell arguments would be concatenated into a single command argument for your Java application, ignoring the original word boundaries.

Refer to the bash or sh manual ... or the POSIX shell spec ... for more information on how the shell handles quoting.