Shell script and java parameter

Question

I have written the script run.sh below for calling a java class :

java -cp . Main $1 $2 $3 $4

Here is my java class (it just displays the args) :

public class Main {
    public static void main(String[] args) {
        for (String arg : args) {
            System.out.println(arg);
        }
    }
}

This is working unless I try to pass a parameter containing a space. For example :

./run.sh This is "a test"

will display :

This
is
a
test

How could I modify my shell script and/or change my parameter syntax to pass the parameter "a test" unmodified ?

Ignacio Vazquez-Abrams · Accepted Answer · 2012-05-14 03:25:08Z

9

Like this:

java -cp . Main "$@"

answered May 14, 2012 at 3:25

Ignacio Vazquez-Abrams

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1 Comment

Tim Pote Over a year ago

It might help if you explained why that works. This is a pretty contrived example. That may not be exactly what he wants to do in the end.

user unknown · Accepted Answer · 2012-05-14 03:42:03Z

5

You have to mask every parameter in the script as well:

java -cp . Main "$1" "$2" "$3" "$4"

Now parameter 4 should be empty, and $3 should be "a test".

To verify, try:

#!/bin/bash
echo 1 "$1"
echo 2 "$2"
echo 3 "$3"
echo 4 "$4"
echo all "$@"

and call it

./params.sh This is "a test" 
1 This
2 is
3 a test
4 
all This is a test

edited May 14, 2012 at 3:42

answered May 14, 2012 at 3:32

user unknown

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2 Comments

Arnaud Over a year ago

The fact that parameter 4 is the empty string is annoying because the java program will receive a wrong number of parameters (4 while the user passed only 3). I think Ignacio's answer is better, thanks anyway.

user unknown Over a year ago

@SuperChafouin: Oh, I see, my test wasn't accurate. I thought the "$@" would consume the masking, after seeing the output of my test, but I guess it is the last echo, which consumes the quotes, just like the Java program does.

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