How to pass argument to java bash script? [duplicate]

Question

I have a simple bash script which runs my java program. Here is it:

run.sh

#!/bin/sh
java -jar target/my-jar-arch.jar

I want to pass arguments to this script, which must pass them to java program:

bash run.sh myArg -key 100

When I try to execute this line my app doesn't get any parameters. How can I fix it?

java -jar target/my-jar-arch.jar "$@"

Dima
– Dima

2014-12-07 15:29:09 +00:00
Commented Dec 7, 2014 at 15:29 — Dima
– Dima, Commented Dec 7, 2014 at 15:29

fge · Accepted Answer · 2014-12-07 15:29:32Z

6

Change your invocation line to:

java -jar target/my-jar-arch.jar "$@"

answered Dec 7, 2014 at 15:29

fge

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4 Comments

Aaron Over a year ago

Oh maybe you know how to make my script executable? If I run it run myArg -p 100 I get -bash: run: command not found. Seams the problem is in my env.

fge Over a year ago

It should be ./run.sh. Unix systems won't automagically "append extensions" or whatever; in fact they don't have the notion of an extension at all.

Aaron Over a year ago

The problem isn't in extension. It is in my .bash_profile I think.

udarakr Over a year ago

is there any possibility to pass few bash variables in to the java -jar command? Eg:- java -jar org.sample.jar $HOST $PORT assume I already have HOST and PORT defined within my bash script.

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