如何在Mac OS X上的bash脚本中更改当前工作目录？

Question

I'm struggling with a little problem, I never wrote a bash script before and it may be a little thing for someone who knows how it's done correctly.

I have the following folder structure (cannot be changed):

MyApp.app
   |_Contents
       |_Java
           |_MyApp.jar
       |_MacOS
           |_launch.sh

The launch.sh script should start the MyApp.jar executable and currently looks as follows:

#!/bin/bash
SCRIPT_DIR=$(cd "$(dirname "$0")/../Java"; pwd)
exec "$SCRIPT_DIR/jre/bin/java" -Xms256m -XX:PermSize=64m -jar "MyApp.jar"

When starting the launch.sh script, the second line should change the current working directory from MyApp.app/Contents/MacOS to MyApp.app/Contents/Java in order to call subsequently java -jar MyApp.jar (3rd line) with its correct root directory as working directory...

But as soon as the MyApp.jar starts, I print out the current working directory within my Java application with:

System.out.println(System.getProperty("user.dir"));

... and it prints out the directory of the launch.sh script, namely MyApp.app/Contents/MacOS instead of MyApp.app/Contents/Java .

Any ideas how to properly change the directory within the bash scripts are highly appreciated : ) ... ?!

Answer 1

You could always work with the cd command (change directory). But a more optimal solution, if you had to change directories, may be using the pushd & popd commands to build a stack.

In this case scenario you could hardcode the file path(s) that the script will have to navigate to during execution :

_java=/Applications/MyApp.app/_Contents/_Java;

Then using pushd and popd (push directory onto stack, pop directory off of stack) :

# Currently in /Applications/MyApp.app/_Contents/_MacOS/
pushd $_java;

# Do what you have to do in the _Java working directory (./ = current working directory)
exec "./jre/bin/java" -Xms256m -XX:PermSize=64m -jar "MyApp.jar";

# Print the working directory and return to bottom of stack
pwd; popd;

Nothing against the usage of cd, but pushd & popd answer the question as well and are more flexible than cd in some situations, especially when you really have to traverse a file hierarchy on several occasions during execution. for more information man pushd , man popd , google or related articles such as : https://unix.stackexchange.com/questions/77077/how-do-i-use-pushd-and-popd-commands

Besides that, even for day-to-day usage, cd is great, but in my case I always move around and it's nice to be in my /Library/LaunchDaemons/ and to simply pushd ~/Documents/Scripts/ ; vim bashp0rn.sh ; popd ; and back in /Library/LaunchDaemons/ to load & launch ( launchctl ).

Now , with that said, you don't have to change to your apps working directory in order to launch the application :

... in order to call subsequently java -jar MyApp.jar (3rd line) with its correct root directory as working directory ...

If that is your sole requirement && if that is your requirement because in this case you are trying to complete the path by assigning pwd 's output to $SCRIPT_DIR , then I would suggest hardcoding the full path and using the eval command instead as such :

#!/bin/bash

app='/Applications/MyApp.app/_Contents/_Java/jre/bin/java -Xms256m -XX:PermSize=64m -jar "MyApp.jar"';

Now from anywhere in the file hierarchy you can evaluate the variable :

eval $app;

Answer 2

Try to do it like this:

cd "$(dirname "$0")/../Java"
SCRIPT_DIR=$(pwd)