I am developing a small Java application using Swing, that is supposed to run on Windows/Linux and MacOS.
Eventually it will ship as a runnable jar with some config files in the same folder. To load them I need the path to the folder the jar is stored in within the program.
There are already a couple of threads like this one or this one.
My problem is, that all the solutions discussed there work fine, when I run the program from within eclipse or call the runnable jar from a terminal like so:
java -jar /path/to/jar/jarfile.jar
However when I click on the jar file in Cinnamon or Gnome (which is what most of the users will know to do), I do not get the correct paths. (MacOS users report the same issue)
Here is what I've tried so far and what the output is when run via double click (all those display the desired path when run from eclipse or a terminal):
ClassLoader.getSystemClassLoader().getResource(".").getPath()
Output: file:/usr/lib/jvm/java-6-openjdk-common/jre/lib/ext/pulse-java.jar!/
SomeClass.class.getProtectionDomain().getCodeSource().getLocation().getPath();
Output: ./
System.getProperty("user.dir");
Output: /home/myusername
Is there any other way to do it or am I doing something wrong when exporting the jar? Any help would be appreciated!
Cheers
Nick
Make it simple, and use a startup script (.bat/.sh file) to run your application. This startup script will get the path of its own location in the filesystem, and pass it as an argument or system property to the Java application. This has the additional advantage of being able to pass other arguments, like the size of the heap, etc.
On windows, %~dp0 is the path of the directory containing the executed bat file. On Unix, you can use $(dirname $0).
You could store all config files as resources in a jar, and copy them to files in home.dir + ".AppName/".
Painful as it is, the Preferences API, Preferences.systemNodeForPackage, seems the wisest alternative, if there is little structured config data. There is an inputStream method for import; your initial config template could be a resource in the jar.
Just get the class path using System.getProperty("java.class.path") and scan it for your ".jar" name. Note that path separators are OS dependent (File.pathSeparator)
Related
This likely isn't possible, but I thought I would ask just to be sure. I have a batch file which starts my java app using relative paths. Ie. i have:
Application\start.bat
Application\lib*.jar
My application creates a configuration file in the Application directory. My batch script uses relative paths to point to the lib directory jars. This has worked really well for me because I can move the program wherever I want and it will just run. However now I would like to be able to call the same app from command line as well not just from a shortcut which has the working directory set to Application. The problem is that I want to be able to call my application from any directory on the command line and right now this doesn't work because the working directory will be different.
I know I can always add another parameter to my app for the working directory but then I still have to create a batch script with a hard coded path to my application. Is there any way around this in Java, for example to get the directory that my main java file is in on top of the working directory? Is there a launcher app perhaps?
I can't bundle my app as a jar because it creates configuration files which I want to be in the same directory as the application.
Consider just changing current dir in start.bat:
#cd /d %~dp0
java ...
This would change it to the folder where script is located.
pushd/popd commands can also be used to preserve current dir for calling script if needed.
Alternatively getClass().getProtectionDomain().getCodeSource().getLocation() can be used from java to get path to jar/classes.
There is also path-independent approach with config path system property:
java "-DconfigDir=%~dp0" ...
There is no portable solution that is going to work in all cases. Indeed, the notion of an application directory does not necessarily make sense; e.g. consider applets and applications launched using WebStart.
If the notion of an application or installation directory does make sense, then the simplest way to support it is to have the application's launcher / launch script pass it to the application Java code via an application specific -D property.
I have written a quick Java wrapper that fires off the NMAP executable and waits for it to finish. I am using Eclipse. I would like to include the NMAP source/executable
in my Java jar file so that I have a self-contained package.
Within Eclipse I have added the NMAP folder hierarchy. Within Eclipse I can see Java firing off the NMAP utility correctly, and waiting for the utility to end before exiting.
So far, so good. I then export a JAR file for with eclipse. I can see the NMAP folder hierarchy present. However when I try to run the JAR file it is having trouble finding nmap.exe. Can a foreign executable be called from with a jar file? if not, what are
my options? If so, why can't it find it within the jar file when it could find it within Eclipse?
Thanks.
You will need extract the .exe and its required support files onto the disk before you can access them as regular files (which execution through standard methods requires, I believe). You can look up examples of how to copy a resource from a jar to a file using getResourceAsStream() on one of your class files in the jar.
Once extracted, you can execute is as you are doing now (you might need to ensure execution rights, etc. based on your OS)
The "execute native program" facility does not understand how to invoke EXE-files inside other files (like ZIP or JAR).
If you want to do this, you must extract the files to a file system location and invoke it there. Due to the diversity of Linux distributions (PowerPC? other library versions etc) you should probably ask the user to install it instead and invoke that instead of bringing your own.
To my knowledge, you cannot execute an executable embedded in a jar file.
One solution would be to embed the executable in the jar file. Then use getClass().getResourceAsStream("/path/to/executable") to retrieve the bytes and output them to a temporary file (File.createTempFile()). On UN*X system, you will have to chmod u+x file before trying to execute. Eventually, you could delete the temp file or create the file once and reuse it everytime and call deleteOnExit().
Of course, this solution implies that you have executable(s) that work on all platforms.
Your solution probably works in eclipse because your "executable" file is not in a jar.
You may also have to be careful of how you are distributing this because Nmap isn't free for use in commercial software.
There is an open source library for Java to call Nmap but it assumes that Nmap is installed on the OS on which you are running your code. That API is Nmap4j and it is on sourceforge.net.
At the moment, in my java program, I am accessing a file that is in the project folder. When I'm loading the file its path is "./src/package/package/file.txt". When I build the program into an executable it dosen't work.
I would prefer the files to be outside of the .jar, but in the same folder, how would I got about this?
Samishal
You can use a relative path if they are in the same folder, such as ./file.txt. That should carry over even with a compiled JAR.
Otherwise, if you're going to be using the same machine and are confident of the placement of the files, you could use an absolute path, however I don't recommend it.
You can use Class.getResourceAsStream to get the InputStream. It works for jar package too. It is not possible to access the file in jar file using File API directly.
InputStream ins = Class.getResourceAsStream("/gigadot/exp/resource.properties");
More details at http://blog.gigadot.net/2010/10/loading-classpath-resources.html
Because the relative path is relative to where your command prompt is when you execute the program, not from where the program lives.
You somehow need to tell the program where to find resources. Most people use a .sh/.bat script to figure out where the script itself lives, and either pass -D flags or set the classpath based on that location.
as a note, I $0 gives you the script as it was run on the command line in linux (it could be relative or absolute), and you can use dirname from there to find it's directory, and alter the java command line.
in windows %~dp0 gives you the directory of the batch script which was run, and you can use that to form your java command line.
Im looking for a solution to create a new folder in the folder, where the .jar file is located.
So if i have a test.jer file in C:\test\temp I would like to be able to create a new folder (newFolder) : C:\test\temp\newFolder.
Anyone know how to do this.
I have tried to use
System.getProperty("user.dir").
But this only gives me the correct, if i double click the jar file.
I need it to be run from a scheduler and then the location is changed.
In general you cannot easily figure out where your JAR file is at runtime. The execution environment of your VM can tell you the user directory (as you said), but that is the directory from which the VM was started. So if you double click your JAR, that's actually the directory where the JAR is. But if you call java on the commandline, for example
c:\some\directory> java -cp c:\temp\test.jar main.Class
then "user.dir" will contain "c:\some\directory".
This lack of information is intentional, because it allows the VM to transparently load classes from different locations.
Your best choice to do what you want would be to pass the directory of choice as a commandline argument. And in the worst case, you could use "user.dir" as default. But what the best solution really is depends on why you want to place the directory where the JAR is.
Without learning new programing languages, can we using Java get .exe (executable windows file) software directly? And is there away to make .jar (Java ARchive) software transform to.exe (executable windows file)?
Would this conversion effect the performance of the software?
One of the important points of Java is that it'll run on any platform (e.g. Windows, Linux, etc) that has a suitable Java Virtual Machine (JVM) installed. A .exe file is compiled to work on only one platform.
What you think you want is to turn a .jar into an .exe so you can launch it easily. What I expect you really want is just an easy way of launching a Java app that anybody can understand, including your parents.
The easiest way of doing this is to create a Windows batch file that launches your Java app. One of line of script, one new file, one new instruction to your users - "Double click on runme.bat"
There are ways of turning a .jar into an .exe, but you should think about whether that's really what you want, and if so, why.
Note: you can launch a .jar in Windows by just double clicking on it, as long as the main class is specified in the manifest. You might want to look into tools like Apache Ant to simplify building .jars and their manifests. Just a thought.
EDIT:
A batch file in Windows is a simple text file that contains commands. You can test the commands by running them in the command prompt (Start -> Run -> cmd). When you run a batch file the commands in it are just fed to the command prompt one at a time.
How to run a Jar from the command prompt: "java -jar myfile.jar"
If you create a batch file (.bat - use Notepad or your favourite text editor) containing "java -jar myfile.jar" (without the quotes!) then double-clicking it will launch the main class specified in the manifest of myfile.jar. myfile.jar has to be in the same folder as the batch file, though.
If you use the command "java -jar lib\myfile.jar" (again, without the quotes) then your batch file will run myfile.jar, which needs to be in a folder called lib that's in the same folder as the batch file. Use this approach when you have a whole load of Jars with your application and you don't want to shove them in the user's face :)
Note that moving the batch file will break it, unless it uses absolute paths to the jar file - e.g. "java -jar C:\dev\myfile.jar". Of course, if you use absolute paths then moving the Jar will break the batch file anyway :)
Also, note that you should be able to run a Jar file just by doubling clicking on it in Windows, as long as the main class is specified in the manifest. Give it a try. Of course, convincing your users that they can double click on it is another matter entirely...
As a final note, if you use a batch file, your users will get a nice command prompt window sitting in the background until your Java app closes. That is, unless you start your batch file command with "start". E.g. "start java -jar myfile.jar". If you have your app configured to log to System.out or System.err, you will still get a command prompt when your app writes to either of those streams, though.
Final final note, in the Linux world the equivalent of batch files are shell scripts.
You can launch a .exe from java with Runtime.exec() or ProcessBuilder.
You can make a .exe launcher with http://launch4j.sourceforge.net/
Yes, it is possible: http://www.excelsior-usa.com/articles/java-to-exe.html
I'm not sure if this is an appropriate answer because I don't work with Java but could you not using a shortcut file pointing to the java runtime and pass the jar as an argument, this
I realise this is not the only way, nor probably the best way but to me it seams a little better than creating a batch or converting to an exe file if all you're looking for is a convienient way to launch a java app
The very simplest way is to write a windows batch (.bat) file which will start the application.
I was using a wrapper for java to make exe files for a pretty long time:
JStart32
It is just a wrapper for *.jar files.
But ask yourself:
Wouldn't an exe kill the purpose of javas platform independency?
Check out jsmooth (free) and exe4j and you might want to read make your swing app go native