I am curious if I can take a .jar file and somehow add it to my classpath so that I can run it from any directory. For example, let's say I have a .jar located at /home/setup/someJar.jar. Is there a way I can run this from another directory (preferably any) so that I do not have to navigate back to /home/setup/ whenever I want to run it?
I tried adding the path to my .bash_profile file by adding :/home/setup to the PATH= line, but didn't work.
I think you probably want an 'alias'.
http://www.linfo.org/alias.html
alias myJarShortcut="java -jar /direct/path/to/nameOfYourJar.jar"
Add this to your .bash_profile and it will be available every time you boot up.
It's possible but not so easy. you have 3 possiblities
Make an alias or symbolic link (create symbolic link)
Start it as deamon
start it as service (create service)
after ready the information above
you can start it like this
service [yourservice] start|stop|restart
You could create a bash script that execute your .jar file in the directory of your choice, with of course the right path to your .jar file.
Related
Hello all :) I love stackoverflow where I always find answers but this time I could not so personally asking... Its bit lengthy please go through it.
I am creating a java application where one of my resource is an exe file which I need to call in the java code. But later I would convert the whole java code to a JAR file... and I would add the JAR file and the exe file for the Setup file for installation process. So when I extract the files I want my JAR file to call the exe file while running... I am doing this all in Eclipse :)
So my doubt comes here which path I should put up in the java code... ? So that it will always call the file the exe file from the same directory where the JAR file is also present.. :)
Any help would be great :) Thank you in advance :)
Assuming that you control the whole installation process; and that you also control the script that later starts a JRE to run your JAR within; my suggestion would be: simply use a property here.
In other words; your installer knows that it copied JAR and EXE to D:\example\ for example. Then just make sure that your JAR is started like:
java -jar D:\example\your.jar -D your_path=D:\example
(this is just meant as example, you would have to work that out, probably the \ in there need some special treatment for example)
Then your application can simply query for that system property "your_path" and take the value from there.
Alternatively, you could try this solution that works "pure java".
I am developing a program that uses a sqlite file as a database. When I compile and test my program I haven't any problem:
Conection c = DriverManager.getConnection(
"jdbc:sqlite:/home/mehdi/my_database.sqite");
as you can see in above, the code shows a direct path that set to database file (this path is only in my system).
So it works fine, but my problem starts when I create an executable jar file of my program, if I create executable jar file and share it with other users, when they run the executable jar it doesn't work.
My first question: how do I set my database path for an executable jar file in my code?
My second question: is it possible for the database to be along side the executable jar? (and i can move my executable jar file with its database)
some_path/my_program_file.jar
some_path/my_database.sqite
The best option is externalizing the database path instead of hardcoding it, and letting the user choose where to put the actual sqlite file.
You could put the sqlite file in the same directory as the program archive, but please note that this does not really help in locating the file from your code, because relative paths are definitely resolved against the JVM (process) working directory, not the location of the JAR:
new File(".").getAbsolutePath()
There are ways to get the location of the JAR:
getClass().getProtectionDomain().getCodeSource().getLocation()
but they are deploy-dependent, so it's not a very robust strategy to rely on.
Most Java program are configured with external files put in well-known locations (like the current JVM directory or a .myapp/conf.properties in the user home), and individual properties can be overridden in the starting command line, as either system properties -Dkey=value or program arguments (there's a library to simplify this)
Java uses relative paths when you do not specify the absolute path. I'm not sure if it works for the DriverManager though. What this means is that if you call jdbc:sqlite:my_database.sqite it should look for my_database.sqite in the folder the jar is executed from. So if the user calls java -jar my_program_file.jar from the folder where the database is it will work. If the user calls the jar from another location it won't. To fix that, you have to grab the path where the jar is located, like so:
return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation().getPath());
This was showcased in another SO post.
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)
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.
How do u move a Jar file to the Start up folder with code? Like within the code either make a Jar file or move it to a different directory with Java.
Edit: So basicly i have a Jar file on the Desktop, I want to move the Jar file to or duplicate the jar file and move that to lets say C:\Program Files (x86) When u run the Jar File
I haven't tested this code, but if I remember correctly you want something along the lines of:
Runtime.getRuntime().exec("cp path/to/jar path/to/destination");
Where the string is the appropriate terminal command for your OS and what you want to do. Different methods surely exist, however I believe this is the easiest way. Here are some example commands:
Copying files:
Windows: cp path\to\jar destination\path
Linux: cp path/to/jar destination/path (note: you may need to prefix this command with sudo if the logged in user doesn't have the proper permissions. This can introduce its own headaches, so tread lightly)
Making a JAR archive:
Windows: jar cf path\to\jar path\to\files
Linux: jar cf path/to/jar path/to/files (again, it is possible you may need the sudo prefix)