I have written a Java program which I package and run from a JAR file. I need to have some user-changeable configuration files which are simply text lines of:
key = value
format. To load these files I used the class described here. When I run my program through Netbeans IDE all works fine as I have included the directory where I store the configuration files in the Project properties.
The problem comes when I build my application into a JAR file. As I want the configuration files to be user-editable I keep them OUTSIDE of the JAR but in the same directory but now when I run my application from the command line it cannot find the configuration files. If I manually add the files to JAR file at the ROOT folder then all is well.
So how can I tell Java to look outside of the JAR for my loadable files? The -classpath option has no effect.
That's because the way you are loading them requires that they be inside the .jar when running from a jar, or inside the project directory if not; it's relying on the classloader to tell it where to find the file.
If you want to open a file outside the .jar, you need to just open it as a File and read it in.
One of the ways we've approached this is to take the external filename as an option on the command line (e.g. java -jar myJar.jar -f filename). This allows you to explicitly state where the file is located. You can then decide whether or not to also look in a default location, or inside the .jar if the file isn't specified on the command line.
I resolved it by referring to this question. I Added the current directory to the MANIFEST file of the jar and it works.
Why is the -classpath option ignored in this case I wonder? Security?
I had the same problem and saw your post, but the answer in the end, was simple.
I have an application deployed via Java Webstart and am building it in Netbeans 7.3.
I have a properties file config.xml that will be updated during run time with user preferences, for instance, "remember my password".
Hence it needs to be external to the jar file.
Netbeans creates a 'dist' folder under the project folder. This folder contains the project jar file and jnlp file. I copied over the config.xml to the dist folder and the properties file was loaded using standard
FileInputStream in = new FileInputStream("config.xml");
testData.loadFromXML(in);
in.close();
Related
Lets say that I built a GUI Application using NetBeans. To run this java application I need to open source code in IDE and then run. I know that I can also run through command prompt.
But how do I start the application independent of IDE. Isn't there some .exe file or something like that, which on double clicking directly runs the application?
If not, how do I generate such a file?
Here you can find how to create .jar in Netbeans: How to create a Jar file in Netbeans
You can run the executable jar on every single computer, on one condition - the system have JRE installed.
If you want to, you can also build the .jar using command line, to do that use the following command:
jar cf jar-file input-file(s)
Description from Oracle doc:
The options and arguments used in this command are:
The c option indicates that you want to create a JAR file. The f
option indicates that you want the output to go to a file rather than
to stdout. jar-file is the name that you want the resulting JAR file
to have. You can use any filename for a JAR file. By convention, JAR
filenames are given a .jar extension, though this is not required. The
input-file(s) argument is a space-separated list of one or more files
that you want to include in your JAR file. The input-file(s) argument
can contain the wildcard * symbol. If any of the "input-files" are
directories, the contents of those directories are added to the JAR
archive recursively. The c and f options can appear in either order,
but there must not be any space between them.
This command will generate a compressed JAR file and place it in the
current directory. The command will also generate a default manifest
file for the JAR archive.
After you build your application look for a folder named "dist" in your project's folder. You should find there a file *.jar which can be run anywhere with double click.
STEPS TO FOLLOW:
create a jar
run the jar
I've been wanting to make executable jar files with java lately. When executing my code with Eclipse it works perfectly. But when I use Eclipse to export the same code as a runnable jar, Most of my jars work except the ones that draw from separate source folders.
The jar will be made but when launched it will try and open and then just say to check to console for possible errors. I try and run the jar through the console with the command "java -jar test.jar". and It says it cannot access the jar. Any Ideas? Btw Im on a macbook pro osX. Thank you!!
picture of where my files are within eclipse
If you have a file you want to store in a jar and access from there, you don't really have a Java File any more. Look at Class.getResourceAsStream() and Class.getResource() - the first can give you an InputStream to the (used-to-be) file, the second returns a URL and can be used for things like images. Note that the file being accessed can be accessed relative to the package/folder location of the class or relative to a classpath root (by putting "/" at the front of the resource name ("/resource/funny.jpg")).
When you execute the jar from a command line, be aware that you have a thing called the "default directory"; it is a folder in which your commands execute by default. If your jar is not in the default directory, you have to specify a valid folder path to your jar to execute it.
First of all, I have read through many S.O. questions regarding this topic and I have tried the things suggested in them.
Here is my situation. I am writing a Java app using the Processing framework and I'm in the final stages where I need to begin thinking about packaging the app. A jar file that is executable from the command line is what I'm attempting to build using the Export feature in Eclipse.
The structure of my project looks like this:
src/
multiple packages/
libs/
jar files and natives
data/
fonts and images
config/
json files
When I export the jar file and uzip the jar to inspect it's contents, I find that the contents of these dirs have been dumped in the top level of the .jar.
Which looks like this:
.jar
packages
jar files
fonts
json files
So, when I attempt to load a config file with something like:
BufferedReader reader = new BufferedReader( new FileReader( path ) );
Everything works just file when I run the app in Eclipse. But the jar file throws a FileNotFoundException.
Many of the questions that I've seen on S.O. regarding problems like these recommend using using class.getClass().getResource() or class.getResourceAsStream(). I have tried both of these using relative paths and just the file name as in:
class.getResource( 'config.json' );
class.getResources( 'cfg/config.json' );
class.getResourceAsStream( '../../config.json' );
All of these methods return null, when run from either Eclipse or the jar using:
java -jar myjarfile.jar
I am also open to using an Ant file. In fact, I'm now using the Ant file generated by the export feature to build the jar. If there is something I can add to that to add the directories into the jar that would be great too.
To reach resources in the root, prepend a / to the path. If not, you do it relative to the current package, which is usually only useful if the resource is physically next to the class in your sources too.
So use class.getResourceAsStream("/config.json"); to reach config.json in the root of the jar.
Also note that jars-inside-jars are not directly supported.
Your .jar file should just include the directories related to the "package" for the compiled code. You might be referencing a .war structure with /lib /WEB-INF etc. and this is different.
If your package structure is:
com.yourco.authentication
And your class in Login
Then your jar should be
/com/
/yourco/
/authentication/
Login.class
you then need the .jar in your classpath for the env to run via command line.
I see you note it works in Eclipse which likely has environment settings and imported required libs, etc. so hard to tell exactly. If your packages/ folder includes the compiled java code, I'm unsure if that'll work when referenced externally, thus suggesting you start your packages in the root folder.
I have written my java codes using eclipse and iam reading a file and processing the codes. The key.txt is present in the same folder as src .
BufferedReader br= new BufferedReader(new FileReader("KEY.txt"));
Now after i compile the jar and place both the jar and file in the same folder and execute iam getting FilenotFoundException. The program works if i run inside Eclipse reading the file perfectly.
The jar file and the key.txt always have to be in the same folder.
Please let me know how can I solve this issue of FilenotFoundException.
The code you have opens a file in the current working directory, which is the directory where you started the Java process from.
This is completely unrelated to where the jar file is located.
How do you execute the jar file?
If the key file is right next to the jar file, this should work:
java -jar theJar.jar
But this will not (because the path to the key is now "test/KEY.txt" ):
java -jar test/theJar.jar
When you run a program in Eclipse, the current working directory is (usually) the project root folder.
An alternative to consider (if the file does not need to be edited by the user) is to put the key file into the jar file. Then it will not get lost, and you can load it via the classloader.
Another good option is to have the user provide the path to the file, via command line parameter or system property.
Or make a batch file / shell script that makes sure that you are always running from the proper directory.
I think you should put these files with class file of the running class, because the current directory is the directory in which the class file is stored, not the source java file.
I have a java desktop app (main project) and another project with a series of packages in NetBeans. Some of the packages use spring for JDBC and IOC.
I am getting the following error when running in debug:
Caused by: java.io.FileNotFoundException: class path resource [config.xml] cannot be opened because it does not exist
Where is the config file supposed to go? Where exactly is the class path? Is it in dist, build, in the root of the project that calls spring, or the main project (the desktop app)?
confused ..
Your classpath is defined when you run your app using the java command. You can specify it using:
java -cp $path my.Main
where $path is your classpath. It is a :-separated (; on windows) list of JAR files and/or directories containing compiled .class files.
If you run your program like:
java -cp configdir my.Main
And put your spring config in configdir (the fully-qualified path) then that should be discovered.
NetBeans: whilst I'm not a netbeans user, it probably offers a number of ways for you to complete the task you want:
In your run configuration (i.e. where you define what class is being run, what the command-line parameters are etc), you will probably be able to add items to the classpath. These might be directories or individual files
In your compiler settings, you can probably tell NetBeans to automatically copy files of a certain type (like properties files, XML config files) from your source locations to where NetBeans puts your class files.
If you put your config.xml file in the directory where NetBeans is compiling your .class files to
Put it in the root folder of you application
if you created your application in a folder called Spring then you should put your file in that folder
Disregard the answer by oxbow_lakes. NetBeans modifies CLASSPATH, so what it is outside the IDE is no measure of what it is inside the IDE.