I'm trying to run my Java program using a batch file, I'm able to run it properly. However, when I run the batch file after inserting the code to read a properties file from the Java program, I'm getting the following error.
Can't find bundle for base name app1, locale en_US
Actually i have a conf folder under which I have this properties folder, then I came to know that I need to keep this conf folder in the class path. But I have actually added it as class folder using Eclipse. However, I'm getting the same error. Please let me know what exactly I need to do for running the Java program using batch file. Using Eclipse I'm able to run the properly.
Thanks,
Balaji.
From the comments your batch script:
#echo off
java -Djava.ext.dirs=lib -classpath ./bin com.myapp.app1.demoprogram
pause
Notice how the /conf/ directory is not listed on the classpath. The easiest way to get it there is to just add it to the -classpath argument being passed to the JVM. Something closer to:
#echo off
java -Djava.ext.dirs=lib -classpath ./bin;./conf com.myapp.app1.demoprogram
pause
This is assuming that /conf/ is in the same directory as bin. You might have to do some tweaking to get the setup to work for you application, but the root problem is that while you added the /conf/ folder to the project classpath in eclipse, you need to do the same thing for the batch script so the JVM can find it
Related
Currently, there is bat file that calls the main class in a jar file. Now I want to run it using eclipse. How do I configure the eclipse to run it?
I have tried the Run > External tools > External tools configuration. But I don't know what to type in...
#echo off
set MODULE2_HOME=%~dp0..
set JAVA_HOME=C:/Program Files/Java/jdk1.6.0_71
set CLASSPATH="%MODULE2_HOME%/classes;%MODULE2_HOME%/lib/*;%MODULE2_HOME%/lib/oracle/*;%MODULE2_HOME%/lib/aspose/*;%MODULE_HOME%/aspose/*"
set SETUP_PROPERTIES="%MODULE2_HOME%\conf\setup.properties"
set PATH=%JAVA_HOME%\bin;%PATH%
java -cp %CLASSPATH% -Dsetup.properties=%SETUP_PROPERTIES% com.module.fast.main.Module2Main %*
How do I configure eclipse to run exactly like this command?
Since it is a batch file which will run in Windows. You can go to the command prompt and run the batch file using .bat. You do not need Eclipse. You can run the java class in eclipse. Still if you want to run, you can check the following screenshot.
While configuring, make sure to give the complete path of batch file. If the bat file needs extra commandline params, you can give the batch file location with space and the command line parameters separated with spaces.
It does not work to run the jar file which contains main class, you have to give the working directory location.
I have trained a model using maltparser version 1.8.1, and I can parse text just fine, but I tried to do it from a non-local directory,
java -Xmx6G -jar /Users/me/maltparser-1.8.1/maltparser-1.8.1.jar -c /Users/me/maltparser-1.8.1/configFile -m parse
and I get the following error:
Couldn't find the MaltParser configuration file: /Users/me/maltparser-1.8.1/configFile.mco
However, the file definitely exists. I copied the text from the error:
$ ls /Users/me/maltparser-1.8.1/configFile.mco
/Users/me/maltparser-1.8.1/configFile.mco
The only way that I can get it to work is if I run java in the directory where the configuration file is. Is there something special that I have to do to use a full path for the configuration file?
It would be possible, I suppose, to cd in a shell script to make this work, but I still want to understand why doing it this way doesn't work.
I was facing the same problem, but then I looked into the source code and found out that you need to provide the relative path of the configuration file from the place you are executing the java command. This is because the path where you have executed the program is prepended to the path of the config file you provide as argument, unless you have mentioned the working directory as argument(in this case put the config in the working directory).
I would like to know can I run a jar file from the command, with the jar file using log4j and ojdbc.jar as well.
The 'main' is located in: nmap_logic.jar.
Within the package containing the 'main' is called: "nn.gmap.logic".
I also use 2 external jar files: log4j.jar & ojdbc.jar.
I have tried running:
java -cp "nmap_logic.jar;log4j.jar;ojdbc.jar" nn.gmap.logic.NNmain
And I get an error that the log4j cannot be initialized.
From the Eclipse environment the application runs fine.
Please let me know how should I execute the command properly.
Thanks.
Try to give the full path to the jars. I believe that there is a difference between what you think is your root folder and what Java thinks about it.
Something like java -cp "c:\myjars\nmap_logic.jar;c:\myjars\log4j.jar;c:\myjars\ojdbc.jar" nn.gmap.logic.NNmain
Btw, you can also do the following: java -cp "c:\myjars\*" nn.gmap.logic.NNmain
I am a bit confused about the process to create a .bat file of a java application. I have exported the executable jar using IDE say Application.jar in C: directory. Then I have written two lines in a .txt file as stated below and saved it as .bat file in the same directory where i have my application.jar. But on double click of the .bat file, the application is not getting executed.
.BAT file code
javac Application.java
java -cp . Application
Note: I have also set the JRE and JDK path in my environment variables till bin path in My Computer properties. But it is not working. Can someone suggest me how can I fix this, because I want to execute my code by doubleclickng on a .bat file. It will be nice if someone can provide me every step I need to follow to accomplish this as I havent ever done this before.
Thanks ,
The first line in your batch file is attempting to compile your program !?
The second line is attempting to run the Application.class file.
What you want if you have produced an executable jar file is:
java -jar Application.jar
But you don't really need the batch file at all. If you double click on the jar file and it runs your program then you can just create a shortcut to it.
Your .bat is just fine. When you double click it might be executing and then closes. This is because your program might not have any UI and it isnt waiting for any input. To verify this take a command prompt and then execute your bat file via that.
In other case I assume that you have a java class called Application and you need to run this via a batch file. In that case if the class have a main method then you just need one line in .bat file
java -cp <the path to class file> Application
So you might be using a javac just to take advantage of class path as current directory. So when you say
javac Application.java
java -cp . Application
It compiles the class to current folder and set that as class path and then execute. This is absolutely file as long as the Application.java doesnt have any third party dependency. But in this case again you need not set -cp to . (current directory will be taken as classpath automatically unless otherwise specificed). So below will also work fine.
javac Application.java
java Application
I support Jurgen reply. If you have an executable jar file and a jre in path then double clicking it will run the application. The META-INF folder inside the jar will have a MANIFEST.MF file which uses a property called Main-Class: to specify the main executing class. And on double clicking this class gets executed. However its only useful if you have a UI. Else it'll also have no effect.
In all these context the Application.jar you mentioned is irrelevant. If that is a third party jar that you need to run the you should include that in -cp argument.
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)