Develop Reference

How to check if file exist after build jlink?

I am trying to check if file exists in project. When I start my application through Intellij idea and check if file exists it's return true but when I create jlink build and start it through .bat and check if file exists it's always return false.
public class App {
public static String getPathToDllTest(String filename) throws URISyntaxException {
return Paths.get(App.class.getResource("files/" + filename + ".txt").toURI()).toString();
}
public static void main(String[] args) throws Exception {
File txt = new File(App.getPathToDllTest("test"));
System.out.println(txt.exists());
}
}
Here is basic Maven project resources structure I use:
src
main
java
com
example
App.java
resources
com
example
files
test.txt
Is there problem in path? How can I fix it?

Java - URI is not hierarchical error appears only in the exported JAR, not in Eclipse

I have a Java program that is able to change the wallpaper taking in input an image using WINAPI.
Everything works fine when I run it inside Eclipse IDE, but when I run the JAR I got the error:
Caused by: java.lang.IllegalArgumentException: URI is not hierarchical
public class Main {
//INIT USER32 for WINAPI
public static interface User32 extends Library {
User32 INSTANCE = (User32) Native.loadLibrary("user32",User32.class,W32APIOptions.DEFAULT_OPTIONS);
boolean SystemParametersInfo (int one, int two, String s ,int three);
}
public static void main(String[] args) throws IOException, URISyntaxException {
//Change wallpaper
System.out.println("Change wallpaper");
URL url = Main.class.getResource("/resources/img.jpg");
File f = new File(url.toURI());
String path = f.getPath();
User32.INSTANCE.SystemParametersInfo(0x0014, 0, path , 1);
}
}
The image is shipped within the JAR, so maybe the error is related to this since the program is not able to correctly read to URL inside the JAR.
Is there a way to solve this?

A jar file is just a compressed file when the resource is bundled as a jar java will be treated as a single file, which means it will not access to your resources.
try using this instead getResourceAsStream(...);

ImportError: No Module named demiso in Jython

I'm new to the Jython and I'm trying to run a python class using Jython in Java. But I'm running into some issues.
The Java class that I've defined:
public class DemistoCalls {
PythonInterpreter interpreter = null;
public DemistoCalls()
{
PythonInterpreter.initialize(System.getProperties(),
System.getProperties(), new String[0]);
this.interpreter = new PythonInterpreter();
}
void execfile( final String fileName )
{
this.interpreter.execfile(fileName);
}
PyInstance createClass( final String className, final String opts )
{
return (PyInstance) this.interpreter.eval(className + "(" + opts + ")");
}
public static void main(String[] args) {
DemistoCalls demistoCalls = new DemistoCalls();
demistoCalls.execfile("C:\\Users\\AlokNath\\Desktop\\Demisto_Project\\demisto-py-master\\demisto\\SimpleConnect.py");
}
}
The SampleConnect.py file that I'm trying to run:
import sys
sys.path.append("C:\Users\AlokNath\Desktop\Demisto_Project\demisto-py-
master\demisto")
import demisto
While Running the java file, I'm getting this error:
File "C:\Users\AlokNath\Desktop\Demisto_Project\demisto-py-master\demisto\SimpleConnect.py", line 3, in <module>
import demisto
ImportError: No module named demisto
Although I've defined the "demisto" module in the system path and checked that the system path in python contains the appropriate path to Jython 2.7.lb2 Library. I'm not sure where am I going wrong. Any help is appreciated.
Regards,
Alok

I found a solution to the import error problem. We need to copy the missing module to the “site-packages” folder under "modeler-installation/lib/jython/Lib". This will resolve the dependency problem.

How to get path to class file in eclipse and linux in java?

In eclipse for windows, when I run
public class HelloWorld {
public static void main(String[] args) {
System.out.println(System.getProperty("user.dir"));
}
}
It gives me the path of the project root folder (which contains the bin folder which has the class file). For example
SampleProject
and the class file is actually located at
SampleProject\bin\myclass.class
But if I run the same program in linux with
javac myclass.java
java myclass
it gives me the directory that has the .class file, which is the same as pwd command. This is what I want in eclipse for windows. I want some code that will give me the path to the class file in both eclipse for windows and linux.
Does anyone know how do this?
Thanks

If I understand you correctly, you'd like a method that retrieves a class' path on disk. This is easily achievable, like so:
public String getClassPath(Class c) {
try {
return c.getProtectionDomain().getCodeSource().getLocation().toURI().getPath();
} catch (URISyntaxException e) {
e.printStackTrace();
}
return null;
}
NOTE this will work even if the class is contained in a jar file. It will return the path to the jar in this case.

The easiest way is to do this:
public class HelloWorld {
public static void main(String[] args) {
System.out.println(HelloWorld.class.getResource("HelloWorld.class"));
}
}

Load a file from a specific directory

I have a specified URL that i cant change i.e.
/opt/local/java/config/npvr.properties
where should i place my file so that the following code can work:
String PROPERTIESFILEPATH1 = "/opt/local/java/config/npvr.properties";
File tmPropertiesFile = new File(PROPERTIESFILEPATH);
Properties properties = new Properties();
if (tmPropertiesFile.exists()) {
.....
}
i have tried placing my file in directory shown below but it didn't work:
My problem is that I can change only the location of property-file without changing the code to solve this problem. Please Help.

The file needs to go in /opt/local/java/config, not [projectdir]/opt/local/java/config. You are putting it in the wrong place

If you are using linux then add this file /opt/local/java/config/ directory location outside of your project.
"OR"
in windows C:\opt\local\java\config\ here.

use getClass().getResourceAsStream to load property file from relative of class
public class Main {
public static void main(String args[]) throws Exception{
String PROPERTIESFILEPATH = "/opt/local/java/config/npvr.properties";
//File tmPropertiesFile = new File(PROPERTIESFILEPATH);
Properties properties = new Properties();
InputStream ins=null;
//ins=new FileInputStream(PROPERTIESFILEPATH);
ins=new Main().getClass().getResourceAsStream(PROPERTIESFILEPATH);
properties.load(ins);
System.out.println(properties.get("Hello"));
}
}