在非Maven项目中如何具有等效的主资源？

Question

I was trying to model my project based on this project at Github.

public class DroolsPlannerPoCApp {
....
private static final String   SOLVER_CONFIG    = "/ServiceDeliverySolverConfig.xml";
....
private Solver createSolver() {
        XmlSolverConfigurer configurer = new XmlSolverConfigurer();
        configurer.configure( SOLVER_CONFIG );
        return configurer.buildSolver();
    }

The file mentioned in SOLVER_CONFIG lies in the resources directory ( see here ).

---

I tried to do the same thing .

package in.co.technovia.sudoku;

import java.util.ArrayList;
....
public class App{
    private static final String SOLVER_CONFIG = "/solver.xml";
    public static void main(String[] args){
    SudokuGenerator sg = new SudokuGenerator();
    ....
    }

    private static Solver createSolver(){

    XmlSolverConfigurer configurer = new XmlSolverConfigurer();
        configurer.configure(SOLVER_CONFIG);
        return configurer.buildSolver();
    }
}

And there was fail.

jesvin@Jesvin-Technovia:~/dev/drools/sudoku$ java in.co.technovia.sudoku.AppException in thread "main" java.lang.IllegalArgumentException: The solver configuration (/solver.xml) does not exist.
    at org.drools.planner.config.XmlSolverConfigurer.configure(XmlSolverConfigurer.java:79)
    at in.co.technovia.sudoku.App.createSolver(App.java:67)
    at in.co.technovia.sudoku.App.main(App.java:43)

I understand that main/resources mean something for Maven builds . But I am building my project manually.

Question: How do I implement the same thing (from resources) in my own project? Where must I place the file and what must I specify in the SOLVER_CONFIG?

---

The function itself is implemented as:

public XmlSolverConfigurer configure(String resource) {
    return configure(getClass().getResourceAsStream(resource));
}

Answer 1

Put the resources on the classpath, for example, in the source directory.

All Maven does is merge the src/main/resources package/directory tree into the target classpath, maintaining the same package/folder layout.

Answer 2

如果手动构建，则需要将资源手动复制到类路径中，即，将XML文件复制到生成的类目录的根路径中。