How to use ResourceBundle

Question

I have a piece of code which tries to read a property file like this:
ResourceBundle bundleHandle = ResourceBundle.getBundle("Example");
Here Example.properties is a file located in another directory which is dynamically constructed
during deployment. So I don't want to put it within the package.
My question is: Will the above code work? When running under debugger, this line always fails to
find the Example.properties even it is there?
Thanks you your help!

Answer 1

Here Example.properties is a file located in another directory which is dynamically constructed The following should provide some examples.
     // Method: demonstratePropertyAbsolute()
     * Demonstrates how a property file is used via absolute naming
     * Because the file name starts with a forward slash this
     * will use an absolute path (in the context of the java application)
     private static final
          String MAIN_FILE_ABS = "/BlueGnome/demos/Properties/DemoPropertiesFile.properties";
     static void demonstratePropertyAbsolute() throws Exception
          aProperties = new Properties();
          // Demonstrate what the name is
          // Notice that we use the Properties for the the class
          // reference.  It does not matter what class we use.
          java.net.URL url = Properties.class.getResource(MAIN_FILE_ABS);
          System.out.println("Absolute: resolved name = " + url);
          // Get a stream so we can read it.
          java.io.InputStream is;
          is = Properties.class.getResourceAsStream(MAIN_FILE_ABS);
          if (is == null)
               throw new Exception("Resource File "
                    + MAIN_FILE_ABS + " not found.");
          // Load the stream
          aProperties.load(is);
          System.out.println("  This is a value from the properties file"
                         + " name = <"
                         + aProperties.getProperty("a.value")
                         + ">");
     // Method: demonstratePropertyPath()
     * Demonstrates how a property file is used via path
     * The file name is basically a system file path (unlike the previous two)
     private static final
          String MAIN_FILE_PATH = "./BlueGnome/demos/Properties/DemoPropertiesFile.properties";
     static void demonstratePropertyPath() throws Exception
          aProperties = new Properties();
          // Demonstrate what the name is
          java.net.URL url = ClassLoader.getSystemClassLoader().getResource(MAIN_FILE_PATH);
          System.out.println("PATH: resolved name = " + url);
          // Get the file.
          // We could just use FileInputStream here.  Then we would have
          // to provide a path that was meaningful to the system like
          //     c:\java\BlueGnome\demos\Properties\DemoPropertiesFile.properties
          // Doing it this way allows us to use generic directory
          // seperators and allows for it to search for the file using
          // the CLASSPATH.  That also allows for it to be in a jar file.
          java.io.InputStream is;
          is = ClassLoader.getSystemClassLoader().getResourceAsStream(MAIN_FILE_PATH);
          if (is == null)
               throw new Exception("Resource File "
                    + MAIN_FILE_PATH + " not found.");
          aProperties.load(is);
          System.out.println("  This is a value from the properties file"
                         + " name = <"
                         + aProperties.getProperty("a.value")
                         + ">");
     }