Re: properties file again

[James Perkins <jrperkinsjr@xxxxxxxxx>]

Actually, now that you say that it's not. Try this.
public class TestClass2 {
    public static void main(String[] args) {
        InputStream myFileIS = null;
        try {
            myFileIS =
SomeClassInYourZipFile.class.getClassLoader().getResourceAsStream(
                    "boss.properties");
            Properties prop = new Properties();
            prop.load(myFileIS);
            System.out.println("loaded");
            for (Map.Entry<Object, Object> type : prop.entrySet()) {
                System.out.println(type.getKey() + "=" + type.getValue());
            }
        } catch (Exception ex) {
            System.out.println(ex);
        } finally {
            try {
                myFileIS.close();
            } catch (Exception e) {
            }
        }
    }
}

--
James R. Perkins


On Tue, May 5, 2009 at 09:53, Lim Hock-Chai
<Lim.Hock-Chai@xxxxxxxxxxxxxxx>wrote:

> I tested it on my local and here is what I see:
> The FileUtil.class.getClassLoader().getResource( "boss.properties")
> returns a File object with the path as
> "file:\C:\jar\Boss.zip!\boss.properties"
> When I try to create a FileInputStream with this File object, it failed
> with "java.io.FileNotFoundException:
> file:\C:\jar\Boss.zip!\boss.properties (The filename, directory name, or
> volume label syntax is incorrect)".
>
> It seems like FileInputStream is not capable of understanding classpath
> variable and/or not capable of reading a file that is in a zip file.
>
> Below is my test program:
> public class Test2 {
>
>        public static void main(String[] args) {
>
>                File myFile = FileUtil.findFile("boss.properties");
>                try {
>                        FileInputStream myFileIS = new
> FileInputStream(myFile);
>                        Properties prop = new Properties();
>                        prop.load(myFileIS);
>                        System.out.println("loaded");
>
>                } catch (Exception ex) {
>                        System.out.println(ex);
>                }
>                System.out.println(myFile);
>         }
> }
>
> "James Perkins" <jrperkinsjr@xxxxxxxxx> wrote in message
> news:<mailman.5204.1241539916.23468.web400@xxxxxxxxxxxx>...
> > You actually shouldn't have to change your code. The using of
> InputStream
> > was just a suggestion. It's always the best to use the highest level
> object
> > you can. It just helps you change from a FileInputStream to say a
> > StringBufferInputStream without changing other code.
> >
> > If you can debug it, I would check to see what the name of the file is
> that
> > is returned and make sure it's in the JAR or at least your classpath.
> The
> > code looks fine, but my guess would be that it's not found in the
> classpath.
> > --
> > James R. Perkins
> >
> >
> > On Tue, May 5, 2009 at 08:42, Lim Hock-Chai
> > <Lim.Hock-Chai@xxxxxxxxxxxxxxx>wrote:
> >
> > > Thanks James, I was hoping that we do not have to change code for
> this.
> > > My project is to move our J2EE application from jBOSS to WAS6.1.
> > >
> > > Below are relevent code that does the setFile:
> > >
> > >        public static File findFile( String fileName ) {
> > >            File file = null;
> > >        URL url = FileUtil.class.getClassLoader().getResource(
> fileName
> > > );
> > >        if( url != null ) {
> > >            String decodedUrl = null;
> > >            try {
> > >                decodedUrl = URLDecoder.decode( url.getPath(),
> "ASCII"
> > > );
> > >            } catch( UnsupportedEncodingException uee ) {
> > >                uee.printStackTrace();
> > >            }
> > >            file = new File( decodedUrl );
> > >        }
> > >        return file;
> > >        }
> > >
> > >
> > > I'll look to see what the different between FileInputStream in
> > > InputStream.
> > >
> > > Thanks
> --
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