Re: properties file again

No problem. I have been down this path before and personally just found the
getResourceAsStream() much easier to use. The other option you can look into
are loading it as a ResourceBundle.

The FileInputStream will read a file in the classpath, but not in a any
compressed file (JAR/ZIP). It expects an actual File object or it will
create it's own File object if you pass the constructor a String. Imagine
FileInputStream as notepad. If you were to open a zip file with notepad it's
not human readable.

--
James R. Perkins


On Tue, May 5, 2009 at 11:41, Lim Hock-Chai
<Lim.Hock-Chai@xxxxxxxxxxxxxxx>wrote:

That work. Thanks James.

Is it safe to say that FileInputStream does not understand classpath and
it does not know how to read a file that is inside of a jar either?

"James Perkins" <jrperkinsjr@xxxxxxxxx> wrote in message
news:<mailman.5245.1241543851.23468.web400@xxxxxxxxxxxx>...

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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