Resource within jar file within jar file

[Fabricio Sanchez]

Hi There,

I have my main jar file containing classes and within that I have another jar file full of images and a library.xml file to manage those images as such:

utilities.jar
- com/esp/textures/JarTextureLibrary.class
- com/esp/textures/textures.jar
- library.xml
- bricks/brick_01.gif
- bricks/brick_02.gif
- etc...

In code I do the following

URL url = com.esp.textures.JarTextureLibrary.class.getResource("textures.jar");

I then modify the URL by:

String urlContext = url.toString() + "!/";
String xmlString = urlContext + "library.xml";
System.out.ptrintln( "xml library location is: " + xmlString );

I then proceed to read the xml file from this newly created xmlString.

Now this doesn't work because of the multiple "!/" that appear on the final resulting URL giving me the following error message:

xml�library�location�is:�jar:file:/C:/current/lib/utilities.jar!/com/esp/textures/textures.jar!/library.xml
JAR entry com/esp/textures/textures.jar!/library.xml not found in C:\current\lib\utilities.jar

Any ideas how I can read not only the library file (library.xml) but also the images contained within it?

Is the only way to extract the textures.jar file locally and then try and access it?

Thank you in advance,

Regards,

Fabricio

[Paul Clapham]

Yes, not having the jar file inside another jar file would make things far easier. And if you made both jar files be in the classpath, you wouldn't have to change your existing code.

[Norm Radder]

Here's an idea. Override/extend the File class with a class that will get its input from memory, say a byte array. Read the nested/contained jar file from the real jar file into the byte array. Then use the ZipFile constructor with your extended File class to process the jar file from memory.