Get a File or URI object for a file inside an archive with Java?

The jar: protocol is a way to build a URI to a resource in a JAR archive: jar:example.com/bar/baz.jar!/path/to/file See the API docs for JarURLConnection: java.sun.com/javase/6/docs/api/java/net/... Between the jar: and! Can be any URL, including a file: URL.

The jar: protocol is a way to build a URI to a resource in a JAR archive: jar:example.com/bar/baz.jar!/path/to/file See the API docs for JarURLConnection: java.sun.com/javase/6/docs/api/java/net/... Between the jar: and! / can be any URL, including a file: URL.

Public List getFilesInJar(String jarName){ List result = new ArrayList(); File jarFile = new File(jarName); JarInputStream jarFile = new JarInputStream(new FileInputStream(jarFile)); JarEntry jarEntry; while ((jarEntry = jarFile.getNextJarEntry())! = null) { result. Add(inputStreamToFile(jarFile.

GetInputStream(jarEntry))); } return result; } for the inputStreamToFile method, google "java inputStream to file", although you might be happy with an InputStream object also, instead of a File object :).

For the actual file data, see ZipFile#getInputStream(ZipEntry). The javadocs for that class explain how it is to be used.

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