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Petterson Mikael · Guest

Hi,

I have unzipped a lot of jar files into /home/eraonel/tmp_data.
Now I need to package them into one big jar.

My problem is how to get the file name without
'/home/eraonel/tmp_data'-part.

I am using the following code to get the file names:

public static void listAllContent(File dir)throws IOException {
System.out.println(dir.getCanonicalPath());

if (dir.isDirectory()) {
String[] children = dir.list();
for (int i=0; i listAllContent(new File(dir, children[i]));
}
}
}

If I put all file names in a List and create a jar file the package will
also contain home.eraonel.tmp_data.se.....

Is it possible to avoid this?

cheers,

//mikael

Christian.Gruen@gmail.com · Guest

Hi there,
if I got you right, you could just add a statement like..

for (int i=0; i if(children[i].startsWith("/home/eraonel/tmp_data")) continue;
listAllContent(new File(dir, children[i]));
}

-- CG

Petterson Mikael schrieb:

Remi Arntzen · Guest

now if I got you right, you could just do this

String fileName = "[filename]";

zipOutputStream.putNextEntry(new ZipEntry(new
File(fileName).getName()));

zipOutputStream.write... write... write...