从相对路径读取（zip）文件

Question

I am trying to read a zip file, which is inside my plugin project. This is the file hierarchy:

my.super.project
  - Referenced Libraries
  - JRE System Library
  - src
  - binary
  - META-INF
  - resources

Inside the resources directory there is a zip file myarchive.zip which I want to load.

This is how I tried to do it:

byte[] buffer = new byte[1024];
try {
    ZipInputStream zis = new ZipInputStream(new FileInputStream(
            "resources/myarchive.zip"));
    // get the zipped file list entry
    ZipEntry ze = zis.getNextEntry();
    while (ze != null) {
        String fileName = ze.getName();
        File newFile = new File(OUTPUT_DIR
                + File.separator + fileName);
        // create all non exists folders
        new File(newFile.getParent()).mkdirs();
        FileOutputStream fos = new FileOutputStream(newFile);
        int len;
        while ((len = zis.read(buffer)) > 0) {
            fos.write(buffer, 0, len);
        }

        fos.close();
        ze = zis.getNextEntry();
    }
    zis.closeEntry();
    zis.close();
} catch (FileNotFoundException e) {
    e.printStackTrace();
} catch (IOException e) {
    e.printStackTrace();
}

The file cannot be found ( FileNotFoundException ), which means, that something is wrong with the relative path.

What is the problem and how can I read the zip file?

Answer 1

I suggest to:

• detect the location of your current main jar
• from this location determine the installation root
• from the location of the installation root determine the path of your ZIP file.

Also consider to use the NIO.2 File API for both working with paths and working with ZIP files (using the ZIP file system provider ).

Answer 2

Depends on how you are running this, is this a standalone or inside tomcat or such?

Try adding a prefix "/" OR thsi should work

URL url = this.getClass().getResource("/resources/myarchive.zip");
ZipInputStream zis = new ZipInputStream(new FileInputStream( url.getFile() );