Question

I am trying to use ZipInputStream to put each file inside the archive in an ArrayList. Can I do this with ZipInputStream?

My main goal is to unzip cbr/cbz files (archives containing only images (jpg/png)), I am trying to put each of these images on a ArrayList so the ZipInputStream to ArrayList was my plan to eventually get them to Bitmaps, but if you can get them to Bitmaps straight from the ZipInputStream would be great!

Answer 1

In the end, doing what I originally planned took too much memory! Instead I ended up just taking one ZipEntry at a time, but only the one I wanted that way it wouldn’t have to loop through each one everytime.

public Bitmap getBitmapFromZip(final String zipFilePath, final String imageFileInZip){
    Bitmap result = null;
try {
    ZipEntry ze = zipfile.getEntry(imageFileInZip);
    InputStream in = zipfile.getInputStream(ze);
    result = BitmapFactory.decodeStream(in);
} catch (IOException e) {
    e.printStackTrace();
} catch (Exception e) {
    e.printStackTrace();
}
return result;

}

Just had to quickly loop through at the beginning to get all the names

public ArrayList<String> unzip() { 
    ArrayList<String> fnames = ArrayList<String>();
    try  { 
        FileInputStream fin = new FileInputStream(_zipFile); 
        ZipInputStream zin = new ZipInputStream(fin); 
        ZipEntry ze = null; 
        while ((ze = zin.getNextEntry()) != null) { 

            if(ze.isDirectory()) { 
            } else { 
            fnames.add(ze.getName()/*fname[fname.length - 1]*/);
            zin.closeEntry(); 
            }          
        } 
        zin.close(); 
    } catch(Exception e) { 
    } 
    return fnames;
}