java zip to binary format and then decompress

Question

I have a task that

1. read a zip file from local into binary message

2. transfer binary message through EMS as String (done by java API)

3. receive transferred binary message as String (done by java API)

4. decompress the binary message and then print it out

The problem I am facing is DataFormatException while decompress the message.

I have no idea which part went wrong.

I use this to read file into binary message:

static String readFile_Stream(String fileName) throws IOException {
    File file = new File(fileName);
    byte[] fileData = new byte[(int) file.length()];
    FileInputStream in = new FileInputStream(file);
    in.read(fileData);
    String content = "";
    System.out.print("Sent message: ");
    for(byte b : fileData)
    {
        System.out.print(getBits(b));
        content += getBits(b);
    }
    in.close();
    return content;
}

static String getBits(byte b)
{
    String result = "";
    for(int i = 0; i < 8; i++)
        result = ((b & (1 << i)) == 0 ? "0" : "1") + result;
    return result;
}

I use this to decompress message:

private static byte[] toByteArray(String input)
{       
    byte[] byteArray = new byte[input.length()/8];
    for (int i=0;i<input.length()/8;i++)
    {
        String read_data = input.substring(i*8, i*8+8); 
        short a = Short.parseShort(read_data, 2);
        byteArray[i] = (byte) a;                    
    }       
    return byteArray;       
}

public static byte[] unzipByteArray(byte[] file) throws IOException {
    byte[] byReturn = null;

    Inflater oInflate = new Inflater(false);
    oInflate.setInput(file);

    ByteArrayOutputStream oZipStream = new ByteArrayOutputStream();
    try {
        while (! oInflate.finished() ){
            byte[] byRead = new byte[4 * 1024];
            int iBytesRead = oInflate.inflate(byRead);
            if (iBytesRead == byRead.length){
                oZipStream.write(byRead);
            }
            else {
                oZipStream.write(byRead, 0, iBytesRead);
            }
        }
        byReturn = oZipStream.toByteArray();
    }
    catch (DataFormatException ex){
        throw new IOException("Attempting to unzip file that is not zipped.");
    }
    finally {
        oZipStream.close();
    }
    return byReturn;
}

The message I got is

java.io.IOException: Attempting to unzip file that is not zipped.
at com.sourcefreak.example.test.TibcoEMSQueueReceiver.unzipByteArray(TibcoEMSQueueReceiver.java:144)
at com.sourcefreak.example.test.TibcoEMSQueueReceiver.main(TibcoEMSQueueReceiver.java:54)

After check, the binary message does not corrupted after transmission. Please help to figure out the problem.

Answer 1

I finally figure out the problem.

The problem is the original file is a .zip file, so I should use zipInputStream to unzip the file before further processing.

public static byte[] unzipByteArray(byte[] file) throws IOException {

    // create a buffer to improve copy performance later.
    byte[] buffer = new byte[2048];
    byte[] content ;

    // open the zip file stream
    InputStream theFile = new ByteArrayInputStream(file);
    ZipInputStream stream = new ZipInputStream(theFile);
    ByteArrayOutputStream output = new ByteArrayOutputStream();

    try
    {

        ZipEntry entry;
        while((entry = stream.getNextEntry())!=null)
        {
            //String s = String.format("Entry: %s len %d added %TD", entry.getName(), entry.getSize(), new Date(entry.getTime()));
            //System.out.println(s);

            // Once we get the entry from the stream, the stream is
            // positioned read to read the raw data, and we keep
            // reading until read returns 0 or less.
            //String outpath = outdir + "/" + entry.getName();

            try
            {
                //output = new FileOutputStream(outpath);

                int len = 0;
                while ((len = stream.read(buffer)) > 0)
                {
                    output.write(buffer, 0, len);
                }
            }
            finally
            {
                // we must always close the output file
                if(output!=null) output.close();                                            
            }
        }            
    }
    finally
    {
        // we must always close the zip file.
        stream.close();            
    }
    content = output.toByteArray();

    return content;     

}

This code work for zip file containing single file inside.