Question

I am trying to write an Android application that will take a picture, put the data (byte[]) in an object along with some metadata, and post that to an AppEngine server where it will get persisted in a datastore as a blob. I don’t really want to save the image as a file on Android (unless it’s absolutely necessary). I searched around for solutions but nothing clear or specific enough came up. My questions are:

1. how do I post the object to my servlet? Specifically how to properly serialize the object and get the serialized output to an HttpPost or something similar.
2. once I persist the blob in the datastore, how will I be able to retrieve it as an image to display on a webpage?

Code examples would be very helpful. Also, if the approach I am taking is too complicated or problematic, please suggest other approaches (e.g. saving image as file, posting to server, then deleting).

Answer 1

You just need to do a Http-FileUpload which in a special case of a POST.
There is no need to uuencode the file.
No need to use a special lib/jar
No need to save the object to disk (regardless that the following example is doing so)

You find a very good explanation of Http-Command and as your special focus “file upload” under

Using java.net.URLConnection to fire and handle HTTP requests

The file upload sample from there follows (watch “send binary file”)
and it is possible to add some companion data either

---

String param = "value";
File textFile = new File("/path/to/file.txt");
File binaryFile = new File("/path/to/file.bin");
String boundary = Long.toHexString(System.currentTimeMillis()); // Just generate some unique random value.
String CRLF = "\r\n"; // Line separator required by multipart/form-data.

URLConnection connection = new URL(url).openConnection();
connection.setDoOutput(true);
connection.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);
PrintWriter writer = null;
try {
    OutputStream output = connection.getOutputStream();
    writer = new PrintWriter(new OutputStreamWriter(output, charset), true); // true = autoFlush, important!

    // Send normal param.
    writer.append("--" + boundary).append(CRLF);
    writer.append("Content-Disposition: form-data; name=\"param\"").append(CRLF);
    writer.append("Content-Type: text/plain; charset=" + charset).append(CRLF);
    writer.append(CRLF);
    writer.append(param).append(CRLF).flush();

    // Send text file.
    writer.append("--" + boundary).append(CRLF);
    writer.append("Content-Disposition: form-data; name=\"textFile\"; filename=\"" + textFile.getName() + "\"").append(CRLF);
    writer.append("Content-Type: text/plain; charset=" + charset).append(CRLF);
    writer.append(CRLF).flush();
    BufferedReader reader = null;
    try {
        reader = new BufferedReader(new InputStreamReader(new FileInputStream(textFile), charset));
        for (String line; (line = reader.readLine()) != null;) {
            writer.append(line).append(CRLF);
        }
    } finally {
        if (reader != null) try { reader.close(); } catch (IOException logOrIgnore) {}
    }
    writer.flush();

    // Send binary file.
    writer.append("--" + boundary).append(CRLF);
    writer.append("Content-Disposition: form-data; name=\"binaryFile\"; filename=\"" + binaryFile.getName() + "\"").append(CRLF);
    writer.append("Content-Type: " +     URLConnection.guessContentTypeFromName(binaryFile.getName()).append(CRLF);
    writer.append("Content-Transfer-Encoding: binary").append(CRLF);
    writer.append(CRLF).flush();
    InputStream input = null;
    try {
        input = new FileInputStream(binaryFile);
        byte[] buffer = new byte[1024];
        for (int length = 0; (length = input.read(buffer)) > 0;) {
            output.write(buffer, 0, length);
        }
        output.flush(); // Important! Output cannot be closed. Close of writer will close output as well.
    } finally {
        if (input != null) try { input.close(); } catch (IOException logOrIgnore) {}
    }
    writer.append(CRLF).flush(); // CRLF is important! It indicates end of binary boundary.

    // End of multipart/form-data.
    writer.append("--" + boundary + "--").append(CRLF);
} finally {
    if (writer != null) writer.close();
}

---

Regarding the second part of your question. When successful uploading the file (I use apache common files), it is not a big deal to deliver a blob as an image.

This is how to accept a file in a servlet

public void doPost(HttpServletRequest pRequest, HttpServletResponse pResponse)
        throws ServletException, IOException {
    ServletFileUpload upload = new ServletFileUpload();

    try {
        FileItemIterator iter = upload.getItemIterator (pRequest);

        while (iter.hasNext()) {
            FileItemStream item = iter.next();

            String fieldName = item.getFieldName();

                InputStream stream = item.openStream();
            ....
                stream.close();
            }
    ...

And this code delivers an image

public void doGet (HttpServletRequest pRequest, HttpServletResponse pResponse)  throws IOException {

    try {
        Blob img = (Blob) entity.getProperty(propImg);

        pResponse.addHeader ("Content-Disposition", "attachment; filename=abc.png");
        pResponse.addHeader ("Cache-Control", "max-age=120");

        String enc = "image/png";
        pResponse.setContentType (enc);
        pResponse.setContentLength (img.getBytes().length);
        OutputStream out = pResponse.getOutputStream ();
        out.write (img.getBytes());
        out.close();

I hope this code fragments help to answer your questions