Android image Upload problem

Question

Hello i am wanting to upload image from android emulator to asp.net server. The code below can communicate the server. When I tried to create a text file to see the data sent from android was successful or not . But no the file data didn't send across to the server. I tried sending the plain text to the server but the file I created on the server didn't print the text.

The code here: HttpURLConnection conn = null;

    String boundary = "==============";

        try
        {   
            String disposition = "Content-Disposition: form-data; name=\"userfile\"; filename=\"" + filename + ".jpg\"";
            String contentType = "Content-Type: application/octet-stream";

            String t1   = "Content-Disposition: form-data; name=\"test\";";
            String t2 = "Content-Type: text/plain";

            // This is the standard format for a multipart request
            StringBuffer requestBody = new StringBuffer();
            /*
            requestBody.append("--"+boundary);
            requestBody.append('\n');
            requestBody.append(disposition);
            requestBody.append('\n');
            requestBody.append(contentType);
            requestBody.append('\n');
            requestBody.append('\n');
            requestBody.append(new String(getByteFromStream(stream)));
            */

            requestBody.append('\n');
            requestBody.append('\n');
            requestBody.append("--"+boundary);
            requestBody.append('\n');
            requestBody.append(t1);
            requestBody.append('\n');
            requestBody.append(t2);
            requestBody.append('\n');
            requestBody.append('\n');
            requestBody.append("basdfsdafsadfsad");
            requestBody.append("--"+boundary+"--");

            // Make a connect to the server
            URL url = new URL(targetURL);
            conn = (HttpURLConnection) url.openConnection();

            // Put the authentication details in the request
           /*
             if (username != null) {

                String usernamePassword = username + ":" + password;
                String encodedUsernamePassword = Base64.encodeBytes(usernamePassword.getBytes());
                conn.setRequestProperty ("Authorization", "Basic " + encodedUsernamePassword);
            }
            */
            conn.setDoOutput(true);
            conn.setDoInput(true);
            conn.setUseCaches(false);
            conn.setRequestMethod("POST");
            conn.setRequestProperty("MIME-Version:", "1.0");
            conn.setRequestProperty("Content-Type", "multipart/mixed; boundary=" + boundary);

            // Send the body
            DataOutputStream dataOS = new DataOutputStream(conn.getOutputStream());
            dataOS.writeBytes(requestBody.toString());
            dataOS.flush();
            dataOS.close();

            // Ensure we got the HTTP 200 response code
            int responseCode = conn.getResponseCode();
            if (responseCode != 200) {
                throw new Exception(String.format("Received the response code %d from the URL %s", responseCode, url));
            }

Is my request body layout not correctly ?

Answer 1

Because you upload two files, so you should see the last format. Good luck for you. 6. Examples

Suppose the server supplies the following HTML:

 <FORM ACTION="http://server.dom/cgi/handle"
       ENCTYPE="multipart/form-data"
       METHOD=POST>
 What is your name? <INPUT TYPE=TEXT NAME=submitter>
 What files are you sending? <INPUT TYPE=FILE NAME=pics>
 </FORM>

and the user types "Joe Blow" in the name field, and selects a text file "file1.txt" for the answer to 'What files are you sending?'

The client might send back the following data:

    Content-type: multipart/form-data, boundary=AaB03x

    --AaB03x
    content-disposition: form-data; name="field1"

    Joe Blow
    --AaB03x
    content-disposition: form-data; name="pics"; filename="file1.txt"
    Content-Type: text/plain

     ... contents of file1.txt ...
    --AaB03x--

If the user also indicated an image file "file2.gif" for the answer to 'What files are you sending?', the client might client might send back the following data:

    Content-type: multipart/form-data, boundary=AaB03x

    --AaB03x
    content-disposition: form-data; name="field1"

    Joe Blow
    --AaB03x
    content-disposition: form-data; name="pics"
    Content-type: multipart/mixed, boundary=BbC04y

    --BbC04y
    Content-disposition: attachment; filename="file1.txt"
    Content-Type: text/plain

    ... contents of file1.txt ...
    --BbC04y
    Content-disposition: attachment; filename="file2.gif"
    Content-type: image/gif
    Content-Transfer-Encoding: binary

      ...contents of file2.gif...
    --BbC04y--
    --AaB03x--

Form-based File Upload in HTML

Answer 2

I am using asp.net for as file handler. Below is the simple Android code for the event you will use to upload the file

    String pathToOurFile = "/mnt/sdcard/sysdroid.png";//this will be the file path        String urlServer = "http://yourdomain/fileupload.aspx";
int bytesRead, bytesAvailable, bufferSize;
    byte[] buffer;
    int maxBufferSize = 1*1024*1024;

    try
    {
    FileInputStream fileInputStream = new FileInputStream(new File(pathToOurFile) );

    URL url = new URL(urlServer);
    connection = (HttpURLConnection) url.openConnection();

    // Allow Inputs & Outputs
    connection.setDoInput(true);
    connection.setDoOutput(true);
    connection.setUseCaches(false);

    // Enable POST method
    connection.setRequestMethod("POST");

    connection.setRequestProperty("Connection", "Keep-Alive");
    connection.setRequestProperty("Content-Type",  "multipart/form-data");
    connection.setRequestProperty("SD-FileName", "sysdroid.png");//This will be the file name

    outputStream = new DataOutputStream( connection.getOutputStream() );
    bytesAvailable = fileInputStream.available();
    bufferSize = Math.min(bytesAvailable, maxBufferSize);
    buffer = new byte[bufferSize];

    // Read file
    bytesRead = fileInputStream.read(buffer, 0, bufferSize);

    while (bytesRead > 0)
    {
        outputStream.write(buffer, 0, bufferSize);
        bytesAvailable = fileInputStream.available();
        bufferSize = Math.min(bytesAvailable, maxBufferSize);
        bytesRead = fileInputStream.read(buffer, 0, bufferSize);
    }

     int serverResponseCode = connection.getResponseCode();
     String serverResponseMessage = connection.getResponseMessage();
     Log.d("ServerCode",""+serverResponseCode);
     Log.d("serverResponseMessage",""+serverResponseMessage);
    fileInputStream.close();
    outputStream.flush();
    outputStream.close();


}
    catch (Exception ex)
    {
        //ex.printStackTrace();
        Log.e("Error: ", ex.getMessage());
    }

So far so good. Lets look into the asp.net code. I used simple 'Web Form' for this. The code behind is

protected void Page_Load(object sender, EventArgs e)
{
    string uploadDir = Server.MapPath("~/images");
    string imgPath = Path.Combine(uploadDir, Request.Headers["SD-FileName"]);

    try{          
        byte[]bytes = new byte[Request.InputStream.Length];
        Request.InputStream.Read(bytes, 0, bytes.Length);
        System.IO.MemoryStream ms = new System.IO.MemoryStream(bytes);
        Bitmap btMap = (Bitmap)System.Drawing.Image.FromStream(ms);
        btMap.Save(imgPath, ImageFormat.Jpeg);
        ms.Close();
    }
    catch (Exception exp)
    {            
        Response.Write(exp.Message);
    }     
}

Hope this will work and you have the knowledge of read/write permissions on both Android's SD card and asp.net folders.

cheers Fahar