Send parameters while uploading file to PHP source

Question

I'm using this class to upload image to php server from my android app.

But I want to send some parameters such as custom file name, user who uploaded the file etc.

Is there any way to send some parameters while uploading the file?

Example: name=newimage&uploadedby=username

private class UploadFileAsync extends AsyncTask<String, Void, String> {

@Override
protected String doInBackground(String... params) {

    try {
        String sourceFileUri = "/mnt/sdcard/abc.png";

        HttpURLConnection conn = null;
        DataOutputStream dos = null;
        String lineEnd = "\r\n";
        String twoHyphens = "--";
        String boundary = "*****";
        int bytesRead, bytesAvailable, bufferSize;
        byte[] buffer;
        int maxBufferSize = 1 * 1024 * 1024;
        File sourceFile = new File(sourceFileUri);

        if (sourceFile.isFile()) {

            try {
                String upLoadServerUri = "http://website.com/abc.php?";

                // open a URL connection to the Servlet
                FileInputStream fileInputStream = new FileInputStream(
                        sourceFile);
                URL url = new URL(upLoadServerUri);

                conn = (HttpURLConnection) url.openConnection();
                conn.setDoInput(true); // Allow Inputs
                conn.setDoOutput(true); // Allow Outputs
                conn.setUseCaches(false); // Don't use a Cached Copy
                conn.setRequestMethod("POST");
                conn.setRequestProperty("Connection", "Keep-Alive");
                conn.setRequestProperty("ENCTYPE",
                        "multipart/form-data");
                conn.setRequestProperty("Content-Type",
                        "multipart/form-data;boundary=" + boundary);
                conn.setRequestProperty("bill", sourceFileUri);

                dos = new DataOutputStream(conn.getOutputStream());

                dos.writeBytes(twoHyphens + boundary + lineEnd);
                dos.writeBytes("Content-Disposition: form-data; name=\"bill\";filename=\""
                        + sourceFileUri + "\"" + lineEnd);

                dos.writeBytes(lineEnd);

                bytesAvailable = fileInputStream.available();

                bufferSize = Math.min(bytesAvailable, maxBufferSize);
                buffer = new byte[bufferSize];

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

                while (bytesRead > 0) {

                    dos.write(buffer, 0, bufferSize);
                    bytesAvailable = fileInputStream.available();
                    bufferSize = Math
                            .min(bytesAvailable, maxBufferSize);
                    bytesRead = fileInputStream.read(buffer, 0,
                            bufferSize);

                }

                dos.writeBytes(lineEnd);
                dos.writeBytes(twoHyphens + boundary + twoHyphens
                        + lineEnd);

                serverResponseCode = conn.getResponseCode();
                String serverResponseMessage = conn
                        .getResponseMessage();

                if (serverResponseCode == 200) {

                    //Toast.makeText(ctx, "File Upload Complete.",
                    //      Toast.LENGTH_SHORT).show();



                }

                fileInputStream.close();
                dos.flush();
                dos.close();

            } catch (Exception e) {

                e.printStackTrace();

            }

        }


    } catch (Exception ex) {

        ex.printStackTrace();
    }
    return "Executed";
}

}

Is there any way to do this with header request, like this:

conn.setRequestProperty("parameters","name=filename&uploadedby=username");

Possible duplicate of Uploading file in php server from android device — Muhammad Omer Aslam
– Muhammad Omer Aslam, Commented Sep 8, 2018 at 21:47

MilMike · Accepted Answer · 2018-09-08 08:29:41Z

1

yes it is possible

I am not a java developer but in PHP you can receive such data in 2 ways: as parameter in the URL which will be available in PHP in the $_GET array Example, change your URL to:

String upLoadServerUri = "http://website.com/abc.php?filename=blabla&anotherparam=1234";

then in PHP: echo $_GET['filename']. $_GET is also available if you are POSTing your request.

or/and you can do this in your POST. Your POST request can contain as many extra params as you need. Here is an example how you could do this.

answered Sep 8, 2018 at 8:29

MilMike

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