Php was unable to receive data sent by Android using HttpURLConnection

I am trying to send some data from Android to server (database) using OutputStreamWriter, but I cannot see the result in the database and Logcat is ok. I tried the solutions posted on stackoverflow but no luck. Here are my codes: -

private class sentFeedback extends AsyncTask<String,Void,String>{
    @Override
    protected String doInBackground(String... params) {
        if(isNetworkAvailable()){
            try{
                URL feedback = new URL("http:.../setFeedback.php");
                HttpURLConnection conn = (HttpURLConnection) feedback.openConnection();
                conn.setDoOutput(true);
                conn.setRequestMethod("POST");
                OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
                String data = getData(params);
                wr.write(data);
                wr.flush();

            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }else{
            return "No internet connection";
        }
        return null;
    }

    public String getData(String... params) throws UnsupportedEncodingException {

        String data = "id" + "=" + URLEncoder.encode(params[0], "UTF-8");
        data += "&" + "rating" + "=" + URLEncoder.encode(params[1], "UTF-8");
        data += "&" + "message" + "=" + URLEncoder.encode(params[2], "UTF-8");
        return data;
    }
}

      

php code: -

$id = $_POST['id'];
$rating = (double)$_POST['rating'];
$message = $_POST['message'];

$feedbackQuery = "INSERT INTO feedback(message, id, rating) VALUES ('$message', '$id', '$rating')";
$feedbackQueryResult = $mysqli->query($feedbackQuery);

$mysqli->close();

      

I am currently using the 000webhost free account. Is this the reason why I can send data back?

Thanks in advance!

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1 answer


in php file Replace tablenName

with dbName.tableName

in query string, for example if db namemydb

$feedbackQuery = "INSERT INTO mydb.feedback(message, id, rating) VALUES ('$message', '$id', '$rating')";

      



and repeat the test, in one project my problem is solved with this

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