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I've been following a youtube tutorial and have come on to a problem, the comments section really didnt help me so here's my problem. The app has a login button when pressed is supposed to start a new activity. My problem is that when entering the correct username and password, the button doesn't do anything but the error message will be displayed if incorrect information is entered. I am unsure what exactly is causing the problem but a lot of the youtube comments have said its the php code, but I cannot find what's wrong with it.

    final EditText etUsername = (EditText) findViewById(R.id.etUsername);
    final EditText etPassword = (EditText) findViewById(R.id.etPassword);
    final TextView tvRegisterLink = (TextView) findViewById(R.id.tvRegisterLink);
    final Button bLogin = (Button) findViewById(R.id.bSignIn);

    tvRegisterLink.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            Intent registerIntent = new Intent(LoginActivity.this, RegisterActivity.class);
            LoginActivity.this.startActivity(registerIntent);
        }
    });

    bLogin.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            final String username = etUsername.getText().toString();
            final String password = etPassword.getText().toString();

            // Response received from the server
            Response.Listener<String> responseListener = new Response.Listener<String>() {
                @Override
                public void onResponse(String response) {
                    try {
                        JSONObject jsonResponse = new JSONObject(response);
                        boolean success = jsonResponse.getBoolean("success");

                        if (success) {
                            String name = jsonResponse.getString("name");
                            int age = jsonResponse.getInt("age");

                            Intent intent = new Intent(LoginActivity.this, UserAreaActivity.class);

                            intent.putExtra("name", name);
                            intent.putExtra("age", age);
                            intent.putExtra("username", username);
                            startActivity(intent);

                        } else {
                            AlertDialog.Builder builder = new AlertDialog.Builder(LoginActivity.this);
                            builder.setMessage("Login Failed")
                                    .setNegativeButton("Retry", null)
                                    .create()
                                    .show();
                        }

                    } catch (JSONException e) {
                        e.printStackTrace();
                    }
                }
            };

            LoginRequest loginRequest = new LoginRequest(username, password, responseListener);
            RequestQueue queue = Volley.newRequestQueue(LoginActivity.this);
            queue.add(loginRequest);
        }
    });
}

` Now here's the code where the loginActivity uses the loginRequest to connect to the database and compare the filled in fields.

private static final String LOGIN_REQUEST_URL = "http://etc.com/Login2.php";
private Map<String, String> params;

public LoginRequest(String username, String password, Response.Listener<String> listener) {
    super(Method.POST, LOGIN_REQUEST_URL, listener, null);
    params = new HashMap<>();
    params.put("username", username);
    params.put("password", password);
}

@Override
public Map<String, String> getParams() {
    return params;
}

The login button is supposed to bring the user to this page called UserAreaActivity.

 Intent intent = getIntent();
    String name = intent.getStringExtra("name");
    String username = intent.getStringExtra("username");
    int age = intent.getIntExtra("age", -1);

    TextView tvWelcomeMsg = (TextView) findViewById(R.id.tvWelcomeMsg);
    EditText etUsername = (EditText) findViewById(R.id.etUsername);
    EditText etAge = (EditText) findViewById(R.id.etAge);

    // Display user details
    String message = name + " welcome to your user area";
    tvWelcomeMsg.setText(message);
    etUsername.setText(username);
    etAge.setText(age + "");

Here is the php code that the loginRequest will call. `

$username = $_POST["username"];
$password = $_POST["password"];

$statement = mysqli_prepare($con, "SELECT * FROM `user` WHERE username= ? AND password = ?");
mysqli_stmt_bind_param($statement, "ss", $username, $password);
mysqli_stmt_execute($statement);

mysqli_stmt_store_result($statement);
mysqli_stmt_bind_result($statement, $userID, $name, $age, $username, $password);

$response = array();
$response["success"] = false;  

while(mysqli_stmt_fetch($statement)){
    $response["success"] = true;  
    $response["name"] = $name;
    $response["username"] = $username;
    $response["age"] = $age;
    $response["password"] = $password;
}

echo json_encode($response);

`

D. Anders
  • 11
  • 3
  • I bet you have an exception but the e.printstacktrace is not showing it see this answer https://stackoverflow.com/questions/7469316/why-is-exception-printstacktrace-considered-bad-practice – Namphibian Oct 03 '17 at 03:58
  • 1
    Can you temporarily add `Log.e("LOG", e.getMessage());` above the `e.printStackTrace();` run the code and login and check what's the error on Android Monitor – makoi07 Oct 03 '17 at 04:31

0 Answers0