I have a very basic register script in Android but Android wont connect to my localhost phpmyadmin database (XAMPP).
I was told it could be my XAMPP settings, or I'm using the wrong URL, or I need a timeout chunk in my script. I've been through a lot of debugging methods, but the only result i got was
/Create User: com.android.volley.NoConnectionError: java.io.EOFException.
And now even that error is gone. I don't get any errors, and my script does nothing.
Here is my CreateUser script;
public class CreateUser extends AppCompatActivity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_create_user);
this.setTitle("Create User");
final EditText username1 = findViewById(R.id.Createusername);
final EditText password1 = findViewById(R.id.CreatePassword);
final Switch isAdmin = findViewById(R.id.isadmin);
final Button createuser = findViewById(R.id.createuserbtn);
if (getIntent().hasExtra("com.example.northlandcaps.crisis_response")){
isAdmin.setVisibility(View.GONE);
}
createuser.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
final String username = username1.getText().toString();
final String password = password1.getText().toString();
if (isAdmin.getShowText()) {
Global.isadmin = "1";
}else{
Global.isadmin ="0";
}
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){
Intent intent = new Intent(CreateUser.this, MainActivity.class);
CreateUser.this.startActivity(intent);
}else{
AlertDialog.Builder builder = new AlertDialog.Builder(CreateUser.this);
builder.setMessage("Register Failed")
.setNegativeButton("Retry",null)
.create()
.show();
}
} catch (JSONException e) {
e.printStackTrace();
}
}
};
int socketTimeout = 250000;//30 seconds - change to what you want
RetryPolicy policy = new DefaultRetryPolicy(socketTimeout, DefaultRetryPolicy.DEFAULT_MAX_RETRIES, DefaultRetryPolicy.DEFAULT_BACKOFF_MULT);
RegisterRequest registerRequest = new RegisterRequest(username,password,Global.isadmin,responseListener);
registerRequest.setRetryPolicy(policy);
RequestQueue queue = Volley.newRequestQueue(CreateUser.this);
queue.add(registerRequest);
}
});
}
Here is my RegisterRequest class;
public class RegisterRequest extends StringRequest {
private static final String REGISTER_REQUEST_URL = "http://10.0.2.2:3306/phptesting/Register.php"; //the address was localhost, then my ipaddress, and now its this code i was told to put in
private Map<String, String> params;
public RegisterRequest(String username, String password, String isAdmin, Response.Listener<String> listener){
super(Method.POST, REGISTER_REQUEST_URL,listener,null);
params = new HashMap<String, String>();
params.put("username",username);
params.put("password",password);
params.put("isAdmin", isAdmin);
}
public Map<String, String> getparams() {
return params;
}
}
and if you need it, here is my PHP script located in htdocs in XAMPP.
<?php
$db_host = 'localhost:3306';
$db_user = 'root';
$db_pass = '';
$db_name = 'test';
$con = mysqli_connect($db_host,'user',$db_pass,$db_name);
if($con){
echo "connection successful";
}else{
echo "connection failed";
}
$age = $_POST["isAdmin"];
$username = $_POST["username"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, "INSERT INTO cresidentials (username,password,isAdmin) VALUES (?, ?, ?)");
if(!$statement) { printf("Prepare failed: %s\n", mysqli_error($con)); }
if(!$statement) { return json_encode(['status'=>'failed','message'=>mysqli_error($con)]); }
mysqli_stmt_bind_param($statement, "sss",$username,$password,$isAdmin);
mysqli_stmt_execute($statement);
if(mysqli_error($statement)) { return json_encode(['status'=>'failed','message'=>mysqli_error($con)]); }
$response = array();
$response["success"] = true;
echo json_encode($response);
?>
I appreciate all help, and I'd love it if someone were to walk with me on trying to create this simple login and register app.
