I'm trying to do a login to my app with a XAMMP server. I'm doing the conexion with the server with Volley, but I don't understand why y have this error. I have looked for many questions in stackoverflow and I can't find out the solution. This is the android code:
public class Login extends AppCompatActivity {
RequestQueue queue;
JsonObjectRequest request;
Map<String, String> map = new HashMap<String, String>();
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(activity_login);
getSupportActionBar().hide();//<!-- Para que no salga la barra superior en todas las activitis-->
}
public void sayHelloThreads(final View view) throws IOException { //log in
EditText USUARIO =
(EditText) findViewById(R.id.Usuario);
String usu = USUARIO.getText().toString();
EditText Password =
(EditText) findViewById(R.id.Pass);
String pass = Password.getText().toString();
// the request queue
queue = Volley.newRequestQueue(this);
// the parameters for the php
// map.put(KEY, VALUE);
map.put("uname", usu);
map.put("psw", pass);
// the JSON request
// JsonObjectRequest(METHOD, URL, JSONOBJECT(PARAMETERS), OK_LISTENER, ERROR_LISTENER);
request = new JsonObjectRequest(
Request.Method.POST, // the request method
"http://192.168.1.41:8080/check-login_APP_voluntarios.php", // the URL
new JSONObject(map), // the parameters for the php
new Response.Listener<JSONObject>() { // the response listener
@Override
public void onResponse(JSONObject response){
// here you parse the json response
String pp="lsjhjh";
}
},
new Response.ErrorListener() { // the error listener
@Override
public void onErrorResponse(VolleyError error) {
String pp="lsjhjh";
/* here you can warn the user that there
was an error while trying to get the json
information from the php */
}
});
// executing the quere to get the json information
queue.add(request);
}
}
And this the code of the php file:
<?php
$con = mysqli_connect('localhost','root','','rot_v2');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ajax_demo");
//$usuario = $_POST['uname']; //uname
//$password = $_POST['psw'];
// decoding the json array
$post = json_decode(file_get_contents("php://input"), true);
// getting the information from the array
// in the android example I've defined only one KEY. You can add more KEYS to your app
$usuario = $post['uname'];
$password = $post['psw'];
// the "params1" is from the map.put("param1", "example"); in the android code
// if you make a "echo $my_value;" it will return a STRING value "example"
$sql="SELECT `voluntarios`.`Password` FROM `voluntarios` WHERE (`voluntarios`.`Usuario` like'%{$usuario}%')";
$result = mysqli_query($con,$sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$passwordres = $row['Password'];
}
//echo "<script>console.log( 'Debug Objects: " . $passwordres . "' );</script>";
if ($password == $passwordres){
//echo "Correcta voluntario";
$response = "OK"
header('Content-Type: application/json');
echo(json_encode($response));
}
else{
$response = "KO"
header('Content-Type: application/json');
echo(json_encode($response);
}
}
else{
$response = "KO2"
header('Content-Type: application/json');
echo(json_encode($response);
}
mysqli_close($con);
/*
if($tipousuario == "corredor"){
echo "Se ha enviado a un corredor";
}if($tipousuario == "voluntario"){
echo "Se ha enviado a un corredor";
}*/
?>
This is a screenshoot of the error:
Can you help me with this trouble, thanks you!!!
