使用MySQL登錄Android頁面時出錯用戶名和密碼

Question

我用MySQL和xampp創建了一個簡單的登錄頁面。 問題是登錄頁面將始終工作並將我重定向到其他活動，雖然我提供了錯誤的用戶名或密碼。 我不確定查詢是錯誤的還是代碼，因為我已經三次檢查它並找不到真正的解決方案。 我正在使用我的Android手機而不是模擬器。 LOGIN_URL “192.168.43.231”是我的筆記本電腦的IP，因為我目前正將手機的數據用作熱點。 謝謝

public class LoginActivity extends AppCompatActivity {

private EditText usernameET, passwordET;
private Button login;
private static String LOGIN_URL = "http://192.168.43.231/Login/login.php";
public static  String KEY_USERNAME = "username";
public static  String KEY_PASSWORD = "password";

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login);

    usernameET = findViewById(R.id.usernameET);
    passwordET = findViewById(R.id.passwordET);
    login = findViewById(R.id.loginBTN);

    login.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            final String username = usernameET.getText().toString().trim();
            final String password = passwordET.getText().toString().trim();

            if (username.isEmpty() || password.isEmpty()) {
                Toast.makeText(LoginActivity.this, "Enter Valid Username & Password", Toast.LENGTH_SHORT).show();
            } else {

                StringRequest stringRequest = new StringRequest(Request.Method.POST, LOGIN_URL,
                        new Response.Listener<String>() {
                            @Override
                            public void onResponse(String response) {
                                try {
                                    JSONObject jsonObject = new JSONObject(response);
                                    boolean responseStatus = jsonObject.getBoolean("success");
                                    if (responseStatus) {
                                        Intent intent = new Intent(LoginActivity.this, MainActivity.class);
                                        LoginActivity.this.startActivity(intent);
                                    } else {
                                        Toast.makeText(LoginActivity.this, "Invalid Username or Password", Toast.LENGTH_SHORT).show();

                                    }
                                } catch (JSONException e) {
                                    e.printStackTrace();
                                    Toast.makeText(LoginActivity.this,"Error" + e.toString(), Toast.LENGTH_LONG).show();
                                }
                            }
                        },
                        new Response.ErrorListener() {
                            @Override
                            public void onErrorResponse(VolleyError error) {
                                Toast.makeText(LoginActivity.this, error.toString(), Toast.LENGTH_LONG).show();
                            }
                        }) {
                    @Override
                    protected Map<String, String> getParams() throws AuthFailureError {
                        Map<String, String> params = new HashMap<String, String>();
                        params.put(KEY_USERNAME, username);
                        params.put(KEY_PASSWORD, password);

                        return params;
                    }
                };
                RequestQueue requestQueue =Volley.newRequestQueue(LoginActivity.this);
                requestQueue.add(stringRequest);
            }
        }
    });
}
}

PHP代碼

<?php
include 'db.php';
if($_SERVER['REQUEST_METHOD']=='POST'){
$username = $_POST['username'];
$password = $_POST['password'];
$sql = "select * from blast_user where username = '$username' and password =  '$password' ";

$check = mysqli_query($con,$sql);
$response = array();
$response["success"]=false;

if($check){
$response["success"] = true;
}
else{
$response["success"]=false;
}

echo json_encode($response);
}

?>

Answer 1

用if(mysqli_num_rows($check) > 0)替換if($check) if(mysqli_num_rows($check) > 0) 。