我用 MySQL 和 xampp 制作了一个简单的登录页面。问题是登录页面将始终有效并将我重定向到其他 Activity ，尽管我输入了错误的用户名或密码。我不确定查询是错误的还是代码，因为我已经三次检查它并没有找到真正的解决方案。我正在使用我的 android 手机而不是模拟器。 LOGIN_URL “192.168.43.231” 是我笔记本电脑的 IP，因为我目前正在使用手机的数据作为热点。谢谢

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 代码

<?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);
}

?>

最佳答案

将 if($check) 替换为 if(mysqli_num_rows($check) > 0)。

关于php - 使用 MySQL 在 Android 登录页面上出现错误的用户名和密码，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/51335676/