Java登录方法:如果细节正确则运行但如果不正确则运行NullPointerException

Question

我写了一个方法来检查输入的用户名和密码,文本文件包含所有用户名和密码.有趣的问题是,如果细节是正确的,那么应用程序会进行,但如果细节不正确,则会返回NullPointerException.我的代码如下:

// Checks whether the inputed details are correct
public boolean isCorrect(String u, String p) {
    boolean check = false;
    String line = null;
    try {
        do {
            line = br.readLine();
            // System.out.println("Checking profile : " + line);
            String[] info = line.split("\t");
            // nested if-statement to improve efficiency over &&
            if (info[0].equals(u)) {
                System.out.println("username found!");
                if (info[1].equals(p)) {
                    System.out.println("password correct!");
                    check = true;
                } else System.out.println("password incorrect!");
            } else System.out.println("username not found!");
        } while (line != null && check == false);

    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

    return check;
}

返回的布尔值输入到主活动中的以下代码中:

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.loginpanel); 
    Button button = (Button) findViewById(R.id.btnLogin);
            Correct = false;



    lm = new LoginModel(this);

    button.setOnClickListener(new View.OnClickListener() {
        public void onClick(View v) {
            EditText textUsername = (EditText) findViewById(R.id.textUsername);
            EditText textPassword = (EditText) findViewById(R.id.textPassword);

            // Collect the login details entered by the user
            String username = textUsername.getText().toString();
            String password = textPassword.getText().toString();

            // Check the application is registering the correct details
            System.out.println(username + "\n" + password);

            Correct = lm.isCorrect(username, password);

            if (Correct) { // if details are correct then start main program
                Intent intent = new Intent (LoginView.this, MenuListView.class);
                startActivity(intent);
            }
            else System.out.println("Login is incorrect...");
        }
    });
}

我不明白为什么如果Correct = true然后它运行没有问题,但如果Correct = false它崩溃程序产生一个FATAL EXCEPTION:main,然后一个NullPointerException - 它不应该只是输出System.out .println.

Answer 1

当你到达文件的末尾时,line是null.
因此,你不能split()在巢行(双关语)中调用.

但是,请勿使用此代码.这是非常错误的,并且一旦发布就会轻易泄露密码.