AlertDialog 中显示的空 EditText 值

Question

我是 Android 开发新手，但遇到了问题。我有 2 个EditText 登录名和密码字段和 1 个Button。当我点击Button 时，我得到AlertDialog 并带有消息“以身份登录”，但我需要它是“以*值登录”进入“登录”@ 987654325@”。 EditText 值始终为空。这只是一个简单的实现，但它不起作用。 我在这里做错了什么？

代码如下：

public class FirstFragment extends Fragment {
    EditText loginText;
    EditText passwText;
    @Override
    public View onCreateView(
            LayoutInflater inflater, ViewGroup container,
            Bundle savedInstanceState
    ) {
        View view2 = inflater.inflate(R.layout.fragment_first, null, false);

        loginText = (EditText)view2.findViewById(R.id.loginText);
        passwText = (EditText)view2.findViewById(R.id.passText);
        // Inflate the layout for this fragment
        return inflater.inflate(R.layout.fragment_first, container, false);
    }

    public void onViewCreated(@NonNull View view, Bundle savedInstanceState) {
        super.onViewCreated(view, savedInstanceState);

        view.findViewById(R.id.button_first).setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick( View view) {
                handleLoginDialog();
            }
        });
    }

    private void handleLoginDialog() {
        AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
        //View view2 = getLayoutInflater().inflate(R.layout.fragment_first, null, false);

        String st = loginText.getText().toString();

        if (loginText != null && passwText != null){
            builder.setMessage("Logged in as " + st);
        }
        else
            builder.setMessage("Login or password was not entered!" );
        builder.show();
    }
}

Answer 1

我想以你做了什么开始回答：

您夸大了 2 个与以下相同的布局：

1. View view2 = inflater.inflate(R.layout.fragment_first, null, false);
2. return inflater.inflate(R.layout.fragment_first, container, false);

我们将第二次膨胀称为view1，它是为Fragment 膨胀的原始视图，也是onViewCreated() 方法中返回的视图。

然后，为view1 的button_first 设置onClickListener。

然后，在handleDialogLogin() 方法中，您尝试获取使用view2 对象创建的EditText 字段的文本值。
这就是问题所在。因为，在应用程序使用期间，您将文本插入到 view1 的 EditText 字段中，但您要求 view2 的 EditText 字段的文本值显然是空的。

正确和干净的方法在这里：

public class FirstFragment extends Fragment {
    EditText loginText;
    EditText passwText;
    @Override
    public View onCreateView(
            LayoutInflater inflater, ViewGroup container,
            Bundle savedInstanceState
    ) {
        // just inflate the layout for this fragment as it is not correct
        // place for declaring and initializing views
        return inflater.inflate(R.layout.fragment_first, container, false);
    }

    // this is the best place for declaring and initializing views
    public void onViewCreated(@NonNull View view, Bundle savedInstanceState) {
        super.onViewCreated(view, savedInstanceState);

        // declare and initialize your EditText here 
        loginText = (EditText)view.findViewById(R.id.loginText);
        passwText = (EditText)view.findViewById(R.id.passText);
        view.findViewById(R.id.button_first).setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick( View view) {
                handleLoginDialog();
            }
        });
    }

    private void handleLoginDialog() {
        AlertDialog.Builder builder = new AlertDialog.Builder(requireContext());

        String st = loginText.getText().toString().trim();
        String pt = passText.getText().toString().trim();

        if (st.isEmpty() && pt.isEmpty())
           builder.setMessage("Login or password was not entered!" );
        else
           builder.setMessage("Logged in as " + st);
        builder.show();
    }
}

Answer 2

也许你可以试试这个，但这不使用onViewCreated

public class FirstFragment extends Fragment {
EditText loginText;
EditText passText;
Button btnLogin;
@Override
public View onCreateView(
        LayoutInflater inflater, ViewGroup container,
        Bundle savedInstanceState
) {
    View view2 = inflater.inflate(R.layout.fragment_first, null, false);

    loginText = (EditText)view2.findViewById(R.id.loginText);
    passText = (EditText)view2.findViewById(R.id.passText);
    btnLogin = (Button) view2.findViewById(R.id.button_first);

    btnLogin.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            handleLoginDialog();
        }
    });

    return view2;
}

private void handleLoginDialog() {
    AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
    //View view2 = getLayoutInflater().inflate(R.layout.fragment_first, null, false);

    String st = loginText.getText().toString().trim();
    String pt = passText.getText().toString().trim();

    if (st.isEmpty() && pt.isEmpty()){
        builder.setMessage("Login or password was not entered!" );
    }
    else
        builder.setMessage("Logged in as " + st);
    builder.show();
}

Answer 3

安全的一面是在handleLoginDialog 方法中将loginText 作为字符串参数发送，并在那里检查字符串是否为空。