Question

我弹出一个对话框，用户输入用户名和密码。出于测试目的，我想给出一条警告消息，其中包含用户在单击“登录”时在EditText中输入的用户名。但是，只要按下登录，我的应用就会崩溃。注意：如果我删除EditText并在单击Login时弹出Toast消息，应用程序不会崩溃。这是我的实施：

public Dialog onCreateDialog(Bundle savedInstanceState){
        AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
        //get the layout inflater
        final LinearLayout loginLayout = (LinearLayout) findViewById(R.id.popup_signin);

        LayoutInflater inflater = getActivity().getLayoutInflater();

        builder.setView(inflater.inflate(R.layout.popup_signin, null))

                .setPositiveButton(R.string.signin, new DialogInterface.OnClickListener(){
                    @Override
                    public void onClick(DialogInterface dialog, int d){
                        //sign in the user...
                        String username = ((EditText) loginLayout.findViewById(R.id.username)).getText().toString();
                        //String password = ((EditText) findViewById(R.id.password)).getText().toString();
                        Toast.makeText(getApplicationContext(), "Hello World "+username, Toast.LENGTH_LONG).show();
                       // new PostLogin(username,password).execute();

                    }

popup_signin.xml

<?xml version="1.0" encoding="utf-8"?>

<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:id="@+id/popup_signin"
    android:orientation="vertical"
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:background="#FFFFFF">

<EditText
    android:id="@+id/username"
    android:inputType="text"
    android:layout_width="match_parent"
    android:layout_height="wrap_content"
    android:layout_marginTop="4dp"
    android:layout_marginLeft="4dp"
    android:layout_marginRight="4dp"
    android:layout_marginBottom="4dp"
    android:typeface="monospace"
    android:hint="enter your ID"/>

</LinearLayout>

logcat的

07-08 01:03:15.812: E/AndroidRuntime(28048): FATAL EXCEPTION: main
07-08 01:03:15.812: E/AndroidRuntime(28048): java.lang.NullPointerException
07-08 01:03:15.812: E/AndroidRuntime(28048):    at com.example.hkuapp.MainActivity$PopUpDialog$2.onClick(MainActivity.java:61)
07-08 01:03:15.812: E/AndroidRuntime(28048):    at com.android.internal.app.AlertController$ButtonHandler.handleMessage(AlertController.java:166)
07-08 01:03:15.812: E/AndroidRuntime(28048):    at android.os.Handler.dispatchMessage(Handler.java:99)
07-08 01:03:15.812: E/AndroidRuntime(28048):    at android.os.Looper.loop(Looper.java:137)
07-08 01:03:15.812: E/AndroidRuntime(28048):    at android.app.ActivityThread.main(ActivityThread.java:4745)
07-08 01:03:15.812: E/AndroidRuntime(28048):    at java.lang.reflect.Method.invokeNative(Native Method)
07-08 01:03:15.812: E/AndroidRuntime(28048):    at java.lang.reflect.Method.invoke(Method.java:511)
07-08 01:03:15.812: E/AndroidRuntime(28048):    at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:786)
07-08 01:03:15.812: E/AndroidRuntime(28048):    at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:553)
07-08 01:03:15.812: E/AndroidRuntime(28048):    at dalvik.system.NativeStart.main(Native Method)

Answer 1

您的EditText为空，因为它是popup_signin的一部分而您是从loginLayout获取的。

因此，在您的情况下，您应该将XML布局扩展为View object，然后在editText中找到view - 然后您可以将视图传递给builder.i.e。变化

 LayoutInflater inflater = getActivity().getLayoutInflater();
 builder.setView(inflater.inflate(R.layout.popup_signin, null))

到

LayoutInflater inflater = getActivity().getLayoutInflater();
final View view = (View) inflater.inflate(R.layout.popup_signin, null)
builder.setView(view)

和Onclick

String username = ((EditText)loginLayout.findViewById(R.id.username)).getText().toString();

到

String username = ((EditText) view.findViewById(R.id.username)).getText().toString();

即。
将您的代码重写为

public Dialog onCreateDialog(Bundle savedInstanceState){
        AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
        //get the layout inflater
        final LinearLayout loginLayout = (LinearLayout) findViewById(R.id.popup_signin);

        LayoutInflater inflater = getActivity().getLayoutInflater();
        final View view = (View) inflater.inflate(R.layout.popup_signin, null)
        builder.setView(view))

                .setPositiveButton(R.string.signin, new DialogInterface.OnClickListener(){
                    @Override
                    public void onClick(DialogInterface dialog, int d){
                        //sign in the user...
                        String username = ((EditText) view.findViewById(R.id.username)).getText().toString();
                        //String password = ((EditText) findViewById(R.id.password)).getText().toString();
                        Toast.makeText(getApplicationContext(), "Hello World "+username, Toast.LENGTH_LONG).show();
                       // new PostLogin(username,password).execute();

                    }

Answer 2

您正试图在loginLayout中找到您的EditText，即为null。膨胀您的布局然后将其分配给loginLayout。像：

LayoutInflater inflater = getActivity().getLayoutInflater();
View loginLayout = inflater.inflate(R.layout.popup_signin, null);

builder.setView(loginLayout);
...

Answer 3

我使用了第一个答案和第二个答案的组合，它起作用了

LayoutInflater inflater = getActivity().getLayoutInflater();
View loginLayout = inflater.inflate(R.layout.popup_signin, null);

eFullName = (AppCompatEditText) loginLayout.findViewById(R.id.eName);

从EditText获取价值的问题

3 个答案: