[英]Problems with getting value from EditText
我彈出一個對話框,用戶在其中輸入用戶名和密碼。 為了進行測試,我想給出一條警告消息,其中包含用戶單擊“登錄”時在EditText中輸入的用戶名。 但是,每按一次登錄,我的應用程序就會崩潰。 注意:如果我刪除EditText並單擊“登錄”后僅彈出一條敬酒消息,則該應用程序不會崩潰。 這是我的實現:
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
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)
您的EditText
為null,因為它是popup_signin
一部分,並且您可以從loginLayout
中獲取它。
因此,在您的情況下,您應該將XML布局添加到View object
,然后在view
找到editText
然后可以將視圖傳遞給editText
。
LayoutInflater inflater = getActivity().getLayoutInflater();
builder.setView(inflater.inflate(R.layout.popup_signin, null))
至
LayoutInflater inflater = getActivity().getLayoutInflater();
final View view = (View) inflater.inflate(R.layout.popup_signin, null)
builder.setView(view)
然后在Onclick中
String username = ((EditText)loginLayout.findViewById(R.id.username)).getText().toString();
至
String username = ((EditText) view.findViewById(R.id.username)).getText().toString();
即
將您的代碼重寫為
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();
}
您試圖在loginLayout中找到您的EditText,這是空的。 放大您的布局,然后將其分配給loginLayout。 喜歡:
LayoutInflater inflater = getActivity().getLayoutInflater();
View loginLayout = inflater.inflate(R.layout.popup_signin, null);
builder.setView(loginLayout);
...
我使用了第一個答案和第二個答案的組合,並且有效
LayoutInflater inflater = getActivity().getLayoutInflater();
View loginLayout = inflater.inflate(R.layout.popup_signin, null);
eFullName = (AppCompatEditText) loginLayout.findViewById(R.id.eName);
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