Android PopupWindow可以显示另一个PopupWindow吗? 可以在同一时间打开多少PopupWindow?只有一个?
第一个PopupWindow正常显示。但是在按钮点击(在第一个PopupWindow内容视图中)我有一个例外:
08-13 16:28:38.682: ERROR/AndroidRuntime(11760): FATAL EXCEPTION: main
android.view.WindowManager$BadTokenException: Unable to add window -- token android.view.ViewRootImpl$W@41286250 is not valid; is your activity running?
at android.view.ViewRootImpl.setView(ViewRootImpl.java:600)
at android.view.WindowManagerImpl.addView(WindowManagerImpl.java:313)
at android.view.WindowManagerImpl.addView(WindowManagerImpl.java:215)
at android.view.WindowManagerImpl$CompatModeWrapper.addView(WindowManagerImpl.java:140)
at android.view.Window$LocalWindowManager.addView(Window.java:537)
at android.widget.PopupWindow.invokePopup(PopupWindow.java:992)
at android.widget.PopupWindow.showAsDropDown(PopupWindow.java:901)
at org.example.qberticus.quickactions.BetterPopupWindow.showLikePopDownMenu(BetterPopupWindow.java:159)
at org.example.qberticus.quickactions.BetterPopupWindow.showLikePopDownMenu(BetterPopupWindow.java:129)
at name.antonsmirnov.android.popup.ui.MainActivity$1$1.run(MainActivity.java:44)
at android.app.Activity.runOnUiThread(Activity.java:4170)
at name.antonsmirnov.android.popup.ui.MainActivity$1.onClick(MainActivity.java:42)
at android.view.View.performClick(View.java:3558)
at android.view.View$PerformClick.run(View.java:14157)
at android.os.Handler.handleCallback(Handler.java:605)
at android.os.Handler.dispatchMessage(Handler.java:92)
at android.os.Looper.loop(Looper.java:137)
at android.app.ActivityThread.main(ActivityThread.java:4514)
at java.lang.reflect.Method.invokeNative(Native Method)
at java.lang.reflect.Method.invoke(Method.java:511)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:790)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:557)
at dalvik.system.NativeStart.main(Native Method)
代码是:
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
bindControls();
initControls();
}
private Button button;
private void bindControls() {
button = (Button) findViewById(R.id.button);
}
private void initControls() {
initButton(button);
}
private void initButton(final Button button) {
button.setOnClickListener(new View.OnClickListener() {
public void onClick(View view) {
final BetterPopupWindow window = new BetterPopupWindow(button);
View popupview = createPopupView();
window.setContentView(popupview);
runOnUiThread(new Runnable() {
public void run() {
window.showLikePopDownMenu();
}
});
}
});
}
private View createPopupView() {
View v = LayoutInflater.from(MainActivity.this).inflate(R.layout.window, null);
Button popupButton = (Button) v.findViewById(R.id.popupbutton);
initButton(popupButton);
return v;
}
答案 0 :(得分:10)
玩完之后我发现了
window.showAtLocation(getWindow().getDecorView(), Gravity.CENTER, x, y);
工作正常,但
window.showAsDropDown(getWindow().getDecorView(), Gravity.CENTER, x, y);
引发异常!如果您使用showAtLocation(view)
与getWindow().getDecorView()
不同的任何视图,您仍然会有例外。
答案 1 :(得分:0)
是的,你可以。但是在查看异常堆栈跟踪之后,似乎你给出了错误的上下文。尝试使用相同的上下文或活动的上下文启动另一个对话框,如果它不是任何父级的子视图,如TabView。