Question

我创建了一个显示弹出窗口的活动。活动停止后，将打印以下消息以记录：

10-22 13:36:05.539: E/WindowManager(14865): android.view.WindowLeaked: Activity qnd.papaya.counter.gui.payment.PaymentActivity has leaked window android.widget.LinearLayout{430ac2a0 V.E..... ......I. 0,0-821,306} that was originally added here
10-22 13:36:05.539: E/WindowManager(14865):     at android.view.ViewRootImpl.<init>(ViewRootImpl.java:346)
10-22 13:36:05.539: E/WindowManager(14865):     at android.view.WindowManagerGlobal.addView(WindowManagerGlobal.java:248)
10-22 13:36:05.539: E/WindowManager(14865):     at android.view.WindowManagerImpl.addView(WindowManagerImpl.java:69)
10-22 13:36:05.539: E/WindowManager(14865):     at android.widget.PopupWindow.invokePopup(PopupWindow.java:1019)
10-22 13:36:05.539: E/WindowManager(14865):     at android.widget.PopupWindow.showAtLocation(PopupWindow.java:850)
10-22 13:36:05.539: E/WindowManager(14865):     at android.widget.PopupWindow.showAtLocation(PopupWindow.java:814)
10-22 13:36:05.539: E/WindowManager(14865):     at qnd.papaya.counter.controller.PaymentPanelController.showPaymentTypesPopup(PaymentPanelController.java:188)
10-22 13:36:05.539: E/WindowManager(14865):     at qnd.papaya.counter.controller.PaymentPanelController.access$2(PaymentPanelController.java:159)
10-22 13:36:05.539: E/WindowManager(14865):     at qnd.papaya.counter.controller.PaymentPanelController$3.onClick(PaymentPanelController.java:197)
10-22 13:36:05.539: E/WindowManager(14865):     at android.view.View.performClick(View.java:4438)
10-22 13:36:05.539: E/WindowManager(14865):     at android.view.View$PerformClick.run(View.java:18422)
10-22 13:36:05.539: E/WindowManager(14865):     at android.os.Handler.handleCallback(Handler.java:733)
10-22 13:36:05.539: E/WindowManager(14865):     at android.os.Handler.dispatchMessage(Handler.java:95)
10-22 13:36:05.539: E/WindowManager(14865):     at android.os.Looper.loop(Looper.java:136)
10-22 13:36:05.539: E/WindowManager(14865):     at android.app.ActivityThread.main(ActivityThread.java:5001)
10-22 13:36:05.539: E/WindowManager(14865):     at java.lang.reflect.Method.invokeNative(Native Method)
10-22 13:36:05.539: E/WindowManager(14865):     at java.lang.reflect.Method.invoke(Method.java:515)
10-22 13:36:05.539: E/WindowManager(14865):     at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:785)
10-22 13:36:05.539: E/WindowManager(14865):     at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:601)
10-22 13:36:05.539: E/WindowManager(14865):     at dalvik.system.NativeStart.main(Native Method)

因此，我想解除活动的onPause（）方法中的弹出窗口。

有没有办法在活动中引用所有可见的PopupWindows？类似的东西：

@Override
protected void onPause() {
    super.onPause();
    findPopupWithTag("MY_POPUP").dismiss(); // <--- IS THERE A SIMILAR METHOD?
}

我不想明确引用弹出窗口来保持我的代码干净简单......

Answer 1

看起来这是不可能的

如果您查看PopupWindow.invokePopup方法，它看起来像这样

private void invokePopup(WindowManager.LayoutParams p) {
    ...
    mWindowManager.addView(mPopupView, p);
    ...
}

this quiestion根据View ID

找不到WindowManager

如何通过ID / tag / ...找到弹出窗口？

1 个答案: