与标志FLAG_ACTIVITY_CLEAR_TOP（机器人）活动 [英] Activity with Flag FLAG_ACTIVITY_CLEAR_TOP (android)

问题描述

我已经堆了一个接一个推出的活动。
但在活动中的一个，我需要推出特定活动，旗 FLAG_ACTIVITY_CLEAR_TOP 。

I have stack of activities launched after one another.
but in one of the activity I need to launch that particular activity with flag FLAG_ACTIVITY_CLEAR_TOP.

所以，这将完成所有的previous活动开始。

So that it will finish all the previous activities and start.

问题我面对，我不希望完成从堆栈中它应该是在新推出的活动，回到presence活动之一。

Problem I am facing that i don't want to finish one of the activity from the stack it should be presence on back of the newly launched activity.

假设我有活动 [A] [B] [C] [D]
我开始研究[D] ，旗活动 FLAG_ACTIVITY_CLEAR_TOP 开始活动后研究[D] 它会破坏所有的活动，我想活动 [B] 来保持在后面跑，当我们preSS返回键上研究[D] 它应该显示活动 [B] 。

Suppose I have activities [A] [B] [C] [D]
I am starting [D] activity with flag FLAG_ACTIVITY_CLEAR_TOP after starting activity [D] it will destroy all the activities, I want activity [B] to be kept running on the back and when we press back key on [D] it should display activity [B].

如何做到这一点？

谢谢，
PP。

Thanks,
PP.

推荐答案

就像@Mudassir说，开始活动[A]。当你调用[b]简单地使用标志来清除所有数据，[B]，然后你可以从[B]调用[D]。无标志;因而具有[B] [D]像你想要的吗？然后对后退按钮preSS你会从[D]。为[B]，而无需任何其它数据？你似乎不以为他要你打电话的研究[D] [B]，旗，和什么的这样做的目的是什么？你会风与刚刚[B]在这一点上。

Like @Mudassir said, start activity [A]. When you call [B] simply use the FLAG to clear all data but [B], and then you can call [D] without flag from [B];thus having [B][D] like you wanted? Then on back button press you will go from [D] to [B], without having any other data?. You seem to not be thinking he wants you to call [B] with flag from [D], and what would the purpose of that be? You'd wind up with just [B] at that point.

-Either方式，你失去[C]前进的道路上。一个解决方案可能是：

-Either way, you lose [C] along the way. A solution might be:

startActivity(A)
[in activity A] startActivity(B) - WITH CLEAR_TOP TO KILL [A]
[in activity B] startActivityForResult(C) - WHEN [C] IS finish() IT WILL COME BACK TO [B]
[in activity B, onActivityResult()] startActivity(D)

这会给你只是[B]放大器; [D]。同时还获得到[C] ....试试吧，它会工作

This will give you JUST [B]&[D] while still getting to [C]....try it, it will work.

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