使用导航组件从Backstack中解开片段

问题描述

我有X，A，B片段，并且我正在使用导航体系结构组件在它们之间进行导航.

I have Fragments X, A, B, and i'm using Navigation architecture component to navigate between them.

特定于片段A，B，但是片段X可以是any(C，D，...);

Fragments A, B specific, but Fragment X can be any(C,D,...);

底部导航中的片段A和B及其导航图标"始终显示在屏幕上，这意味着用户可以随时从任何片段(包括A和B)转到A或B:

Fragments A and B from Bottom Navigation and their "navigations icons" always on the screen, it means user can go to A or B anytime from any Fragment(include A and B):

X -> A -> B

X -> B -> A

A -> B -> X

A -> B -> A

//another ways

我在这种情况下的问题:

My problem about this case:

X -> A -> B -> A -> B -> ?

如果用户从X开始，到达?并通过返回"按钮开始返回，则他会多次抛出A，B:

If user started from X, reached ? and begin to go back by "back" button, he goes throw A,B several times:

用户已按下:

? -> B -> A -> B -> A -> X

但是我想从堆栈中排除"碎片，如果它们已经存在的话:

But I want "to exclude" fragments from backstack if they already on it:

? -> A -> B -> X

如果用户导航:

X -> A -> B -> A

我想看:

A -> B -> X

不是:

A -> B -> A -> X

我正在尝试使用Pop To做它，但是它只能使我回到一个具体的片段上.我需要返回开始的X片段，而不是硬编码.包容性和单身"顶部不是这个问题.

I'm trying to do it with Pop To, but it can return me on one one concrete Fragment only. I need to return on my started X Fragment, not hardcoded. Inclusive and Single top is not about it.

我不确定我可以使用基本的导航组件来做到这一点，所以我需要您的建议.如果我做不到，应该使用哪种方式?有什么好的做法吗?

I'm not sure i can do it with basic Navigation component, so i need your advice. If i can't do it, what way should i use? Is there any good practices about it?

UPD:

我正在使用全局点在片段之间导航.这就是我的导航样子:

I'm using global points to navigate between Fragments. It's how my navigation looks like:

右边的底部链是X，我可以使用底部导航从其中任何一个导航到不被搅动的片段.它是单项活动应用，用于我仅使用的导航方式:

The right|bottom chain is X, i can navigate from any of it to to not chanied fragments using bottom navigation. It's Single Activity app, to navigate i'm using just:

   //Using global points
   findNavController(R.id.host).navigate(R.id.toLibrary)

推荐答案

以下解决方案使用片段可见性技术来管理片段事务onBackPressed().

Following Solution uses Fragment Visibility Technique in order to manage Fragments Transaction onBackPressed().

第一步，当每个片段称为&为了能够识别哪些片段已经被添加到Backstack中而被调用.通过下面的代码，我们为将被称为&的片段分配标签.交易.

In the first step, We assign a tag to each fragment when it's called & invoked in order to be able to recognize which fragments already have been added to Backstack.by code below, we assign a tag to a fragment that is going to be called & transacted.

 fragmentManager.beginTransaction()
.replace(R.id.FragmentHolder, Fragment_A OR Fragment_B OR Fragment_ANY, "A or B or ANY")
.addToBackStack(tag).commit();

请记住，您必须为要调用的任何片段分配一个标签.

在第二个中，我们将处理public void onBackPressed().

您必须忽略super.onBackPressed();，因为我们不想默认 onBackPressed函数影响Backstack(按原样)，而我们想处理Backstack自己.

You MUST ignore super.onBackPressed(); as becuase, we don't want to defualt onBackPressed function to impact Backstack(as it is) and we are want to handle Backstack ourselves.

你在这里

@Override
public void onBackPressed() {
    if(isVisible("A")) { //Go some where else you wanna go }
    else if(isVisible("B")){ //Go some where else you wanna go }
    else if(isVisible("X or any"){ //Go some where else you wanna go }
    else { //Go default page }
}


public boolean isVisible(String tag){
    try {
        Fragment fragment = fragmentManager.findFragmentByTag(tag);
        if (fragment != null && fragment.isVisible())
            return true;
        else
            return false;
    } catch (Exception e) {
        e.printStackTrace();
        return false;
    }
 }

按下后退按钮时，我们将检查哪个片段已经可见，并将用户重定向到相应的片段.

When back button is pressed, We check which fragment is already visible and do redirect user to corresponding Fragment.

例如:A - > B或A -> X或X -> B

我正在将此技术用于大型应用程序，并且一切都很好. AF受助人

I am using this Technique for a big application and all good. Mr.AF grantee

这篇关于使用导航组件从Backstack中解开片段的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！