使用导航组件按“后退"按钮向上导航 [英] Navigate up by back button with Navigation Component

问题描述

我使用导航组件从一个片段导航到另一个片段.但是，当用户按下后退"按钮时，我想导航回到第一个片段.但是它继续显示第二个片段.这是我的nav_graph:

I use navigation components to navigate from one fragment to another. However, when the user press the back button, I want to navigate back to first fragment. But it keep showing the second fragment. This is my nav_graph:

<?xml version="1.0" encoding="utf-8"?>
<navigation xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:app="http://schemas.android.com/apk/res-auto"
    xmlns:tools="http://schemas.android.com/tools"
    android:id="@+id/nav_graph"
    app:startDestination="@id/fragment1">
    <fragment
            android:id="@+id/fragment2"
            android:name="com.myapp.ui.fragments.Fragment2"
            android:label="fragment_2" />
    <fragment
            android:id="@+id/fragment1"
            android:name="com.myapp.ui.fragments.Fragment1"
            android:label="fragment_1">

        <action
                android:id="@+id/action_fragment1_to_fragment2"
                app:destination="@id/fragment2"
                app:enterAnim="@anim/fragment_fade_enter"
                app:exitAnim="@anim/fragment_fade_exit"
                app:popUpTo="@id/fragment1" />
    </fragment>
</navigation>

这就是我在Fragment1-Class的代码中触发导航的方式:

And this is how I trigger the navigation in the code of my Fragment1-Class:

 viewModel.itemSelected.observe(viewLifecycleOwner) {
        navigate(it)
    }

....

fun navigate(id: Long){
    val bundle = Bundle()
    bundle.putLong("itemid", id)    
    getNavController().navigate(R.id.action_fragment1_to_fragment2, bundle)

]

修改:更正了XML中的startDestination.

Corrected startDestination in XML.

Edit2:添加了更多代码.

推荐答案

您正在使用LiveData进行事件.LiveData始终会缓存设置的值，因此，当您返回Fragment1时，您会再次观察LiveData并再次获得相同的值，从而再次导航(

You're using a LiveData for an event. LiveData always caches the set value, so when you return to your Fragment1, you observe the LiveData again and get the same value a second time, causing you to navigate() yet again.

请参见此博客发布以获取更多信息和替代方法.

See this blog post for more information and alternatives.

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