所以,我有一个片段,我们称之为片段 A,我需要它作为我的 sharedviewmodel 的主要宿主.
So, I have one fragment, lets call it Fragment A, in which I need it to be the main host of my sharedviewmodel.
所以我是这样的
class FragmentA:Fragment() {
private val model: SharedViewModel by viewModels()
}
现在,在 Fragment A 中,我设置了 viewpager2 适配器
Now, inside Fragment A, I set viewpager2 Adapter
private fun setupViewPagerWithTabLayout(
productList: MutableList<Producto>,
drinkList: MutableList<Producto>
) {
val fragmentList = listOf(
FragmentProducts.newInstance(productList),
FragmentProducts.newInstance(drinkList))
viewPager2.adapter = MyViewPageAdapter(requireActivity(),fragmentList)
val tabLayoutMediator = TabLayoutMediator(tabLayout, viewPager2,
TabLayoutMediator.TabConfigurationStrategy { tab, position ->
when (position) {
0 -> {
tab.text = "Option1"
}
1 -> {
tab.text = "Option2"
}
}
})
tabLayoutMediator.attach()
}
在这里我实例化了我的 FragmentProduct,它位于视图页面的 Fragment A 中,但我的适配器需要一个 FragmentActivity
Here I instantiate my FragmentProduct which is inside Fragment A in a viewpager, but my adapter needs a FragmentActivity
class MyViewPageAdapter(fragmentActivity: FragmentActivity,private val fragmentList: List<FragmentProducts>) :
FragmentStateAdapter(fragmentActivity) {
override fun getItemCount(): Int {
return fragmentList.size
}
override fun createFragment(position: Int): Fragment {
return fragmentList[position]
}
}
所以,从 FragmentA 中包含的 FragmentProducts,我需要与这个视图模型共享数据,但是当我这样做是为了将数据从这个 FragmentA 共享到 FragmentA
So, from FragmentProducts that is contained inside FragmentA, I need to share data with this viewmodel, but when I do this to share data from this fragment to FragmentA
class FragmentProducts:Fragment(){
private val model: SharedViewModel by viewModels ({requireParentFragment()})
}
我收到以下错误
java.lang.IllegalStateException: Fragment FragmentProducts{6953da8}(3a2f5cbb-113d-4ed2-8f85-56e04fcea356) f0} 不是子片段,它直接附在com.example.MainActivity@dcc2679
java.lang.IllegalStateException: Fragment FragmentProducts{6953da8} (3a2f5cbb-113d-4ed2-8f85-56e04fcea356) f0} is not a child Fragment, it is directly attached to com.example.MainActivity@dcc2679
所以,这告诉我 FragmentProducts 引用了我的 MainActivity,这就是逻辑,因为适配器的实例在 FragmentActivity 需要 requireActivity()
so , this is telling me that FragmentProducts references to my MainActivity, and thats logic because the instance of the adapter takes requireActivity() for FragmentActivity at
viewPager2.adapter = MyViewPageAdapter(requireActivity(),fragmentList)
所以,我需要传递我的 FragmentA 而不是 requireActivity() 作为该 viewpager 片段的主机,但是在我的 ViewPagerAdapter 中,我不能传递 FragmentStatePagerAdapter 或其他东西来传递我的 FragmentA 作为该 viewpager 的上下文
So, I need to pass instead of requireActivity() my FragmentA as the host of this viewpager fragment but at my ViewPagerAdapter, I cant pass FragmentStatePagerAdapter or something else to pass my FragmentA as the context for that viewpager
由于我使用 viewpager2,有没有办法让 FragmentA 托管 FragmentProducts viewpager,这样我就可以像这样在我的 FragmentProducts 中获取它
Since Im using viewpager2, is there a way to let FragmentA host FragmentProducts viewpager so I can get it in my FragmentProducts like this
private val model: SharedViewModel by viewModels ({requireParentFragment()})
谢谢
你正在使用 FragmentStateAdapter(fragmentActivity)
.不要那样做.将您的父 Fragment 传递给 MyViewPageAdapter
并将 that 传递给 FragmentStateAdapter
.
You're using FragmentStateAdapter(fragmentActivity)
. Don't do that. Pass your parent Fragment to MyViewPageAdapter
and pass that to FragmentStateAdapter
.
class MyViewPageAdapter(
parentFragment: Fragment,
private val fragmentList: List<FragmentProducts>
) : FragmentStateAdapter(parentFragment) {
这篇关于来自 viewpager2 和我的父片段的 SharedViewModel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跟版网!