使用安全参数在片段抛出错误之间传递 arraylist of object

Question

我正在开发一个应用程序，我需要将 ArrayList model object 传递给下一个片段，但它会抛出错误。 我尝试了很多文件，但没有任何效果请指导我为什么会这样。我正在通过 firebase 检索数据，并想在我的 recyclerview 适配器上设置这个 ArrayList。 我怎样才能做到这一点？？ 请指导我

来自片段 A 的代码

   ArrayList<RecyclerviewModel> mydatalist = new ArrayList<>();
                HomeFragmentDirections.ActionHomeToSearchDataFragment dr = HomeFragmentDirections.actionHomeToSearchDataFragment(mydatalist);
                Navigation.findNavController(view).navigate(dr);

来自片段 B 的代码（接收片段）

 ArrayList<RecyclerviewModel> mylist;
    if (getArguments()!=null)
    {

        SearchDataFragmentArgs args = SearchDataFragmentArgs.fromBundle(getArguments());
        mylist = args.getMylist();
        adapter = new SearchRecycleAdapter(mylist, getContext());
        binding.searchDataRecyclerview.setAdapter(adapter);
        
    }

MODEL CLASS

public class RecyclerviewModel implements Parcelable {

String name;
String email;
String desc;
String book;
List<String> images;
String uid;
String searching;


public RecyclerviewModel() {
}

public RecyclerviewModel(String name, String email, String desc, String book, List<String> images, String uid,String searching) {
    this.name = name;
    this.email = email;
    this.desc = desc;
    this.book = book;
    this.images= images;
    this.uid = uid;
    this.searching = searching;
}

protected RecyclerviewModel(Parcel in) {
    name = in.readString();
    email = in.readString();
    desc = in.readString();
    book = in.readString();
    images = in.createStringArrayList();
    uid = in.readString();
    searching = in.readString();
}

public static final Creator<RecyclerviewModel> CREATOR = new Creator<RecyclerviewModel>() {
    @Override
    public RecyclerviewModel createFromParcel(Parcel in) {
        return new RecyclerviewModel(in);
    }

    @Override
    public RecyclerviewModel[] newArray(int size) {
        return new RecyclerviewModel[size];
    }
};

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public String getEmail() {
    return email;
}

public void setEmail(String email) {
    this.email = email;
}

public String getDesc() {
    return desc;
}

public void setDesc(String desc) {
    this.desc = desc;
}

public String getBook() {
    return book;
}

public void setBook(String book) {
    this.book = book;
}

public List<String> getImages() {
    return images;
}

public void setImages(List<String> images) {
    this.images = images;
}


public String getUid() {
    return uid;
}

public void setUid(String uid) {
    this.uid = uid;
}


public String getSearching() {
    return searching;
}

public void setSearching(String searching) {
    this.searching = searching;
}

@Override
public int describeContents() {
    return 0;
}

@Override
public void writeToParcel(Parcel parcel, int i) {
    parcel.writeString(name);
    parcel.writeString(email);
    parcel.writeString(desc);
    parcel.writeString(book);
    parcel.writeStringList(images);
    parcel.writeString(uid);
    parcel.writeString(searching);
}

}

导航图 抛出的错误

error: incompatible types: ArrayList<RecyclerviewModel> cannot be converted to RecyclerviewModel

Answer 1

您应该创建另一个空的 class，它扩展 ArrayList，如下所示：

@Parcelize
public class RecyclerModels extends ArrayList<RecyclerViewModels> implements Parcelable

然后将导航参数的类型设置为此 class。

<argument
  android:name="mylist"
  app:argType="your.package.RecyclerModels"/>

你的RecyclerViewModel也应该是Parcelable

Answer 2

如果你不知道，你正试图在某个地方做这样的事情：

ArrayList<RecyclerviewModel> arr=...;
RecyclerviewModel rm=arr;

所以，再次检查你的代码。≈≈

您可以使用 class 和接口来传递数据（这不是最好的方法，但我使用它）

public class Helper{
public static Sender sender;
interface Sender {
public void Send(Object object);
}
}

在片段 2（接收者）中：

Helper.sender = new Helper.Sender(){
public void Send(Object ob){
//What it should do on recieve ?
//e.g : RecyclerviewModel r=(RecyclerviewModel)ob;
}
}

在片段 1（发件人）中：

//example:
Helper.sender.Send(RecyclerviewModel);

我希望这能帮到您：）