Question

我想从我的适配器按钮显示一个DialogFragment。我已经在适配器本身内部创建了一个DialogFragment类，我希望它在我按下按钮时向我显示对话框，但是出现此错误：

java.lang.ClassCastException: android.app.Application cannot be cast to androidx.appcompat.app.AppCompatActivity
        at com.example.karate_manager.Adapters.AdapterMarket$1.onClick(AdapterMarket.java:106)

这是我的适配器：

public class AdapterMarket extends ArrayAdapter{
private FragmentManager fm;
    Context context;
    int item_Layaut;
    ArrayList<Karateka> data;
    ApiUtils apiUtils;

  public AdapterMarket(@NonNull Context context, int resource, @NonNull ArrayList objects, FragmentManager fm) {
        super(context, resource, objects);
        this.context = context;
        this.item_Layaut = resource;
        this.data = objects;
        this.fm = fm;
    }

    public void setData(MarketResponse data) {
        if(data!=null){
            this.data = data.getKaratekas();
            this.notifyDataSetChanged();
        }
    }

    @NonNull
    @Override
    public View getView(int position, @Nullable View convertView, @NonNull ViewGroup parent) {
        if (convertView == null) {
            LayoutInflater layoutInflater = LayoutInflater.from(context);
            convertView = layoutInflater.inflate(item_Layaut, parent, false);
        }

        String value = String.valueOf(data.get(position).getValue());
        Button buttonValue = convertView.findViewById(R.id.item_button_value_karateka);
        buttonValue.setText(value);

        popupBidKarateka(buttonValue);

        return convertView;
    }

    public void popupBidKarateka(Button buttonValue){
        buttonValue.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {
                FragmentActivity activity = (FragmentActivity)(context);
                fm = activity.getSupportFragmentManager();
                DialogFragment newFragment = new BidKaratekaDialogFragment();
                newFragment.show(fm, "bid" );
            }
        });
    }
}

这是片段中的引用：

     FragmentManager fragmentManager = getActivity().getSupportFragmentManager();
        adapterMarket = new AdapterMarket(getActivity().getApplicationContext(), R.layout.item_market_layout, marketResponse.getKaratekas(), fragmentManager);

Answer 1

               public void onClick(View view) {
                Application activity1 = (Application) view.getContext(); // bad

                FragmentActivity activity = (FragmentActivity)(context); // context must come from FragmentActivity

//我已经假设FragmentActivity是调用适配器的类。

                FragmentManager fm = activity.getSupportFragmentManager();
                DialogFragment newFragment = new BidKaratekaDialogFragment();
                newFragment.show(fm, "bid" );
            }
        });

一个聪明的人说..

即使您处于活动状态，上下文也不总是活动。它可以是一个应用程序，也可以是另一个上下文的包装。它的将上下文强制转换为活动几乎总是错误的。如果你绝对需要一个，您应该将Activity作为参数传递，而不是上下文。或者更好，传递支持片段经理而不是活动，因为这就是您所需要的

public AdapterMarket(@NonNull Context context, FragmentManager fm, int resource, @NonNull ArrayList objects) {
        super(context, resource, objects);
        this.context = context;
        this.fm = fm;
        this.item_Layaut = resource;
        this.data = objects;
    }

//例如，Afragment正在调用适配器。

AFragment.class
FM fm = getActivity.getSupportFM/getFM();
AdapterMarket market = new AdapterMarket(someContext, fm, ....)

在Adapter Android中从按钮显示DialogFragment时出现问题

1 个答案: