java - Toast 项目选择 - AdapterView/ArrayList

Question

我创建了一个使用 LayoutInflater 的 Adapter arraylist，以便图像保持比例。我正在尝试对所选项目标题(即“牛肉”)进行简单的 toast 。一切都按预期工作，除了我从 toast 中得到输出，例如

test com.sweatboxbbq.www.sweatboxbbq.Kansas City$MyAdaptor$Item@42aa9398

我知道我正在请求一个字符串值并且我得到了它，但是从代码中，我不知道如何得到我想要的。我只想捕获标题，这样我就可以把它干杯了(现在再做其他事情)。

示例 items.add(new Item( beef, R.drawable.mine_beef)); 我会得到一个字符串值“beef”

也许我可以通过某种方式识别这些数组项的标题，然后使用 onclick 方法将它们从那里拉出来？

下面是我的Java代码

public class KansasCity extends ActionBarActivity {

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.kansas_city);

    final GridView gridView = (GridView) findViewById(R.id.meat_choice);
    gridView.setAdapter(new MyAdapter(this));


    //on click starts here
    gridView.setOnItemClickListener(new AdapterView.OnItemClickListener(){

        public void onItemClick(AdapterView<?> MyAdapter, View v, int position, long id)
        {
            String flavorPicked = "" +
                    String.valueOf(MyAdapter.getItemAtPosition(position));
            Toast.makeText(KansasCity.this, "Test " + flavorPicked,
                    Toast.LENGTH_LONG).show();
        }
    });

}


//gridview
public class MyAdapter extends BaseAdapter {
    public List<Item> items = new ArrayList<Item>();
    private LayoutInflater inflater;


    public MyAdapter(Context context) {
        inflater = LayoutInflater.from(context);

        items.add(new Item( "Beef", R.drawable.mine_beef));
        items.add(new Item("Chicken", R.drawable.mine_chicken));
        items.add(new Item("Swine", R.drawable.mine_pork));
        items.add(new Item("Fish", R.drawable.mine_pork));
        items.add(new Item("Turkey", R.drawable.mine_pork));
        items.add(new Item("Sauce", R.drawable.mine_pork));

    }

    @Override
    public int getCount() {
        return items.size();
    }

    @Override
    public Object getItem(int i) {
        return items.get(i);
    }

    @Override
    public long getItemId(int i) {
        return items.get(i).drawableId;
    }

    @Override
    public View getView(int i, View view, ViewGroup viewGroup) {
        View v = view;
        ImageView picture;
        TextView name;

        if (v == null) {
            v = inflater.inflate(R.layout.gridview_item, viewGroup, false);
            v.setTag(R.id.picture, v.findViewById(R.id.picture));
            v.setTag(R.id.text, v.findViewById(R.id.text));
        }

        picture = (ImageView) v.getTag(R.id.picture);
        name = (TextView) v.getTag(R.id.text);

        Item item = (Item) getItem(i);

        picture.setImageResource(item.drawableId);
        name.setText(item.name);

        return v;


    }

    private class Item {
        final String name;
        final int drawableId;

        Item(String name, int drawableId) {
            this.name = name;
            this.drawableId = drawableId;
        }



    }


}

   }

感谢您的帮助。

Answer 1

您可以覆盖 Item 类中的 toString() 并使其返回您要显示的字符串，或者使成员 name Item 公开，并像使用它一样

String flavorPicked = ((Item)MyAdapter.getItemAtPosition(position)).name;