GridView的Android ImageAdapter显示相同（加倍）的图像

我的问题很简单：当我点击显示GridView（一切都很好）时，我按下后退键，然后再次点击GridView。然后在GridView中有两个图像。例如，如果GridView中第一次有5张照片，第二次有10张照片。我认为问题在于我的ImageAdapter：

public class ImageAdapter extends BaseAdapter {

    private List<Item> items = new ArrayList<Item>();
    private LayoutInflater inflater;

    public ImageAdapter(Context context, List<Item> items) {
        inflater = LayoutInflater.from(context);
        this.items = items;
    }

    @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).drawable;
    }

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

        if (view == null) {

            v = inflater.inflate(R.layout.squareimageview, null, 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) items.get(i);

            picture.setImageResource(item.getDrawable());
            name.setText(item.name);
        } else {
            v = view;
            System.out.println("Cia");
        }



        return v;

    }

}

项目类别：

public class Item {
    String name;
    int drawable;
    String text;

    public int getDrawable() {
        return drawable;
    }

    public void setDrawable(int drawable) {
        this.drawable = drawable;
    }

    public Item(String name, String text, int id) {
        this.name = name;
        this.text = text;
        this.drawable = id;
    }

    public String getText() {
        return text;
    }

    public void setText(String text) {
        this.text = text;
    }

    public String getName() {
        return name;
    }

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

}

---

返回时，您很可能没有完成活动，因此gridview永远不会被终止，并再次获得相同的一批图像（我相信是在onresume（）回调中）

如果是这种情况，请确保完成（）覆盖onBackPressed（）的活动，或重新创建ListView every onResume（）调用：

然后添加数据

然而，为了确定问题，我们需要更多的代码

>

好的，这就是问题所在：

public static List<Item> items = new ArrayList<Item>();

---

首先，由于未清除用作数据保持器的

ArrayList

，因此会获得双图像

您有一些解决方案：

• 您可以在添加项目之前清除ArrayList

  items.clear()
  items.add(new Item("Name1", "Text1", R.drawable.pic_12));
  items.add(new Item("Name2", "Text2", R.drawable.pic_13));
  items.add(new Item("Name3", "Text3", R.drawable.pic_14));
  items.add(new Item("Name4", "Text4", R.drawable.pic_15));
  items.add(new Item("Name5", "Text5", R.drawable.pic_2));
  ImageAdapter imaadpter=new ImangeAdapter(this,items);
  gridView.setAdapter(imaadpter);
  

• 在添加项目之前，请检查数组中是否有项目

  if( items.isEmpty() ) {
      items.add(new Item("Name1", "Text1", R.drawable.pic_12));
      items.add(new Item("Name2", "Text2", R.drawable.pic_13));
      items.add(new Item("Name3", "Text3", R.drawable.pic_14));
      items.add(new Item("Name4", "Text4", R.drawable.pic_15));
      items.add(new Item("Name5", "Text5", R.drawable.pic_2));
  }
  ImageAdapter imaadpter=new ImangeAdapter(this,items);
  gridView.setAdapter(imaadpter);
  

• 当您返回时，调用

  items.clear（）

  ，那么您肯定不会得到双重项目

NB:您的代码中还有一个相当大的错误，您没有注意到它，因为您没有足够的项需要滚动或重用网格视图项

在getView方法中，必须在if-else语句之外设置值，以便在重用视图时更新其内容

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

    if (view == null) {
        v = inflater.inflate(R.layout.squareimageview, null, 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);            
    } else {
        v = view;
        System.out.println("Cia");
    }

    Item item = (Item) items.get(i);

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


    return v;

}

---

问题不在这里。。问题是你可以使用这个BaseAdapter。好的，现在看看我的代码。你是否正在清除列表（Items.clear）！！！！？？在哪里设置适配器或通知适配器？否。我可以将其放置在哪里？还有别的吗？

public List<Item> items = new ArrayList<Item>();

items.clear();
items.add(new Item("Name1", "Text1", R.drawable.pic_12));
        items.add(new Item("Name2", "Text2", R.drawable.pic_13));
        items.add(new Item("Name3", "Text3", R.drawable.pic_14));
        items.add(new Item("Name4", "Text4", R.drawable.pic_15));
        items.add(new Item("Name5", "Text5", R.drawable.pic_2));

        GridView gridView = (GridView) findViewById(R.id.gridview);
        gridView.setAdapter(new ImageAdapter(this, items));

items.clear()
items.add(new Item("Name1", "Text1", R.drawable.pic_12));
items.add(new Item("Name2", "Text2", R.drawable.pic_13));
items.add(new Item("Name3", "Text3", R.drawable.pic_14));
items.add(new Item("Name4", "Text4", R.drawable.pic_15));
items.add(new Item("Name5", "Text5", R.drawable.pic_2));
ImageAdapter imaadpter=new ImangeAdapter(this,items);
gridView.setAdapter(imaadpter);

if( items.isEmpty() ) {
    items.add(new Item("Name1", "Text1", R.drawable.pic_12));
    items.add(new Item("Name2", "Text2", R.drawable.pic_13));
    items.add(new Item("Name3", "Text3", R.drawable.pic_14));
    items.add(new Item("Name4", "Text4", R.drawable.pic_15));
    items.add(new Item("Name5", "Text5", R.drawable.pic_2));
}
ImageAdapter imaadpter=new ImangeAdapter(this,items);
gridView.setAdapter(imaadpter);

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

    if (view == null) {
        v = inflater.inflate(R.layout.squareimageview, null, 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);            
    } else {
        v = view;
        System.out.println("Cia");
    }

    Item item = (Item) items.get(i);

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


    return v;

}