与朋友风格listview的单词

Question

我有一个安卓游戏，可以加载用户对他朋友的“游戏”列表。类似于“与朋友一起玩游戏”

为了重新创建这种类型的视图，我最初使用的是listview，但由于有不同的元素，即游戏，部分（他们的移动 - 你的移动）等，我认为我不能纯粹地做这种布局列表视图？

有没有人做过类似的事情？我可能只是使用滚动视图并动态构建布局，但我担心的是，如果用户有很多游戏，性能将受到影响。

有人可以提供建议吗？

Answer 1

您应该覆盖listview适配器中的getView()方法。 它负责显示每个项目的布局。你将能够做到这样的事情：

if(position == YOUR_MOVE_POSITION) {

     // here hide ordinary elements, and set the ones you need to be visible
}

在此查看详细信息：link

示例:(我如何看待解决方案）

此方法位于适配器类中（用于填充ListView，因此您需要使用以下代码覆盖它：

public View getView(int position, View convertView, ViewGroup parent) {

  final View v;

  if(position == 1) {  // here come instructions for 'header' no.1
    v = createViewFromResource(position, convertView, parent, mResource);

    // the widget you want to show in header:
    ImageView yourMoveImage = (ImageView) v.findViewById(R.id.your_move);

    // and here come widgets you don't want to show in headers:
    ImageView otherWidget = (ImageView) v.findViewById(R.id.other_widget);

    // then you set the visibility:
    yourMoveImage.setVisibility(View.VISIBLE);    // here is the key
    otherWidget.setVisibility(View.GONE);       // it may also be View.INVISIBLE (look up the official docs for details)   


  } else {

    if(position == 5){  // here come instructions for 'header' no.1


      v = createViewFromResource(position, convertView, parent, mResource);

      // the widget you want to show in header:
      ImageView theirMoveImage = (ImageView) v.findViewById(R.id.their_move);

      // and here come widgets you don't want to show in headers:
      ImageView otherWidget = (ImageView) v.findViewById(R.id.other_widget); 

      // then you set the visibility:
      yourMoveImage.setVisibility(View.VISIBLE);    // here is the key
      otherWidget.setVisibility(View.GONE);       // it may also be View.INVISIBLE (look up the official docs for details)   

    } else {

      // if it is the regular item, just show it as desribed in your XML:
      v = createViewFromResource(position, convertView, parent, mResource);
    }
  }

  return v;
}