Question

我有一个应用程序，我接受输入，例如没有用户，并在列表视图方面填充一些修复名称，电话等。我使用了viewholder来优化我的所有列表视图。除了列表视图加载userdata之外，所有列表视图都可以正常工作。

enter image description here

从图中可以看出，第一行和最后一行都已加载，但中间的两行未加载，并且需要花费大量时间才能加载。我提供给每一行的信息最初是相同的，留给用户修改。这是我用于此的代码。

public View getView(final int position, View convertView,final ViewGroup parent)
{
    View rowview = convertView;
    if(rowview == null)
    {
        ViewHolder viewholder = new ViewHolder();
        final LayoutInflater inflater = (LayoutInflater)context.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
        rowview = inflater.inflate(R.layout.list_skeleton_people_expenses, parent, false);
        viewholder.tv1 = (TextView)rowview.findViewById(R.id.name);
        viewholder.tv2 = (TextView)rowview.findViewById(R.id.destination);
        viewholder.tv3 = (TextView)rowview.findViewById(R.id.start);
        viewholder.tv4 = (TextView)rowview.findViewById(R.id.phno);
        viewholder.rowview = rowview;
        rowview.setTag(viewholder);
    }
    else
    {   
        ViewHolder holder = (ViewHolder) rowview.getTag();
        UserData temp = objects.get(position);
        holder.tv1.setText(temp.name);

        holder.tv2.setText("Destination: "+temp.loc);

        holder.tv3.setText("Start: "+start);

        String phone_txt = objects.get(position).phno;
        phone_txt = "("+phone_txt.substring(0, 3)+") "+phone_txt.substring(3, 6)+"-"+phone_txt.substring(6, 10);

        holder.tv4.setText(phone_txt);
    }
    return rowview;
}

static class ViewHolder
{
    TextView tv1,tv2,tv3,tv4;
    View rowview;
}

提前致谢。

Answer 1

您需要在if else语句之外填充文本视图。中间的那些被第一个案件抓住了，而且房产没有被填满。

public View getView(final int position, View convertView,final ViewGroup parent)
{
    View rowview = convertView; 
    ViewHolder viewholder;

    if (rowview == null) {
        viewholder = new ViewHolder();
        final LayoutInflater inflater = (LayoutInflater)context.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
        rowview = inflater.inflate(R.layout.list_skeleton_people_expenses, parent, false);
        viewholder.tv1 = (TextView)rowview.findViewById(R.id.name);
        viewholder.tv2 = (TextView)rowview.findViewById(R.id.destination);
        viewholder.tv3 = (TextView)rowview.findViewById(R.id.start);
        viewholder.tv4 = (TextView)rowview.findViewById(R.id.phno);
        viewholder.rowview = rowview;
        rowview.setTag(viewholder);     
    } else {
        viewholder = (ViewHolder) convertView.getTag();
    }

    UserData temp = objects.get(position);

    holder.tv1.setText(temp.name);

    holder.tv2.setText("Destination: "+temp.loc);

    holder.tv3.setText("Start: "+start);

    String phone_txt = objects.get(position).phno;
    phone_txt = "("+phone_txt.substring(0, 3)+") "+phone_txt.substring(3, 6)+"-"+phone_txt.substring(6, 10);

    holder.tv4.setText(phone_txt);
    return rowview;
}

另外，随机的一些建议，尝试使用StringBuilder而不是将一堆字符串与'+'运算符连接起来，因为效率非常低。每次使用'+'时，Java都会在后台创建一个新的StringBuilder对象，将字符串附加在一起，因为字符串不可变。所以不妨只创建一个StringBuilder并在其上附加要连接的字符串，而不是使用“+”四到五次，并强制它创建垃圾收集器必须清理的更多不必要的对象。

快乐的编码！

Viewholder为Listview加载了太多时间

1 个答案: