java - 单击 ListView 上的按钮会显示相同的输出

Question

public View getView(int position, View convertView, ViewGroup parent) {
    // TODO Auto-generated method stub
    inflater = (LayoutInflater)context.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
    View itemView = inflater.inflate(R.layout.joinlistviewitem, parent, false);
    resultp = data.get(position);
    name = (TextView) itemView.findViewById(R.id.personname);
    name.setText(resultp.get(Joinlistview.RANK));   
    Button deletejoin=(Button)itemView.findViewById(R.id.delete);
    deletejoin.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View v) {
            //resultp.get("userId");            
            // TODO Auto-generated method stub
            Log.e("delete clicked", "delete clicked");
            Toast.makeText(context,"clicked"+resultp.get(Joinlistview.RANK), Toast.LENGTH_SHORT).show();            
        }
    });

    return itemView;
}

我使用上面的代码来单击 ListView 上的按钮。如果我单击按钮，它会播种用户名及其位置。但它在所有位置都显示相同的名称。??如何解决这个问题??

Answer 1

使用 getTag/setTag 方法获取 onClick 中的当前行值，如下所示:

Button deletejoin=(Button)itemView.findViewById(R.id.delete);
deletejoin.setTag(resultp.get(Joinlistview.RANK));

点击按钮:

Toast.makeText(context,"clicked "+ v.getTag().toString(),
                                   Toast.LENGTH_SHORT).show();