ListVIew第一item里的Button状态的问题
@Overridepublic View getView(int position, View convertView, ViewGroup arg2) {
if (convertView == null) {
convertView = LayoutInflater.from(mContext).inflate(R.layout.update_item_layout, null);
textUpdate = (TextView)convertView.findViewById(R.id.update);
TextView textName = (TextView)convertView.findViewById(R.id.name);
bean = (Bean)getItem(position);
textName.setText(bean.name);
if (bean.status == 0) {
textUpdate.setText("升级");
} else if (bean.status == 1) {
textUpdate.setText("暂停");
}
final int positions = position;
textUpdate.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
if (bean.status == 0) {
textUpdate.setText("升级");
Log.d("test", "----->status = 0");
} else if (bean.status == 1) {
textUpdate.setText("暂停");
Log.d("test", "----->status = 1");
}
}
});
}
return convertView;
}
怎么点击能改变其状态值,并在UI上显示出来。 listview button --------------------编程问答-------------------- 你这个有效果吗? --------------------编程问答-------------------- 这样是不会响应的,写各接口,注册一个回调才可以,用settag把position传过来
补充:移动开发 , Android