Question

I got a ContentProvider that serves some content e.g. filters. Those will be rendered in a ListView. Because the filter has many fields I needed to create an own View for the list items. The fields are mapped in a class that extends CursorAdapter.

public void bindView(View view, Context context, Cursor cursor) {

    TextView searchPattern = (TextView) view.findViewById(R.id.tv_searchpattern);
    TextView searchType = (TextView) view.findViewById(R.id.tv_searchtype);

    int type = cursor.getInt(FilterProvider.SEARCH_TYPE_COLUMN);
    [...]
}

@Override
public View newView(Context context, Cursor cursor, ViewGroup parent) {
    final View view = inflater.inflate(R.layout.group_list_item, parent, false);
    return view;
}

But now I am wondering how I could "carry" the content uris along with the list items. So that I later can retrieve them later easily to operate (e.g. update, delete) on the item?

利用观点是否是一个好的想法?

    view.setId(cursor.getInt(FilterProvider.KEY_COLUMN));

还是我完全走上错误的道路? 难道我需要担心的是,在Sqlite,Integer实际上是很长的?

Answer 1

顺便说一句,我不喜欢这样的看法。似乎没有好的想法。我可以说,我有这样说的任何困难证据,但治疗者适应者正在处理这些看法,他们说,今后一些平台版本并没有对儿童的看法有任何重要意义?

然而,对于你想要的是什么,有一个完美的方法:

view.setTag(Object someObject);

http://developer.android.com/vis/android/view/View.html#setTag(java.lang.Object

因此,就你而言,你可以:

view.setTag(new Integer(cursor.getInt(FilterProvider.KEY_COLUMN)));

之后,你将:

Integer key = (Integer) view.getTag();

我希望这一帮助!

友情链接