C# 列表视图项绑定
我得到了一个场景,需要用音频文件名绑定android xamarin中的ListView。 它应该只显示音频名称的前5个字符。我通过拆分它来实现这一点,它工作得很好C# 列表视图项绑定,c#,android,.net,mono,xamarin,C#,Android,.net,Mono,Xamarin,我得到了一个场景,需要用音频文件名绑定android xamarin中的ListView。 它应该只显示音频名称的前5个字符。我通过拆分它来实现这一点,它工作得很好 adapter = new ArrayAdapter<string> ( this , Resource.Layout.list_item , Resource.Id.audio_names , lstSearchResult.ToArray () ); listResult.Adapter = adapte
adapter = new ArrayAdapter<string> ( this , Resource.Layout.list_item , Resource.Id.audio_names , lstSearchResult.ToArray () );
listResult.Adapter = adapter;
任何建议/替代vl都值得赞赏最好的办法是创建自定义适配器。 首先,为一行创建自己的布局。称之为“SongListRow”(或任何其他名称) 然后为您的歌曲创建一个类:
public class Song {
public string Name { get; set; }
public string File { get; set; }
}
然后创建适配器:
public class SongListAdapter : BaseAdapter<Song> {
private List<Song> _items;
private Activity _context;
public SongListAdapter(Activity context, List<Song> songs)
{
this._items = songs;
this._context = context;
}
public override Song this[int position]
{
get { return this._items[position]; }
}
public override int Count
{
get { return this._items.Count; }
}
public override long GetItemId(int position)
{
return position;
}
public override View GetView(int position, View convertView, ViewGroup parent)
{
var item = this._items[position];
View view = convertView;
//If there is nothing to reuse, then create view from your row layout
if (view == null)
view = this._context.LayoutInflater.Inflate(Resource.Layout.SongListRow, null);
view.FindViewById<TextView>(Resource.Id.SongTitle).Text = item.Name;
return view;
}
}
现在创建歌曲的通用列表,创建适配器并将歌曲列表传递给适配器的构造函数。然后你可以这样做:
listResult.Adapter = this._songsAdapter;
listResult.ItemClick += (s,e) => {
var fileName = this._songsAdapter[e.Position].File;
// Play your song
}
将发送者的DataContext设置为对象的实例,并从中获取完整数据非常感谢…节省了我的时间:)如果在列表中添加、删除某个项目,或只是更改了某个项目,则不会更新listResult(UI)。如何应对?
private SongListAdapter _songsAdapter;
listResult.Adapter = this._songsAdapter;
listResult.ItemClick += (s,e) => {
var fileName = this._songsAdapter[e.Position].File;
// Play your song
}