我在设备上创建了一系列歌曲。
对于每个项目,在我的ListView中显示艺术家姓名和歌曲名称。我希望当选择一首歌时,开始播放,我该怎么办?
StorageFolder musicLibrary = KnownFolders.MusicLibrary;
IReadOnlyList<StorageFile> musica = await musicLibrary.GetFilesAsync();
if (musica != null)
{
List<Testo> song = new List<Testo>();
{
foreach (StorageFile storage in musica)
{
MusicProperties musicProp = await storage.Properties.GetMusicPropertiesAsync();
song.Add(new Testo
{
NomeArtista = musicProp.Artist,
NomeCanzone = musicProp.Title,
Anno = (int)musicProp.Year,
});
}
}
}
private async void TestiCanzone_ItemClick(object sender, ItemClickEventArgs e)
{
Testo NuovoTesto = e.ClickedItem as Testo;
}
我还创建了一个MediaElement
<MediaElement x:Name="AudioPlay" Source="" AutoPlay="True"/>
答案 0 :(得分:0)
您需要将listview的SelectedValue
绑定到viewmodel上的某个属性:
<ListView SelectedItem="{Binding SelectedAudio, Mode=TwoWay}"/>
在您的viewmodel上,您需要处理更改属性:
public string SelectedCustomMusic
{
get
{
return this.selectedCustomMusic;
}
set
{
if (value != null)
{
this.selectedCustomMusic = value;
this.MusicSource = this.selectedCustomMusic;
base.OnPropertyChanged();
}
}
}
然后将this.MusicSource
绑定到Source
的{{1}}。