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我正在Android上构建音乐播放器应用程序,我无法在设备上快速加载歌曲列表。它在大约6.8-7秒内加载577首歌曲,这太长了。有小费吗?慢速音乐加载算法
我试图获得以下信息: 宋
- 歌曲名称
- 歌曲艺术家
- 专辑名称
- 宋ID
- 无论是铃声或通知(以忽略它)
她e是我目前的算法:
public static ArrayList<Song> getSongList(Activity activity, String artistBound, Album albumBound) {
long start = System.currentTimeMillis();
ArrayList<Song> songList = new ArrayList<>();
ContentResolver musicResolver = activity.getContentResolver();
Uri musicUri = android.provider.MediaStore.Audio.Media.EXTERNAL_CONTENT_URI;
Cursor musicCursor = musicResolver.query(musicUri, null, null, null, null);
if(musicCursor!=null && musicCursor.moveToFirst()){
//get columns
int titleColumn = musicCursor.getColumnIndex
(android.provider.MediaStore.Audio.Media.TITLE);
int idColumn = musicCursor.getColumnIndex
(android.provider.MediaStore.Audio.Media._ID);
int artistColumn = musicCursor.getColumnIndex
(android.provider.MediaStore.Audio.Media.ARTIST);
int albumIdColumn = musicCursor.getColumnIndex
(MediaStore.Audio.Media.ALBUM_ID);
int ringColumn = musicCursor.getColumnIndex(MediaStore.Audio.Media.IS_RINGTONE);
int notifColumn = musicCursor.getColumnIndex(MediaStore.Audio.Media.IS_NOTIFICATION);
//add songs to list
do {
long thisId = musicCursor.getLong(idColumn);
String thisTitle = musicCursor.getString(titleColumn);
Artist thisArtist = new Artist(musicCursor.getString(artistColumn));
if(artistBound != null && !thisArtist.getName().equals(artistBound)) continue;
long albumId = musicCursor.getLong(albumIdColumn);
Cursor albumCursor = activity.getContentResolver().query(
MediaStore.Audio.Albums.EXTERNAL_CONTENT_URI,
new String[]{MediaStore.Audio.Albums.ALBUM_ART, MediaStore.Audio.Albums.ALBUM},
MediaStore.Audio.Albums._ID + " = ?",
new String[]{Long.toString(albumId)},
null
);
boolean queryResult = albumCursor.moveToFirst();
String albumCover = null;
String albumName = null;
Album album = null;
if (queryResult) {
albumCover = albumCursor.getString(0);
albumName = albumCursor.getString(1);
album = new Album(albumName, albumCover);
}
albumCursor.close();
if(musicCursor.getInt(ringColumn) > 0 || musicCursor.getInt(notifColumn) > 0) {
} else {
if(albumBound == null || albumBound.getName().equals(album.getName())) {
songList.add(new Song(thisId, thisTitle, thisArtist, album));
}
}
}
while (musicCursor.moveToNext());
}
Collections.sort(songList);
Log.d("app", "Got " + songList.size() + " songs in: " + (System.currentTimeMillis() - start) + "ms");
return songList;
}
的问题是,我想在开始时加载这个名单,作为音乐播放器的第一个画面是歌曲列表。但就目前而言,应用程序在试图加载时挂在此屏幕上。
编辑:删除搜索相册信息的代码使其运行得非常快。我怎样才能优化专辑信息搜索?
缓存结果,只查找新文件? –
也许在一个异步任务上运行它,并使用publishProgress(Song song),在那里你做songList.add,这样UI保持响应,并立即开始显示歌曲 – Linxy