如何从歌曲中排序专辑？
所以我的问题是，当我尝试对专辑进行排序时，专辑标题和专辑封面是错误的。
我尝试对专辑ID进行排序，但这并不能解决问题，因为专辑ID显然与艺术排序无关。
当我忽略排序时，一切都是正确的，但是当我尝试对它们进行排序时，专辑名称与专辑封面不匹配。
如何将片段中的专辑排序？
在这里您可以找到我的代码。
提前致谢，
文斯
// Columns I'll retrieve from the song table String[] columns = {
SONG_ID,
SONG_TITLE,
SONG_ARTIST,
SONG_ALBUM,
SONG_ALBUMID,
SONG_FILEPATH,
};
// Limits results to only show music files.
//
// It's a SQL "WHERE" clause - it becomes `WHERE IS_MUSIC=1`.
//
final String musicsOnly = MediaStore.Audio.Media.IS_MUSIC + "=1";
// Querying the system
cursor = resolver.query(musicUri, columns, musicsOnly, null, null);
if (cursor != null && cursor.moveToFirst())
{
do {
// Creating a song from the values on the row
Song song = new Song(cursor.getInt(cursor.getColumnIndex(SONG_ID)),
cursor.getString(cursor.getColumnIndex(SONG_FILEPATH)));
song.setTitle (cursor.getString(cursor.getColumnIndex(SONG_TITLE)));
song.setArtist (cursor.getString(cursor.getColumnIndex(SONG_ARTIST)));
song.setAlbumID (cursor.getInt(cursor.getColumnIndexOrThrow(SONG_ALBUMID)));
song.setAlbum (cursor.getString(cursor.getColumnIndexOrThrow(SONG_ALBUM)));
// Using the previously created genre and album maps
// to fill the current song genre.
String currentGenreID = songIdToGenreIdMap.get(Long.toString(song.getId()));
String currentGenreName = genreIdToGenreNameMap.get(currentGenreID);
song.setGenre(currentGenreName);
// Adding the song to the global list
songs.add(song);
}
while (cursor.moveToNext());
}
else
{
// What do I do if I can't find any songs?
}
cursor.close();
public ArrayList<String> getArtists() {
ArrayList<String> artists = new ArrayList<String>();
for (Song song : songs) {
String artist = song.getArtist();
if ((artist != null) && (! artists.contains(artist)))
artists.add(artist);
}
// Making them alphabetically sorted
Collections.sort(artists, new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
return o1.compareTo(o2);
}
});
return artists;
}
/**
* Returns an alphabetically sorted list with all the
* albums of the scanned songs.
*
* @note This method might take a while depending on how
* many songs you have.
*/
public ArrayList<String> getAlbums() {
ArrayList<String> albums = new ArrayList<String>();
for (Song song : songs) {
String album = song.getAlbum();
if ((album != null) && (! albums.contains(album)))
albums.add(album);
}
public class Song implements Serializable {private long id;
private String data;
private String title = "";
private String artist = "";
private int albumid = -1;
private String album = "";
private String genre = "";
public Song(long songId, String songData){
this.id = songId;
this.data = songData;
}
public long getId(){
return id;
}
public String getData(){return data;}
//Optional meta data
public void setTitle(String title){
this.title = title;
}
public String getTitle() {
return title;
}
public void setArtist(String artist){
this.artist = artist;
}
public String getArtist() {
return artist;
}
public int getAlbumID() {
return albumid;
}
public void setAlbumID(int albumid) { this.albumid = albumid; }
public void setAlbum(String album){
this.album = album;
}
public String getAlbum() { return album; }
public void setGenre(String genre) {
this.genre = genre;
}
public String getGenre() {
return genre;
}
}
回答：
首先，我不确定在存储按歌曲存储的返回值时为什么要尝试按专辑排序（请参见上面的@Usman Rafi），但是。
ArrayList<Song> Albums = new Arraylist<>();我尝试对专辑ID进行排序，但这并不能解决问题，因为专辑ID显然与艺术排序无关。
专辑封面Uri可以写成：
ContentUris.withAppendedId(Uri.parse("content://media/external/audio/albumart"), cursor.getInt(cursor.getInt(cursor.getColumnIndexOrThrow(SONG_ALBUMID))));
因此，专辑封面和album_id实际上是密不可分的。
所以我的问题是当我尝试对专辑进行排序时…
在查询的选择变量中使用MediaStore.Audio.Media.IS_MUSIC +“ = 1）GROUP BY（” +
MediaStore.Audio.Media.ALBUM …
这将返回唯一的专辑名称（也将仅返回专辑中的一首歌曲），如果在您的媒体存储数据库中重复该专辑（通过从同一专辑中获得几首歌曲）重复该专辑，则只会添加与您的查询匹配的第一个实例到您的光标。
对专辑进行排序…
使用排序顺序对相册返回的光标行进行排序；我个人使用sql的字母顺序对其进行排序（符号，数字，a，b，c…。）
您应在此处注意，排序是区分大小写的，除非您指定“ COLLATE NOCASE”
编写查询并对其进行排序，我将使用以下代码：
String[] projection = {"DISTINCT " + MediaStore.Audio.Media.ALBUM_ID, MediaStore.Audio.Media._ID,
MediaStore.Audio.Media.TITLE,
MediaStore.Audio.Media.ARTIST,
MediaStore.Audio.Media.DATA,
MediaStore.Audio.Media.ALBUM,
MediaStore.Audio.Media.IS_MUSIC};
String selection = MediaStore.Audio.Media.IS_MUSIC +
"=1 ) GROUP BY (" + MediaStore.Audio.Media.ALBUM_ID;
String sort = MediaStore.Audio.Media.ALBUM + " COLLATE NOCASE ASC";
Cursor cursor = context.
getContentResolver().
query(MediaStore.Audio.Artists.EXTERNAL_CONTENT_URI,
projection,
selection,
null,
sort);
之后，您可以简单地在光标中移动，将每一行添加到所构建的数据对象中，无需进一步排序，并且事物应该以正确的顺序进行。
我个人只是遍历
if(cursor != null && cursor.getCount >0){ while(cursor.moveToNext()){
//create new song item and add the album information you need
Song album = new Song(cursor.getInt(cursor.getColumnIndex(MediaStore.Audio.Media._ID)),
cursor.getString(cursor.getColumnIndex(MediaStore.Audio.Media.DATA)));
album.setAlbumId(cursor.getInt(cursor.getColumnIndex(MediaStore.Audio.Media.ALBUM_ID)));
album.setAlbumId(cursor.getInt(cursor.getColumnIndex(MediaStore.Audio.Media.ALBUM)));
//add the Song item to the global arraylist
Albums.add(album)
}
}
cursor.close();
您现在可以按数组列表中的位置访问排序的专辑信息…您可以使用我在顶部显示的Uri构建器了解专辑封面…
像这样
Song Album = Albums.get(position); imageView.setURI(ContentUris.withAppendedId(Uri.parse("content://media/external/audio/albumart"),
Album.getAlbumID());
我希望这对您有用。
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