節點/快速目錄瀏覽

Question

我試着整天來解決這個問題：

我有子目錄的目錄，FE：

| Music Artist 1 
| - Album Nr 1 
| -- Track 1 
| -- Track 2 
| -- ... 
| - Album Nr 2 
| -- Track 1 
| -- Track 2 
| -- ... 
| Music Artist 2 
| - Album Nr 1 
| -- Track 1 
| -- Track 2 
| -- ... 

現在，我將這些目錄循環 - 與添加所有細節到一個數組/對象。 因此，它應該是這樣的：

[ { artist: Music Artist 1, album { title: Album Nr1, songs: { title: Track 1 } ... } ] 

讓所有目錄名/文件是沒有問題的。我只是不明白如何創建數組:(

在先進的感謝

編輯： 這是我的嘗試：http://pastebin.com/vWnbvu5m

Answer 1

您可以創建artist對象和push()每一個你創建成的陣列。同樣，album和song可以是對象push()編成對應附連到它們的父對象陣列。

var artists = []; 
// for each artist we have 
    var artist = {}; 
    artist.name = 'Music Artist 1'; 
    artist.albums = []; 
    // for each album we have 
     var album = {}; 
     album.title = 'Album Nr1' 
     album.songs = []; 
     // for each song that we have 
      var song = {}; 
      song.title = 'Track 1'; 
      album.songs.push(song); 
     // end song loop 
     artist.albums.push(album); 
    // end album loop 
    artists.push(artist) 
// end artist loop 

如果您隨後需要JSON格式的此信息，則可以使用JSON解析器對其進行解析。或者，您可以編程讀取每個artist的數據，方法是循環訪問artists陣列。

// returns name of first artist in array 
artists[0].name; 

// returns title of first album by first artist in respective arrays 
artists[0].albums[0].title; 

// returns title of first song in first album by first artist in respective arrays 
artists[0].albums[0].songs[0].title;