使用LINQ来元组的XML元素？

Question

我有一个这样的XML：

<?xml version="1.0" encoding="Windows-1252"?> 
<!--XML Songs Database.--> 
<Songs> 
    <Song><Name>My Song 1.mp3</Name><Year>2007</Year><Genre>Dance</Genre><Bitrate>320</Bitrate><Length>04:55</Length><Size>4,80</Size></Song> 
    <Song><Name>My Song 2.mp3</Name><Year>2009</Year><Genre>Electro</Genre><Bitrate>192</Bitrate><Length>06:44</Length><Size>8,43</Size></Song> 
    <Song><Name>My Song 3.mp3</Name><Year>2008</Year><Genre>UK Hardcore</Genre><Bitrate>128</Bitrate><Length>05:12</Length><Size>4,20</Size></Song> 
</Songs> 

我想存储的元素放进一个数组或类似容易阅读他们的东西。

我认为元组的Lis将是一个很好的集合“容器”，如果不是，那么我可以听到建议。

我做了转换，但没有使用LINQ，我想简化下面创建的元组使用LINQ的列表，而不是使用FOR +选择情况下的代码，所以我需要帮助重写/改善这个代码：

Dim Name As String = String.Empty 
Dim Year As String = String.Empty 
Dim Genre As String = String.Empty 
Dim Bitrate As String = String.Empty 
Dim Length As String = String.Empty 
Dim Size As String = String.Empty 

Dim SongsList As New List(Of Tuple(Of String, String, String, String, String, String)) 
Dim Elements As IEnumerable(Of XElement) = XDocument.Load(xmlfile).Descendants() 

For Each Element As XElement In Elements 

    Select Case Element.Name 

     Case "Name" 
      Name = Element.Value 
     Case "Year" 
      Year = Element.Value 
     Case "Genre" 
      Genre = Element.Value 
     Case "Bitrate" 
      Bitrate = Element.Value 
     Case "Length" 
      Length = Element.Value 
     Case "Size" 
      Size = Element.Value 
      SongsList.Add(Tuple.Create(Name, Year, Genre, Bitrate, Length, Size)) 

    End Select 

Next 

，并阅读的歌曲我这样做：

For Each song As Tuple(Of String, String, String, String, String, String) In SongsList 

    MsgBox(String.Format("Name:{1}{0}Year:{2}{0}Genre:{3}{0}Bitrate:{4}{0}Length:{5}{0}Size:{6}", _ 
         Environment.NewLine, _ 
         song.Item1, song.Item2, song.Item3, song.Item4, song.Item5, song.Item6)) 

    ' Output: 
    ' 
    ' Name:My Song 1.mp3 
    ' Year:2007 
    ' Genre:Dance 
    ' Bitrate:320 
    ' Length:04:55 
    ' Size:4,80 

Next 

Answer 1

这是那么容易，因为

Dim xml = <?xml version="1.0" encoding="Windows-1252"?> 
      <!--XML Songs Database.--> 
      <Songs> 
       <Song><Name>My Song 1.mp3</Name><Year>2007</Year><Genre>Dance</Genre><Bitrate>320</Bitrate><Length>04:55</Length><Size>4,80</Size></Song> 
       <Song><Name>My Song 2.mp3</Name><Year>2009</Year><Genre>Electro</Genre><Bitrate>192</Bitrate><Length>06:44</Length><Size>8,43</Size></Song> 
       <Song><Name>My Song 3.mp3</Name><Year>2008</Year><Genre>UK Hardcore</Genre><Bitrate>128</Bitrate><Length>05:12</Length><Size>4,20</Size></Song> 
      </Songs> 

Dim result = from song in xml.<Songs>.<Song> 
      select Tuple.Create(song.<Name>.Value, 
           song.<Year>.Value, 
           song.<Genre>.Value, 
           song.<Bitrate>.Value, 
           song.<Length>.Value, 
           song.<Size>.Value) 

---

但不是怪物Tuple，你也可以使用匿名类型。这很容易为

Dim result = from song in xml.<Songs>.<Song> 
      select new with {song.<Name>.Value, 
           song.<Year>.Value, 
           song.<Genre>.Value, 
           song.<Bitrate>.Value, 
           song.<Length>.Value, 
           song.<Size>.Value} 

这样你就可以像用一个有意义的名称访问Name，Year，Genre等：

For Each song In result 
    Console.WriteLine(String.Format("Name:{1}{0}Year:{2}{0}Genre:{3}{0}Bitrate:{4}{0}Length:{5}{0}Size:{6}", _ 
            Environment.NewLine, _ 
            song.Name, song.Year, song.Genre, song.Bitrate, song.Length, song.Size)) 
Next 

，而不是无意义Item1，Item2，Item3等