XML désérialisation (Universal App) avec sérialisé vb.net

Question

Je sérialisation une classe (PSD) en utilisant vb.net dans une application Windows Forms:

Dim objStreamWriter As New System.IO.StreamWriter("e:\test.xml") 
Dim x As New System.Xml.Serialization.XmlSerializer(PSD.GetType) 
x.Serialize(objStreamWriter, PSD) 
objStreamWriter.Close() 

Je dois deserialze cette classe dans une application universelle de Windows . Ce est mon code à ce jour:

Dim picker As Windows.Storage.Pickers.FileOpenPicker = New Windows.Storage.Pickers.FileOpenPicker 
picker.ViewMode = Windows.Storage.Pickers.PickerViewMode.Thumbnail 
picker.FileTypeFilter.Add(".xml") 
Dim File As Windows.Storage.StorageFile = Await picker.PickSingleFileAsync 
Dim Fl = Await File.OpenAsync(Windows.Storage.FileAccessMode.Read) 
Dim inStream As Stream = Fl.AsStreamForRead() 
Dim PSD As New PlatinaStammdaten 
Dim serializer = New XmlSerializer(GetType(List(Of PlatinaStammdaten)), New Type() {GetType(PlatinaStammdaten)}) 
PSD = serializer.Deserialize(inStream) 

Ce que je reçois est l'erreur "Il y a une erreur dans le document XML (2, 2)."

Ceci est le début du fichier XML généré par:

<?xml version="1.0" encoding="utf-8"?> 
<Stammdaten xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> 
    <Lose> 
    <Los> 
     <Name>Sonstige</Name> 
     <KZ>000</KZ> 
    </Los> 
    <Los> 
     <Name>LOS 1; Allgemein/Planung</Name> 
     <KZ>110</KZ> 
    </Los> 
    ... 

Quelle erreur est dans mon code (écriture ou la lecture) qui conduit à cette erreur?

Answer 1

Je l'ai travaille maintenant. Ceci est mon code désérialisation maintenant:

Dim picker As Windows.Storage.Pickers.FileOpenPicker = New Windows.Storage.Pickers.FileOpenPicker 
     picker.FileTypeFilter.Add(".xml") 
     Dim File As Windows.Storage.StorageFile = Await picker.PickSingleFileAsync 
     Dim Fl = Await File.OpenAsync(Windows.Storage.FileAccessMode.Read) 
     Dim FileZiel As Windows.Storage.StorageFile = Await File.CopyAsync(FolderZiel, File.Name, Windows.Storage.NameCollisionOption.ReplaceExisting) 
     Dim Fl2 = Await FileZiel.OpenAsync(Windows.Storage.FileAccessMode.ReadWrite) 
     Dim inStream2 As Stream = Fl2.AsStreamForRead() 
     Dim sr As StreamReader = New StreamReader(inStream2) 
     Dim PSD As New PlatinaStammdaten 
     Dim x As New XmlSerializer(PSD.GetType) 
     PSD = x.Deserialize(sr) 
     sr.Dispose() 

Answer 2

On dirait que vous essayez de désérialiser une liste en une seule instance

Dim PSD en tant que nouveau PlatinaStammdaten Dim sérialiseur = New XmlSerializer (GetType (List (Of PlatinaStammdaten)), Nouveau Type() {GetType (PlatinaStammdaten)}) PSD = serializer.Deserialize (InStream)

Essayez de changer PSD à

Dim PSD As List(of PlatinaStammdaten) 

puisque nous ne pouvons pas voir le xml complet ou la classe que vous mettez le XML en elle i Difficile de dire si vous rencontrerez d'autres erreurs

Answer 3

Le code ci-dessous fonctionne

Imports System.Xml 
Imports System.Xml.Serialization 
Imports System.IO 

Module Module1 
    Const FILENAME As String = "c:\temp\test.xml" 
    Sub Main() 
     Dim xs As XmlSerializer = New XmlSerializer(GetType(PlatinaStammdaten)) 
     Dim reader As XmlTextReader = New XmlTextReader(FILENAME) 
     Dim PSD As PlatinaStammdaten = CType(xs.Deserialize(reader), PlatinaStammdaten) 

    End Sub 

End Module 
<XmlRoot("Stammdaten")> _ 
Public Class PlatinaStammdaten 
    Private m_lose As Lose 
    <XmlElement("Lose")> _ 
    Public Property _lose As Lose 
     Get 
      Return m_lose 
     End Get 
     Set(ByVal value As Lose) 
      m_lose = value 
     End Set 
    End Property 
End Class 

<Serializable(), XmlRoot("Lose")> _ 
Public Class Lose 
    Private m_los As List(Of Los) 
    <XmlElement("Los")> _ 
    Public Property _los As List(Of Los) 
     Get 
      Return m_los 
     End Get 
     Set(ByVal value As List(Of Los)) 
      m_los = value 
     End Set 
    End Property 

End Class 

<Serializable(), XmlRoot("Los")> _ 
Public Class Los 
    Private m_Name As String 
    <XmlElement("Name")> _ 
    Public Property name As String 
     Get 
      Return m_Name 
     End Get 
     Set(ByVal value As String) 
      m_Name = value 
     End Set 
    End Property 

    Public m_KZ As Integer 
    <XmlElement("KZ")> _ 
    Public Property KZ As Integer 
     Get 
      Return m_KZ 
     End Get 
     Set(ByVal value As Integer) 
      m_KZ = value 
     End Set 
    End Property 

End Class 

'<Lose> 
' <Los> 
' <Name>Sonstige</Name> 
' <KZ>000</KZ> 
' </Los> 
' <Los> 
' <Name>LOS 1; Allgemein/Planung</Name> 
' <KZ>110</KZ> 
' </Los> 
'</Lose>