C＃反序列化XML到对象

Question

在将某些xml反序列化为C＃中的对象时出现问题。

，我收到的错误...

xmlns=''> was not expected. 

，我收到生成我的类是如下的XSD ...

<?xml version="1.0" encoding="UTF-8"?> 
<xs:schema targetNamespace="xml.AAAAAAA.com/commerce/apres-vente_technique/assistance" xmlns:pgp="xml.AAAAAAA.com/commerce/apres-vente_technique/assistance" xmlns:xs="http://www.w3.org/2001/XMLSchema" elementFormDefault="qualified" attributeFormDefault="unqualified"> 
    <xs:element name="ListeAvisRemboursements"> 
     <xs:annotation> 
      <xs:documentation>Liste des avis de remboursements</xs:documentation> 
     </xs:annotation> 
     <xs:complexType> 
      <xs:sequence maxOccurs="unbounded"> 
       <xs:element name="AvisRemboursement" type="pgp:AvisRemboursementType"/> 
      </xs:sequence> 
     </xs:complexType> 
    </xs:element> 
    <xs:complexType name="AvisRemboursementType"> 
     <xs:annotation> 
      <xs:documentation>Avis de remboursement lié à une DC</xs:documentation> 
     </xs:annotation> 
     <xs:sequence> 

（剪断）

的我试图导入的文件如下：

<?xml version="1.0" encoding="UTF-8"?> 
<ListeAvisRemboursements xmlns:ast="xml.AAAAAAA.com/commerce/apres-vente_technique/assistance"> 
    <ast:AvisRemboursement NumeroDT="3826961" CodeRA="020545G01" NumeroDC="1"> 
     <ast:DateTraitement>2010-06-22</ast:DateTraitement> 
     <ast:MontantDC>25.0</ast:MontantDC> 
     <ast:MontantMO>0.0</ast:MontantMO> 
     <ast:SommeAD>25.0</ast:SommeAD> 
     <ast:MontantPR>0.0</ast:MontantPR> 
     <ast:SommePR>0.0</ast:SommePR> 
     <ast:FraisGestion>0.0</ast:FraisGestion> 
     <ast:NombreHeuresTotalRemboursees>0</ast:NombreHeuresTotalRemboursees> 
     <ast:Etat>C</ast:Etat> 
     <ast:NoteCredit>319984</ast:NoteCredit> 
     <ast:Imputation>030</ast:Imputation> 
     <ast:ListInterventionsPR/> 
     <ast:ListInterventionsMO/> 
    </ast:AvisRemboursement> 

（剪断）

我认为正在发生的事情是，当净试图derserialize的XML，它击中其中包含的第一行“的xmlns：AST”和关于它的投诉。据我了解，.Net会尝试将属性映射到目标类中的公共属性（并且它不会找到一个名为xmlns的对象），或者我如何处理名称空间时出现问题。

我的反序列化代码如下所示：

XmlDocument _Doc = new XmlDocument(); 
    _Doc.Load(@"C:\inputfile.xml"); 

    XmlSerializer _XMLSer = new XmlSerializer(typeof(ListeAvisRemboursements)); 
    ListeAvisRemboursements _X = (ListeAvisRemboursements)_XMLSer.Deserialize(new StringReader(_Doc.OuterXml)); 

我也试图加入一个命名空间管理XML文档的各种组合..

XmlNamespaceManager _Ns = new XmlNamespaceManager(_Doc.NameTable); 
_Ns.AddNamespace("ast", "xml.AAAAAAA.com/commerce/apres-vente_technique/assistance"); 

我知道有一种方法，我可以用它来告诉净什么命名空间接受。

对于解决这个问题会有帮助。

---在带班片断请求（抱歉，应该包括之前），这与创建XSD.EXE更新---

/// <remarks/> 
    [System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "4.0.30319.1")] 
    [System.SerializableAttribute()] 
    [System.Diagnostics.DebuggerStepThroughAttribute()] 
    [System.ComponentModel.DesignerCategoryAttribute("code")] 
    [System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true, Namespace = "xml.AAAAAAA.com/commerce/apres-vente_technique/assistance")] 
    [System.Xml.Serialization.XmlRootAttribute(Namespace = "xml.AAAAAAA.com/commerce/apres-vente_technique/assistance", IsNullable = false)] 
    public partial class ListeAvisRemboursements 
    { 

     private AvisRemboursementType[] avisRemboursementField; 

     /// <remarks/> 
     [System.Xml.Serialization.XmlElementAttribute("AvisRemboursement")] 
     public AvisRemboursementType[] AvisRemboursement 
     { 
      get 
      { 
       return this.avisRemboursementField; 
      } 
      set 
      { 
       this.avisRemboursementField = value; 
      } 
     } 
    } 

    /// <remarks/> 
    [System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "4.0.30319.1")] 
    [System.SerializableAttribute()] 
    [System.Diagnostics.DebuggerStepThroughAttribute()] 
    [System.ComponentModel.DesignerCategoryAttribute("code")] 
    [System.Xml.Serialization.XmlTypeAttribute(Namespace = "xml.AAAAAAA.com/commerce/apres-vente_technique/assistance")] 
    public partial class AvisRemboursementType 
    { 

     private System.DateTime dateTraitementField; 

     private double montantDCField; 

     private double montantMOField; 

     private double sommeADField; 

     private double montantPRField; 

Answer 1

如果没有一个完整的XSD/XML，或（或者）你的C＃类，我们不能重现。但是从xml开始工作起来，这对我来说很好。意思是：在您发布的代码/数据中，错误不是（据我所知）。你能发布一个更完整（可重现）的例子吗？

public class ListeAvisRemboursements 
{ 
    private readonly List<AvisRemboursement> items = new List<AvisRemboursement>(); 
    [XmlElement("AvisRemboursement", Namespace = "xml.AAAAAAA.com/commerce/apres-vente_technique/assistance")] 
    public List<AvisRemboursement> Items { get { return items; } } 
} 
public class AvisRemboursement 
{ 
    [XmlAttribute] public string NumeroDT {get;set;} 
    [XmlAttribute] public string CodeRA {get;set;} 
    [XmlAttribute] public string NumeroDC {get;set;} 
    public DateTime DateTraitement { get; set; } 
    public decimal MontantDC { get; set; } 
    public decimal MontantMO { get; set; } 
    public decimal SommeAD { get; set; } 
    public decimal MontantPR { get; set; } 
    public decimal SommePR { get; set; } 
    public decimal FraisGestion { get; set; } 
    public int NombreHeuresTotalRemboursees { get; set; } 
    public string Etat { get; set; } 
    public string NoteCredit { get; set; } 
    public string Imputation { get; set; } 
} 
static void Main() 
{ 
    var ser = new XmlSerializer(typeof(ListeAvisRemboursements)); 
    var wrapper = (ListeAvisRemboursements)ser.Deserialize(new StringReader(xml)); 
    // inspect wrapper.Items etc 
} 

也工作正常：

var ser = new XmlSerializer(typeof(ListeAvisRemboursements)); 
using (var reader = XmlReader.Create("inputfile.xml")) 
{ 
    var wrapper = (ListeAvisRemboursements)ser.Deserialize(reader); 
} 

和：

XmlDocument _Doc = new XmlDocument(); 
_Doc.Load("inputfile.xml"); 
var ser = new XmlSerializer(typeof(ListeAvisRemboursements)); 
var wrapper = (ListeAvisRemboursements)ser.Deserialize(new StringReader(_Doc.OuterXml)); 

和

XmlDocument _Doc = new XmlDocument(); 
_Doc.Load("inputfile.xml"); 
var ser = new XmlSerializer(typeof(ListeAvisRemboursements)); 
var wrapper = (ListeAvisRemboursements)ser.Deserialize(new XmlNodeReader(_Doc.DocumentElement)); 

Answer 2

下面是我在用的（对不起我有点晚了党）：

Public Function Serialize(Of YourXMLClass)(ByVal obj As YourXMLClass, 
                 Optional ByVal omitXMLDeclaration As Boolean = True, 
                 Optional ByVal omitXMLNamespace As Boolean = True) As String 

     Dim serializer As New XmlSerializer(obj.GetType) 
     Using memStream As New MemoryStream() 
      Dim settings As New XmlWriterSettings() With { 
        .Encoding = Encoding.UTF8, 
        .Indent = True, 
        .OmitXmlDeclaration = omitXMLDeclaration} 

      Using writer As XmlWriter = XmlWriter.Create(memStream, settings) 
       Dim xns As New XmlSerializerNamespaces 
       If (omitXMLNamespace) Then xns.Add("", "") 
       serializer.Serialize(writer, obj, xns) 
      End Using 

      Return Encoding.UTF8.GetString(memStream.ToArray()) 
     End Using 
    End Function 

Public Function Deserialize(Of YourXMLClass)(ByVal obj As YourXMLClass, ByVal xml As String) As YourXMLClass 
     Dim result As YourXMLClass 
     Dim serializer As New XmlSerializer(GetType(YourXMLClass)) 

     Using memStream As New MemoryStream() 
      Dim bytes As Byte() = Encoding.UTF8.GetBytes(xml.ToArray) 
      memStream.Write(bytes, 0, bytes.Count) 
      memStream.Seek(0, SeekOrigin.Begin) 

      Using reader As XmlReader = XmlReader.Create(memStream) 
       result = DirectCast(serializer.Deserialize(reader), YourXMLClass) 
      End Using 

     End Using 
     Return result 
    End Function