8

在 C# 中将某些 xml 反序列化为对象时遇到问题。

我收到的错误是...

xmlns=''> was not expected.

我收到的用于生成我的课程的 XSD 如下...

<?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）

我尝试导入的文件如下：

<?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 .Net 尝试对 xml 进行反序列化时，它会遇到包含“xmlns:ast”的第一行并抱怨它。据我了解，.Net 将尝试将属性映射到目标类中的公共属性（它不会找到一个名为 xmlns 的属性。或者我处理名称空间的方式有问题。

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));

我还尝试了将命名空间管理器添加到 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");

我知道有一种方法可以用来告诉 .Net 接受哪些命名空间。

对于这个问题的一些帮助将非常有用。

--- 根据要求更新了类片段（抱歉之前应该包含）这是用 xsd.exe 创建的 ---

 /// <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;
4

2 回答 2

20

如果没有完整的 xsd / xml 或（或者）您的 C# 类，我们将无法重现。但是从 xml 向上工作，这对我来说很好；含义：错误不在您发布的代码/数据中（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
}

也适用于：

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));
于 2010-11-03T09:42:14.500 回答
0

这是我正在使用的（抱歉我迟到了）：

  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
于 2015-04-01T18:48:21.337 回答