当xml有多个相同元素时，如何将xml元素与C#模型绑定？

Question

我无法将我的 c# 模型与这种 xml 文件绑定。因为有多个相同的元素。 这是我的 XML 文件的示例。

  <Type>
       <Loop LoopId="2100A" Name="MEMBER NAME">
         <PER>
           <!--Contact Function Code-->
           <PER01>IP<!--Insured Party--></PER01>
           <PER02 />
           <!--Communication Number Qualifier-->
           <PER03>HP<!--Home Phone Number--></PER03>
           <!--Communication Number-->
           <PER04>6235834409</PER04>
         </PER>
       </Loop>
       <Loop LoopId="2100C" Name="MEMBER MAILING ADDRESS">
         <NM1>
           <!--Entity Identifier Code-->
           <NM101>31<!--Postal Mailing Address--></NM101>
           <!--Entity Type Qualifier-->
           <NM102>1<!--Person--></NM102>
         </NM1>
       </Loop>
       <Loop LoopId="2100G" Name="RESPONSIBLE PERSON">
         <PER>
           <!--Contact Function Code-->
           <PER01>IP<!--Insured Party--></PER01>
           <PER02 />
           <!--Communication Number Qualifier-->
           <PER03>HP<!--Home Phone Number--></PER03>
           <!--Communication Number-->
           <PER04>6235834409</PER04>
         </PER>
         <LM>
           <!--Contact Function Code-->
           <LM01>RP<!--Responsible Person--></LM01>
           <LM02 />
         </LM>
       </Loop>
   </Type>

我已使用以下代码进行绑定。但是代码不起作用，导致数据绑定混乱。

        [XmlElement(ElementName = "Loop")]
        public L3_L1_MemberName L3_L1_MemberName { get; set; }

        [XmlElement(ElementName = "Loop")]
        public L3_L2_MemberMailingAddress L3_L2_MemberMailingAddress { get; set; }

        [XmlElement(ElementName = "Loop")]
        public L3_L3_ResponsiblePerson L3_L3_ResponsiblePerson { get; set; }

Answer 1

基于您的 XML，模型看起来像

[XmlRoot(ElementName = "PER")]
public class PER
{
    [XmlElement(ElementName = "PER01")]
    public string PER01 { get; set; }
    [XmlElement(ElementName = "PER02")]
    public string PER02 { get; set; }
    [XmlElement(ElementName = "PER03")]
    public string PER03 { get; set; }
    [XmlElement(ElementName = "PER04")]
    public string PER04 { get; set; }
}

[XmlRoot(ElementName = "Loop")]
public class Loop
{
    [XmlElement(ElementName = "PER")]
    public PER PER { get; set; }
    [XmlAttribute(AttributeName = "LoopId")]
    public string LoopId { get; set; }
    [XmlAttribute(AttributeName = "Name")]
    public string Name { get; set; }
    [XmlElement(ElementName = "NM1")]
    public NM1 NM1 { get; set; }
    [XmlElement(ElementName = "LM")]
    public LM LM { get; set; }
}

[XmlRoot(ElementName = "NM1")]
public class NM1
{
    [XmlElement(ElementName = "NM101")]
    public string NM101 { get; set; }
    [XmlElement(ElementName = "NM102")]
    public string NM102 { get; set; }
}

[XmlRoot(ElementName = "LM")]
public class LM
{
    [XmlElement(ElementName = "LM01")]
    public string LM01 { get; set; }
    [XmlElement(ElementName = "LM02")]
    public string LM02 { get; set; }
}

[XmlRoot(ElementName = "Type")]
public class Type
{
    [XmlElement(ElementName = "Loop")]
    public List<Loop> Loop { get; set; }
}

这是DeSerialize XML to Object 的逻辑

XmlSerializer serializer = new XmlSerializer(typeof(Type));
string xml = File.ReadAllText("XMLFile3.xml");
using (TextReader reader = new StringReader(xml))
{
    var results = (Type)serializer.Deserialize(reader);
    foreach (var item in results.Loop)
    {
        Console.WriteLine($"{item.LoopId} {item.Name}");
        if (item.PER != null)
        {
            Console.WriteLine($"PER:{Regex.Replace(item.PER.PER01, @"\s+", "")}-{Regex.Replace(item.PER.PER02, @"\s+", "")}-{Regex.Replace(item.PER.PER03, @"\s+", "")}-{Regex.Replace(item.PER.PER04, @"\s+", "")}");
        }                      

        if (item.NM1 != null)
        {
            Console.WriteLine($"NM1:{Regex.Replace(item.NM1.NM101, @"\s+", "")}-{Regex.Replace(item.NM1.NM102, @"\s+", "")}");
        }

        if (item.LM != null)
        {
            Console.WriteLine($"LM:{Regex.Replace(item.LM.LM01, @"\s+", "")}-{Regex.Replace(item.LM.LM02, @"\s+", "")}");
        }
    }
}

输出

2100A MEMBER NAME
PER:IP--HP-6235834409
2100C MEMBER MAILING ADDRESS
NM1:31-1
2100G RESPONSIBLE PERSON
PER:IP--HP-6235834409
LM:RP-