在 C# 中反序列化 XML [重复]

Posted 2023-02-22

【中文标题】在 C# 中反序列化 XML [重复]

【英文标题】：Deserialise XML in C# [duplicate]

【发布时间】：2013-04-28 03:44:35

【问题描述】：

我有以下 XML

<getAvailability>
  <result>
    <arrival_date>2013-05-05</arrival_date>
    <block>
    <block_id>80884788</block_id>
    </block>
    <id>230802</id>
   </result>
   <result>
    <arrival_date>2013-05-05</arrival_date>
    <block>
    <block_id>419097</block_id>
    </block>
    <id>98121</id>
  </result>
</getAvailability>

如何反序列化它并将其放入 C# 中的类中？非常感谢。另外这里是我的代码；

XmlSerializer serializer = new XmlSerializer(typeof(getAvailability));
getAvailability output;

using (StringReader reader = new StringReader(xmlSource))

    output = (getAvailability)serializer.Deserialize(reader);


public class Result

    [XmlElement("block_id")]
    public string sBlockID  get; set; 

    [XmlElement("arrival_date")]
    public DateTime dArrivalDate  get; set; 

    [XmlElement("id")]
    public int iID get; set; 


[XmlRoot("getAvailability")]
public class getAvailability

     [XmlArray("result")]
     [XmlArrayItem("block", typeof(Result))]
     public Result[] Result  get; set; 

如果您需要任何进一步的信息，请告诉我。

【问题讨论】：

这里似乎有人问了一个很好的答案***.com/questions/364253/…

我按照***.com/questions/364253/… 的示例进行操作，但由于某种原因，我的程序只读取了 XML 文件的最后一个元素

你能发布一个行为不端的小示例程序吗？

刚刚用代码更新了我原来的帖子

【参考方案1】：

最简单的方法是按照下面的代码示例构建结果类，以便轻松地将其映射到您提供的 XML 示例。我添加了一些提供有关某些方法的信息的 cmets。

using System;
using System.IO;
using System.Xml.Serialization;

namespace Test

    public class Program
    
        public static void Main(string[] args)
        

            Result result1 = new Result
                                 
                                     arrival_date = new DateTime(2013, 05, 05),
                                     block = new Result.Block  block_id = 80884788 ,
                                     id = 230802
                                 ;
            Result result2 = new Result
                                 
                                     arrival_date = new DateTime(2013, 05, 05),
                                     block = new Result.Block  block_id = 419097 ,
                                     id = 98121
                                 ;
            Results results = new Results  result = new Result[2] ;
            results.result[0] = result1;
            results.result[1] = result2;

            WriteSettingsAsXml("D:\\test.xml", typeof(Results), results, true);

            Results gA = (Results)ReadSettingsFromXml("D:\\test.xml", typeof(Results));
        

        // This `Result` class below maps to a single result in the XML you provided. This class is used in the `Results` class to obtain the needed XML structure.   

        public class Result
        
            public class Block
            
                public Int32 block_id  get; set; 
            

            public DateTime arrival_date  get; set; 
            public Block block  get; set; 
            public Int32 id  get; set; 
        

        [Serializable()]
        [XmlRootAttribute("getAvailability", Namespace = "", DataType = "", IsNullable = false)]
        public class Results
        
            [XmlElement("result")]
            public Result[] result  get; set; 
        

        /// Library methods that saves/reads any passed/retrieved object into/from a xml file at specified location

        public static void WriteSettingsAsXml(string destinationPath, Type objectType, object objectValue, bool hideNamespaces)
        
            XmlSerializer serializer = new XmlSerializer(objectType);
            using (TextWriter writer = new StreamWriter(destinationPath))
            
                if (hideNamespaces)
                
                    XmlSerializerNamespaces hiddenNamespaces = new XmlSerializerNamespaces();
                    hiddenNamespaces.Add("", "");
                    serializer.Serialize(writer, objectValue, hiddenNamespaces);
                
                else
                    serializer.Serialize(writer, objectValue);
                writer.Close();
            
        
        public static object ReadSettingsFromXml(string xmlFilePath, Type objectType)
        
            XmlSerializer serializer = new XmlSerializer(objectType);
            using (FileStream fileStream = new FileStream(xmlFilePath, FileMode.Open))
            
                return serializer.Deserialize(fileStream);
            
        
    

【讨论】：

感谢它完美运行。

我用下面的例子给出了两个元素 '2013-05-05808847882308022013-05-0541909798121 ' 但是由于某种原因，当我从实时源尝试更大的数据集时，它只返回第一个节点？任何可能导致此问题的想法