如何将XML文件转换为C#对象？

huangapple
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英文:

How to convert xml file to C# object?

问题

我正在尝试将这个 XML 文件转换为 C# 对象，但是不起作用。我无法获取 Group 标签之间的文本。类的结构允许我获取 Group 标签内部的子元素，但是当 Group 标签内只有一个文本时，我无法选择它。

以下是 XML 结构和我的类：

XML 文件：

  1. <?xml version="1.0" encoding="utf-8"?>
  2. <Main>
  3.   <Group>
  4. 	  <NV>BasicPars</NV>
  5. 	  <Property>
  6. 		  <NV>Name</NV>
  7. 		  <Value>wzorzec</Value>
  8. 	  </Property>
  9. 	  <Property>
  10. 		  <NV>UseFixedSizes</NV>
  11. 		  <Value>False</Value>
  12. 	  </Property>
  13.   </Group>
  14.   <Group>EventPars</Group>
  15.   <Group>Registers</Group>
  16. </Main>

我的类：

  1. [XmlRoot(ElementName="Property")]
  2. public class Property { 
  4. 	[XmlElement(ElementName="NV")]
  5. 	public string NV { get; set; } 
  7. 	[XmlElement(ElementName="Value")]
  8. 	public string Value { get; set; } 
  9. }
  11. [XmlRoot(ElementName="Group")]
  12. public class Group { 
  14. 	[XmlElement(ElementName="NV")]
  15. 	public string NV { get; set; } 
  17. 	[XmlElement(ElementName="Property")]
  18. 	public List<Property> Property { get; set; } 
  19. }
  21. [XmlRoot(ElementName="Main")]
  22. public class Main { 
  24. 	[XmlElement(ElementName="Group")]
  25. 	public List<Group> Group { get; set; } 
  26. }
英文:

I'm trying to convert this xml file to C# object but it doesn't work. I can't to get text between Group tags. The classes structure let me to get childs inside Group tag, but when Group tag has just one text inside, I can't select it.

I give the XML structure and my classes.

XML file:

  1. &lt;?xml version=&quot;1.0&quot; encoding=&quot;utf-8&quot;?&gt;
  2. &lt;Main&gt;
  3.   &lt;Group&gt;
  4. 	  &lt;NV&gt;BasicPars&lt;/NV&gt;
  5. 	  &lt;Property&gt;
  6. 		  &lt;NV&gt;Name&lt;/NV&gt;
  7. 		  &lt;Value&gt;wzorzec&lt;/Value&gt;
  8. 	  &lt;/Property&gt;
  9. 	  &lt;Property&gt;
  10. 		  &lt;NV&gt;UseFixedSizes&lt;/NV&gt;
  11. 		  &lt;Value&gt;False&lt;/Value&gt;
  12. 	  &lt;/Property&gt;
  13.   &lt;/Group&gt;
  14.   &lt;Group&gt;EventPars&lt;/Group&gt;
  15.   &lt;Group&gt;Registers&lt;/Group&gt;
  16. &lt;/Main&gt;

My classes:

  1. [XmlRoot(ElementName=&quot;Property&quot;)]
  2. public class Property { 
  4. 	[XmlElement(ElementName=&quot;NV&quot;)] 
  5. 	public string NV { get; set; } 
  7. 	[XmlElement(ElementName=&quot;Value&quot;)] 
  8. 	public string Value { get; set; } 
  9. }
  11. [XmlRoot(ElementName=&quot;Group&quot;)]
  12. public class Group { 
  14. 	[XmlElement(ElementName=&quot;NV&quot;)] 
  15. 	public string NV { get; set; } 
  17. 	[XmlElement(ElementName=&quot;Property&quot;)] 
  18. 	public List&lt;Property&gt; Property { get; set; } 
  19. }
  21. [XmlRoot(ElementName=&quot;Main&quot;)]
  22. public class Main { 
  24. 	[XmlElement(ElementName=&quot;Group&quot;)] 
  25. 	public List&lt;Group&gt; Group { get; set; } 
  26. }

答案1

得分: 3

首先，只需要在根对象（即Main）上使用XmlRoot。从其他对象中删除它。

为了捕获元素内部的文本，可以使用[XmlText]（参考链接：https://learn.microsoft.com/en-us/dotnet/standard/serialization/attributes-that-control-xml-serialization）。

修正后的类如下：

  1. public class Property { 
  3.     [XmlElement(ElementName="NV")] 
  4.     public string NV { get; set; } 
  6.     [XmlElement(ElementName="Value")] 
  7.     public string Value { get; set; } 
  8. }
  10. public class Group { 
  12.     [XmlElement(ElementName="NV")] 
  13.     public string NV { get; set; } 
  15.     [XmlElement(ElementName="Property")] 
  16.     public List<Property> Property { get; set; }
  17. 	
  18.     [XmlText]
  19.     public string InnerText { get; set; }
  20. }
  22. [XmlRoot(ElementName="Main")]
  23. public class Main { 
  25.     [XmlElement(ElementName="Group")] 
  26.     public List<Group> Group { get; set; } 
  27. }

这样可以正确进行反序列化（参考链接：https://dotnetfiddle.net/RrBTfC）。

英文:

First, you only need XmlRoot on the root object (that is, Main). Remove it from the others.

In order to capture the text inside an element, use [XmlText].

The fixed classes are:

  1. public class Property { 
  3.     [XmlElement(ElementName=&quot;NV&quot;)] 
  4.     public string NV { get; set; } 
  6.     [XmlElement(ElementName=&quot;Value&quot;)] 
  7.     public string Value { get; set; } 
  8. }
  10. public class Group { 
  12.     [XmlElement(ElementName=&quot;NV&quot;)] 
  13.     public string NV { get; set; } 
  15.     [XmlElement(ElementName=&quot;Property&quot;)] 
  16.     public List&lt;Property&gt; Property { get; set; }
  17. 	
  18. 	[XmlText]
  19.     public string InnerText { get; set; }
  20. }
  22. [XmlRoot(ElementName=&quot;Main&quot;)]
  23. public class Main { 
  25.     [XmlElement(ElementName=&quot;Group&quot;)] 
  26.     public List&lt;Group&gt; Group { get; set; } 
  27. }

Which deserializes correctly.

答案2

得分: 0

public class Serializer
{
public T Deserialize(string input) where T : class
{
System.Xml.Serialization.XmlSerializer ser = new System.Xml.Serialization.XmlSerializer(typeof(T));

  1.     using (StringReader sr = new StringReader(input))
  2.     {
  3.         return (T)ser.Deserialize(sr);
  4.     }
  5. }
  7. public string Serialize<T>(T ObjectToSerialize)
  8. {
  9.     XmlSerializer xmlSerializer = new XmlSerializer(ObjectToSerialize.GetType());
  11.     using (StringWriter textWriter = new StringWriter())
  12.     {
  13.         xmlSerializer.Serialize(textWriter, ObjectToSerialize);
  14.         return textWriter.ToString();
  15.     }
  16. }

}

英文: 
  1. public class Serializer
  2.     {       
  3.         public T Deserialize&lt;T&gt;(string input) where T : class
  4.         {
  5.             System.Xml.Serialization.XmlSerializer ser = new System.Xml.Serialization.XmlSerializer(typeof(T));
  7.             using (StringReader sr = new StringReader(input))
  8.             {
  9.                 return (T)ser.Deserialize(sr);
  10.             }
  11.         }
  13.         public string Serialize&lt;T&gt;(T ObjectToSerialize)
  14.         {
  15.             XmlSerializer xmlSerializer = new XmlSerializer(ObjectToSerialize.GetType());
  17.             using (StringWriter textWriter = new StringWriter())
  18.             {
  19.                 xmlSerializer.Serialize(textWriter, ObjectToSerialize);
  20.                 return textWriter.ToString();
  21.             }
  22.         }
  23.     }

huangapple
  • 本文由 发表于 2023年7月27日 15:10:45
  • 转载请务必保留本文链接：https://go.coder-hub.com/76777271.html
  • c#
  • xml
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