如何反序列化這個Xml文件？

Question

我有這個Xml文件，我需要反序列化它回到一個類。問題是：考慮到Xml元素（RowInfo）的計數不是恆定的，這個類的正確結構/設計是什麼？ The Xml文件：

<?xml version="1.0"?> 
<SomeObject xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" > 
    <Layers> 
<Layer Id="0"> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1</RowInfo> 
     <RowInfo>1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1</RowInfo> 
     <RowInfo>1,1,1,0,0,1,1,1,1,1,1,1,1,1,0,0,1,0,0,1</RowInfo> 
     <RowInfo>1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1</RowInfo> 
    </Layer> 
    </Layers> 
</SomeObject> 

感謝您的幫助。 謝謝。

編輯1：也考慮到（層）可能包含多個圖層。

Answer 1

這應該工作，只要你想：

public class SomeObject 
{ 
    public List<Layer> Layers { get; set; } 
} 

public class Layer 
{ 
    [XmlAttribute] 
    public int Id { get; set; } 

    [XmlElement("RowInfo")] 
    public List<RowInfo> RowInfos { get; set; } 
} 

public class RowInfo 
{ 
    [XmlText] 
    public string Info { get; set; } // you'll need to parse the list of ints manually 
} 

唯一不同的是編碼，但你應該能夠解決它。

Answer 2

如果RowInfo的計數不固定，請在班級中使用List。

Answer 3

我會說使用LINQ-to-Xml。讓你的對象的構造函數，可以採取xlement然後做沿着

public class YourObject() 
{ 
public IEnumerable<Layer> Layers { get; set; } 
public int Id { get; set; } 

    public YourObj(XElement x) 
    { 
     this.Id = int.Parse(x.Attribute("Id").ToString()); 
     this.Layers = from layer in x.Elements("Layer") 
        select new Layer(layer); 
    } 
} 



var objs = (from c in XElement.Load("your.xml").Elements("layer") 
      select new YourObject(c)).ToList() ; 

Answer 4

類似下面的行 東西 - 名稱將需要改變，以保護無辜：

類結構之後評論

public class SomeObject 
{ 
    public List<Layer> Layers {get;set;} 
} 

public class Layer 
{ 
    public int Id {get;set;} 
    public List<RowInfo> RowInfos {get;set;} 
} 

public class RowInfo 
{ 
    public List<Row> Rows {get;set;} 
} 

public class Row 
{ 
    public int RowData {get;set;} 
} 

Answer 5

XmlSerializer serializer = new XmlSerializer(typeof(SomeObject)); 

可以使用XmlSerializer與跟隨着G代碼和反序列化調用:)

//------------------------------------------------------------------------------ 
// <auto-generated> 
//  This code was generated by a tool. 
//  Runtime Version:2.0.50727.4952 
// 
//  Changes to this file may cause incorrect behavior and will be lost if 
//  the code is regenerated. 
// </auto-generated> 
//------------------------------------------------------------------------------ 

using System.Xml.Serialization; 

// 
// This source code was auto-generated by xsd, Version=2.0.50727.3038. 
// 


/// <remarks/> 
[System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "2.0.50727.3038")] 
[System.SerializableAttribute()] 
[System.Diagnostics.DebuggerStepThroughAttribute()] 
[System.ComponentModel.DesignerCategoryAttribute("code")] 
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType=true)] 
[System.Xml.Serialization.XmlRootAttribute(Namespace="", IsNullable=false)] 
public partial class SomeObject { 

    private SomeObjectLayers layersField; 

    /// <remarks/> 
    public SomeObjectLayers Layers { 
     get { 
      return this.layersField; 
     } 
     set { 
      this.layersField = value; 
     } 
    } 
} 

/// <remarks/> 
[System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "2.0.50727.3038")] 
[System.SerializableAttribute()] 
[System.Diagnostics.DebuggerStepThroughAttribute()] 
[System.ComponentModel.DesignerCategoryAttribute("code")] 
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType=true)] 
public partial class SomeObjectLayers { 

    private SomeObjectLayersLayer layerField; 

    /// <remarks/> 
    public SomeObjectLayersLayer Layer { 
     get { 
      return this.layerField; 
     } 
     set { 
      this.layerField = value; 
     } 
    } 
} 

/// <remarks/> 
[System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "2.0.50727.3038")] 
[System.SerializableAttribute()] 
[System.Diagnostics.DebuggerStepThroughAttribute()] 
[System.ComponentModel.DesignerCategoryAttribute("code")] 
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType=true)] 
public partial class SomeObjectLayersLayer { 

    private decimal[] rowInfoField; 

    private int idField; 

    private bool idFieldSpecified; 

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

    /// <remarks/> 
    [System.Xml.Serialization.XmlAttributeAttribute()] 
    public int Id { 
     get { 
      return this.idField; 
     } 
     set { 
      this.idField = value; 
     } 
    } 

    /// <remarks/> 
    [System.Xml.Serialization.XmlIgnoreAttribute()] 
    public bool IdSpecified { 
     get { 
      return this.idFieldSpecified; 
     } 
     set { 
      this.idFieldSpecified = value; 
     } 
    } 
} 

Answer 6

簽出一個叫做XSD.exe的工具，它帶有visual studio，可以讓你從xml文件生成代碼。

您應該在visual studio旁邊的程序文件菜單上看到它。