如何使用XMLSerializer生成XML？

Question

我想用XMLSerializer來生成XML。我有一個正在被其他類繼承的抽象Base類。

public abstract class Base 
    { 
     public string Name {get; set;} 
     public int ID {get; set;} 

     public Base(string Name, int ID) 
     { 
     this.Name = Name; 
     this.ID = ID; 
     } 
    } 



public class HR: Base 
    {  
     public HR(string Name, int ID): Base(Name,ID) 
     { 
     } 
    } 

    public class IT : Base 
    { 
     public IT(string Name, int ID): Base(Name,ID) 
     { 
     } 
    } 

我不知道如何生成格式

<Employee> 
<HR> 
    <Name> </Name> 
    <ID> </ID> 
</HR> 
<IT> 
    <Name> </Name> 
    <ID> </ID> 
</IT> 
</Employee> 

我的含糊不清的問題道歉的XML。我從來沒有使用過XMLSerializer，也不知道如何繼續使用它。任何幫助將不勝感激。

感謝

Answer 1

當我讀到你的xml，你似乎想序列化一個Employee列表。 如果您的列表是類的成員（不是直接序列化列表），我有一個解決方案。

public abstract class Employee 
{ 
    public string Name { get; set; } 
    public int ID { get; set; } 
    public Employee(string Name, int ID) 
    { 
     this.Name = Name; 
     this.ID = ID; 
    } 
} 

public class HR : Employee 
{ 
    public HR() : base(null, 0) { } // default constructor is needed for serialization/deserialization 
    public HR(string Name, int ID) : base(Name, ID) { } 
} 
public class IT : Employee 
{ 
    public IT() : base(null, 0) { } 
    public IT(string Name, int ID) : base(Name, ID) { } 
} 

public class Group 
{ 
    [XmlArray("Employee")] 
    [XmlArrayItem("HR",typeof(HR))] 
    [XmlArrayItem("IT",typeof(IT))] 
    public List<Employee> list { get; set; } 

    public Group() 
    { 
     list = new List<Employee>(); 
    } 
} 
class Program 
{ 
    static void Main(string[] args) 
    { 
     Group grp = new Group(); 
     grp.list.Add(new HR("Name HR", 1)); 
     grp.list.Add(new IT("Name IT", 2)); 

     XmlSerializer ser = new XmlSerializer(typeof(Group)); 
     ser.Serialize(Console.Out, grp); 
    } 
} 

，輸出是：

<?xml version="1.0" encoding="ibm850"?> 
<Group xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> 
    <Employee> 
    <HR> 
     <Name>Name HR</Name> 
     <ID>1</ID> 
    </HR> 
    <IT> 
     <Name>Name IT</Name> 
     <ID>2</ID> 
    </IT> 
    </Employee> 
</Group> 

非常相似，所需輸出，除外多了一個元素在根「本集團」。

使用相同的XmlSerializer（typeof（Group））反序列化也應該可以。

Answer 2

添加[Serializable接口]註釋您要序列化類。

 [System.Serializable] 
     public class Base 
     { 
     public string Name { get; set; } 
     public int ID { get; set; } 

     public Base(string Name, int ID) 
      { 
      this.Name = Name; 
      this.ID = ID; 
      } 
     } 

要序列化XML格式，使用下面的代碼：

 System.Xml.Serialization.XmlSerializer Serializer = new System.Xml.Serialization.XmlSerializer(typeof(Base)); 
     Base Foo = new Base(); 
     string xmldata = ""; 

     using (var stringwriter = new System.IO.StringWriter()) 
     { 
      using (System.Xml.XmlWriter xmlwriter = System.Xml.XmlWriter.Create(stringwriter)) 
      { 
       Serializer.Serialize(xmlwriter, Foo); 
       xml = stringwriter.ToString(); // Your XML 
      } 
     } 

從XML序列化回到你的基本對象，使用下面的代碼：

 System.IO.MemoryStream FooStream = new System.IO.MemoryStream(System.Text.Encoding.UTF8.GetBytes(xml)); 
     Base Foo; 
     Foo = (Base)Serializer.Deserialize(FooStream); 

Answer 3

我想你需要使用XmlType屬性來確保您的元素顯示爲<HR>和<IT>而不是<employee xsi:type="HR">。下面的工作演示：

public abstract class Employee 
{ 
    public string Name { get; set; } 
    public string ID { get; set; } 

    public Employee(string Name, string ID) 
    { 
     this.Name = Name; 
     this.ID = ID; 
    } 
} 

public class HR : Employee 
{ 
    public HR(string Name, string ID) : base(Name, ID) 
    { 
    } 


    public HR() : base("No name", "No ID") 
    { 

    } 
} 

public class IT : Employee 
{ 
    public IT(string Name, string ID) : base(Name, ID) 
    { 
    } 

    public IT() : base("No name", "No ID") 
    { 

    } 
} 

我添加了序列化程序的默認（無參數）構造函數。

然後，你必須有某種包裝對象來處理的Employee個清單：

public class Employees 
{ 
    [XmlElement(typeof(IT))] 
    [XmlElement(typeof(HR))] 
    public List<Employee> Employee { get; set; } //It doesn't really matter what this field is named, it takes the class name in the serialization 
} 

接下來，您可以使用串行代碼從我的評論生成XML：

var employees = new Employees 
{ 
    Employee = new List<Employee>() 
    { 
     new IT("Sugan", "88"), 
     new HR("Niels", "41") 
    } 
}; 

var serializer = new XmlSerializer(typeof(Employees)); 
var xml = ""; 

using (var sw = new StringWriter()) 
{ 
    using (XmlWriter writer = XmlWriter.Create(sw)) 
    { 
     serializer.Serialize(writer, employees); 
     xml = sw.ToString(); 
    } 
} 
Console.WriteLine(xml); 

（ommitted爲清楚起見命名空間）

這將返回以下XML：

<?xml version="1.0" encoding="UTF-8"?> 
<Employees xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> 
    <IT> 
     <Name>Sugan</Name> 
     <ID>88</ID> 
    </IT> 
    <HR> 
     <Name>Niels</Name> 
     <ID>41</ID> 
    </HR> 
</Employees>