嵌套的XML序列化C＃

Question

我需要使用序列化来创建嵌套的Xml。我有两个列表中的数据必须生成由Number过滤的嵌套xml。

名单：

List<Person> personList = new List<Person>(); 
personList.Add(new Person{ 
Number = 1, 
Name = "Dean" 
}); 

personList.Add(new Person{ 
    Number = 2, 
Name = "Mike" 
}); 

List<Home> homeList= new List<Home>(); 
    homeList.Add(new Home{ 
Number = 2, 
City= "Paris", 
State = "France" 
}); 

homeList.Add(new Home{ 
Number = 1, 
City= "London", 
State = "England" 
}); 

所以，下次我有我使用的序列化类：

public class CreateXML 
{ 
[XElement(ElementName = "Home")] 
List<Home> homeList= new List<Home>(); 
[XElement(ElementName = "Person")] 
List<Person> personList = new List<Person>(); 
} 

方法创建XML：

public void Serialize(CreateXML list) 
    { 
     XmlSerializer serializer = new XmlSerializer(typeof(CreateXML)); 
     using (TextWriter writer = new StreamWriter(@"D:\XmlTEST.txt")) 
     { 
      serializer.Serialize(writer, list); 
     } 
    } 

现在首先生成一个XML文件与所有个人数据，然后与家庭数据。

实际输出：

<Person> 
    <number>1<number> 
    <name>Dean</name> 
    </Person> 
    <Person> 
    <number>2<number> 
    <name>Mike</name> 
    </Person> 
    <Home> 
    <number>1</number> 
    <city>London</city> 
    <state>England</state> 
    </Home> 
    <number>2</number> 
    <city>Paris</city> 
    <state>London</state> 
</Home> 

通缉输出继电器：

<Person> 
    <number>1<number> 
    <name>Dean</name> 
    <Home> 
    <number>1</number> 
    <city>London</city> 
    <state>England</state> 
    </Home> 
</Person> 
<Person> 
    <number>2<number> 
    <name>Mike</name> 
    <Home> 
    <number>2</number> 
    <city>Paris</city> 
    <state>France</state> 
    </Home> 
</Person> 

有什么建议？

Answer 1

你想一定Home与Person院长相关联：

personList.Add(new Person { 
    Number = 1, 
    Name = "Dean" 
}); 

homeList.Add(new Home{ 
    Number = 1, 
    City= "London", 
    State = "England" 
}); 

在上面的示例实体是： 

<Person> 
    <number>1</number> 
    <name>Dean</name> 
    <Home> 
    <number>1</number> 
    <city>London</city> 
    <state>England</state> 
    </Home> 
</Person> 

然而，在你的代码中没有这两个实体联系起来完全不相关 - 你需要以某种方式将它们联系起来。你可以涉及这些entites的

一种方式是通过在Person类型定义Home属性：

public class Person 
{ 
    public string Name { get; set; } 
    public int Number { get; set; } 
    public Home Home { get; set; } 
} 

然后沟homeList和分配Home实例Person实例：

List<Person> personList = new List<Person>(); 
personList.Add(new Person 
{ 
    Number = 1, 
    Name = "Dean", 
    Home = new Home 
    { 
     Number = 1, 
     City = "London", 
     State = "England" 
    } 
}); 

输出：

<personList> 
    <Person> 
    <Name>Dean</Name> 
    <Number>1</Number> 
    <Home> 
     <Number>2</Number> 
     <City>Paris</City> 
     <State>France</State> 
    </Home> 
    </Person> 
    <Person> 
    <Name>Mike</Name> 
    <Number>2</Number> 
    <Home> 
     <Number>1</Number> 
     <City>London</City> 
     <State>England</State> 
    </Home> 
    </Person> 
</personList> 

更新

基于您的评论，你可以动态地分配给Home属性是这样的：

foreach (var home in homeList) 
{ 
    foreach (var person in personList) 
    { 
     if (home.Number == person.Number) 
     { 
      person.Home = home; 
     } 
    } 
} 

裸记住，如果没有相应的门牌号码，该House属性将保持null。