從多個類生成XML？

我還有一個關於讓我的XML序列化整齊的問題，我似乎無法得到正確的答案。我的配置文件如下：從多個類生成XML？

namespace SMCProcessMonitor 
{ 
[Serializable()] 
[XmlRoot("Email-Settings")] 
    public class Config 
{  
     [XmlElement("Recipient")] 
     public string recipient; 
     [XmlElement("Server-port")] 
     public int serverport; 
     [XmlElement("Username")] 
     public string username; 
     [XmlElement("Password")] 
     public string password; 
     [XmlElement("Program")] 
     public List<Programs> mPrograms = new List<Programs>(); 
     public string serialId; 
    } 

    public class Email 
    { 

      public string Recipient 
      { 
       get 
       { 
        return SMCProcessMonitor.ConfigManager.mConfigurations.recipient; 
       } 
       set 
       { 
        SMCProcessMonitor.ConfigManager.mConfigurations.recipient = value; 
       } 
      } 

      public int ServerPort 
      { 
       get 
       { 
        return SMCProcessMonitor.ConfigManager.mConfigurations.serverport; 
       } 
       set 
       { 
        SMCProcessMonitor.ConfigManager.mConfigurations.serverport = value; 
       } 
      } 
      public string Username 
      { 
       get 
       { 
        return SMCProcessMonitor.ConfigManager.mConfigurations.username; 
       } 
       set 
       { 
        SMCProcessMonitor.ConfigManager.mConfigurations.username = value; 
       } 
      } 
     public string Password { get; set; } 

    } 
     [Serializable()] 
    public class Programs 
{ 
     [XmlElement("Filename")] public string mFileName { get; set; } 
     [XmlElement("Filepath")]public string mFilePath { get; set; } 
} 

     public class Database 
     { 
      public string mSerial { get; set; } 
     } 
     }

理想我想做的事就是有這三個類（電子郵件設置，數據庫和程序）都有自己的標籤，像這樣

<Config xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"    xmlns:xsd="http://www.w3.org/2001/XMLSchema"> 
<email-settings> 
    <Recipient>sadh</Recipient> 
    <Server-port>23</Server-port> 
    <Username>lkms</Username> 
    <Password>kmkdvm</Password> 
</email-settings> 
<Program> 
    <Filename>MerlinAlarm.exe</Filename> 
    <Filepath>D:\Merlin\Initsys\Merlin\Bin\MerlinAlarm.exe</Filepath> 
</Program> 
<database-settings> 
    <serialId>1</serialId> 
</database-settings> 
</Config>

但相反，我得到的東西，類似於此：

<Config xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> 
<Recipient>blah</Recipient> 
<Server-port>1111</Server-port> 
<Username>blah</Username> 
<Password>blah</Password> 
<Program> 
<Filename>chrome.exe</Filename> 
<Filepath> 
C:\Users\Shane\AppData\Local\Google\Chrome\Application\chrome.exe 
</Filepath> 
</Program> 
<serialId>1234</serialId> 
</Config>

對不起，是這樣的麻煩，但現在這是在做我的螺母和IM肯定有我的思念here..can一些基本的邏輯有人給我了一些p-關於如何以上面指定的格式獲取這個XML？謝謝，Shane。

編輯：我的序列化類。

namespace SMCProcessMonitor 
{ 
public class ShanesXMLserializer 
{ 
    private string mFileAndPath; 
    public Config mConfigurations = null; 
    public Config mConfigurationsProgram = null; 



    public ShanesXMLserializer(string inFileAndPath) 
    { 
     mFileAndPath = inFileAndPath; 
     mConfigurations = new Config(); 

    } 

    public bool Write() 
    { 
     try 
     { 
      XmlSerializer x = new XmlSerializer(mConfigurations.GetType()); 
      StreamWriter writer = new StreamWriter(mFileAndPath); 
      x.Serialize(writer, mConfigurations); 
      writer.Close(); 
      return true; 
     } 
     catch (Exception ex) 
     { 
      MessageBox.Show("Exception found while writing: " + ex.Message); 
     }; 

     return false; 
    } 

    public bool Read() 
    { 
     try 
     { 
      XmlSerializer x = new XmlSerializer(typeof(Config)); 
      StreamReader reader = new StreamReader(mFileAndPath); 
      mConfigurations = (Config)x.Deserialize(reader); 
      reader.Close(); 
      return true; 
     } 
     catch (Exception ex) 
     { 
      MessageBox.Show("Exception found while reading: " + ex.Message); 
     }; 

     return false; 
    } 

    public Config GetConfigEmail 
    { 
     get 
     { 
      return mConfigurations; 
     } 
    } 


}

}

編輯2：我的新的配置文件： @Craig - 我使用這個配置文件，它是像你說的，但即時通訊仍然沒有得到想要的XML，後顯示我的配置類。

  using System; 
      using System.Collections.Generic; 
      using System.Linq; 
      using System.Xml.Serialization; 
      using System.Text;

命名空間SMCProcessMonitor { [序列化（）]

 public class Config 
{  

     public string recipient; 
     public int serverport; 
     public string username; 
     public string password; 
     public List<Programs> mPrograms = new List<Programs>(); 
     public string serialId; 
    [XmlElement("email-settings")] 
     public Email Email { get; set; } 
     public Programs Programs { get; set; } 
    [XmlElement("database-settings")] 
     public Database Database { get; set; } 


    } 

    public class Email 
    { 
     [XmlElement("Recipient")] 
     public string Recipient { get; set; } 
      [XmlElement("Server-port")] 
     public int ServerPort { get; set; } 
     [XmlElement("Username")] 
     public string Username { get; set; } 
     [XmlElement("Password")] 
     public string Password { get; set; } 

    } 
     [Serializable()] 
    public class Programs 
    { 
     [XmlElement("Filename")] public string mFileName { get; set; } 
     [XmlElement("Filepath")]public string mFilePath { get; set; } 
    } 

    public class Database 
    { 
     [XmlElement("SerialID")] 
     public string mSerial { get; set; } 
    } 
    }

但我仍然得到：

<Config xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> 
    <recipient>shane</recipient> 
    <serverport>23</serverport> 
    <username>oid</username> 
    <password>jidj</password> 
    <mPrograms/> 
    </Config>

來源

2012-03-23 Shane.C

你能寫出你用於序列化的代碼段嗎？ – daryal 2012-03-23 12:20:57

這是我的序列化代碼。請注意，由於將變量移動到其各自的類，因此在寫入時告訴我「對象引用未設置爲對象的實例」時，現在遇到錯誤。 – 2012-03-23 13:56:45

@Shane'Shamus'Coulter使用你添加到你的問題中的序列化代碼，將使用我提供的'class'修改來輸出你想要的XML。爲了測試它，你可以逐字從我的答案中複製這些類，填充一個'Config'對象並通過你發佈的序列化代碼運行它。你需要改變的唯一事情就是你的財產獲得者和制定者在適用的情況下。 – Craig 2012-03-23 14:14:02

這會給你所需的輸出：

public class Config 
{ 
    [XmlElement("email-settings")] 
    public Email Email { get; set; } 

    public Program Program { get; set; } 

    [XmlElement("database-settings")] 
    public Database Database { get; set; } 
} 

public class Email 
{ 
    public string Recipient { get; set; } 

    [XmlElement("Server-port")] 
    public int ServerPort { get; set; } 

    public string Username { get; set; } 
    public string Password { get; set; } 
} 

public class Program 
{ 
    public string Filename { get; set; } 

    public string Filepath { get; set; } 
} 

public class Database 
{ 
    public string serialId { get; set; } 
}

這是一個控制檯應用程序，它將序列化一個對象到一個文件併產生你正在尋找的確切的XML。只需將其複製並粘貼到控制檯應用程序中，然後從中取出即可。

using System; 
using System.Collections.Generic; 
using System.Linq; 
using System.Text; 
using System.Xml.Serialization; 
using System.IO; 

namespace ConsoleApplication1 
{ 
    class Program 
    { 
     static void Main(string[] args) 
     { 
      var config = new Config 
      { 
       Email = new Email 
       { 
        Recipient = "sadh", 
        ServerPort = 23, 
        Username = "lkms", 
        Password = "kmkdvm" 
       }, 
       Program = new Programs 
       { 
        Filename = "MerlinAlarm.exe", 
        Filepath = @"D:\Merlin\Initsys\Merlin\Bin\MerlinAlarm.exe" 
       }, 
       Database = new Database 
       { 
        serialId = "1" 
       } 
      }; 

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

      var textWriter = new StreamWriter(@"C:\config.xml"); 
      serializer.Serialize(textWriter, config); 
      textWriter.Close(); 

      Console.Read(); 
     } 
    } 

    #region [Classes] 

    public class Config 
    { 
     [XmlElement("email-settings")] 
     public Email Email { get; set; } 

     public Programs Program { get; set; } 

     [XmlElement("database-settings")] 
     public Database Database { get; set; } 
    } 

    public class Email 
    { 
     public string Recipient { get; set; } 

     [XmlElement("Server-port")] 
     public int ServerPort { get; set; } 

     public string Username { get; set; } 
     public string Password { get; set; } 
    } 

    public class Programs 
    { 
     public string Filename { get; set; } 

     public string Filepath { get; set; } 
    } 

    public class Database 
    { 
     public string serialId { get; set; } 
    } 

    #endregion 
}

來源

2012-03-23 12:31:17 Craig

我應該將聲明添加到其各自的類，或將它們留在配置？ – 2012-03-23 13:50:02

@ Shane'Shamus'Coulter把它們留在你的'Config'類中。這將爲您提供您在問題中提到的格式的XML。在你的問題中，你在'Config''類中有'List '，但是這在你提供的所需的XML輸出中沒有描述。 – Craig 2012-03-23 14:03:23

請參閱編輯2. – 2012-03-23 14:12:54

我打算在這裏提出一個完全橫向的方法，只是爲了簡單和維護的緣故。

如果您使用源XML文件並生成了XSD架構，該怎麼辦？

例如：

xsd.exe MyXMLFile1.xml

這將生成一個XML架構文件（MyXMLFile1.xsd）。就拿模式，並生成類（再次使用xsd.exe）：

xsd.exe /c MyXMLFile1.xsd

這將生成一個序列化的保證POCO，您可以使用前進。類名稱和屬性可能與您當前的POCO中的名稱和屬性不匹配，但它會生成預期的XML以及從XML中反序列化。

增加的好處是，未來只需修改源XML文件，然後運行這兩個命令來維護POCO。

希望可以幫助...

來源

2012-03-23 14:26:37 code4life

從多個類生成XML？

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