CODEWITHSUNDEEP
c#xmlxml-serialization
nologo
nologo

Reputation: 6278

xml duplicate node help required

i need to replicate the following in XML, but unsure how to do this:

<FAMILY>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
</FAMILY>

i'm using a very basic example to explain what i need assistance with. the XML is generated from parsing a serialisable class:

[Serializable]
[XmlRoot("FAMILY")]
public class FamilyBlock
{
public string NAME { get; set; }
public int AGE { get; set; }
public DateTime? DOB { get; set; }

 public FamilyBlock(string name, int age, DateTime? dob)
{
 NAME = name;
 AGE = age;
 DOB = dob;
}
}

I attemped to resolve the problem with a list an object but i get the following (the addition of the object name - this i don't need).

<FAMILY>
<MEMBER>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
</MEMBER>
<MEMBER>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
</MEMBER>
<MEMBER>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
</MEMBER>
<MEMBER>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
</MEMBER>
</FAMILY>

i'm sure this is a simple problem but i really dont have much knowledge of xml

Upvotes: 1

Views: 389

Answers (3)

nologo
nologo

Reputation: 6278

Additionally, i refactored it slightly. I created class named FamilyMember:

public class FamilyMember
{
    public FamilyMember()
    {

    }
    public FamilyMember(string name, string age, string dob)
    {
        NAME =name;
        // etc etc
    }

    [XmlElement("NAME")]
    public string NAME { get; set; }

    [XmlElement("AGE")]
    public string AGE { get; set; }

    [XmlElement("DOB")]
    public string DOB { get; set; }
    }
}

so my XML class which i want to serialise now looks like:

[Serializable]
[XmlRoot("RESY")]
public class FamilyBlock: IXmlSerializable
{
    public List<FamilyMember> FAMILYMEMBERS{ get; set; }

    public FamilyBlock()
    {

    }

    public FamilyBlock(string name, int age, DateTime? dob)
    {
        var familyMembers = new List<FamilyMember>  // etc etc
     ....
    }

    public void WriteXml(XmlWriter writer)
    {

        foreach (var item in FAMILYMEMBERS)
        {
            writer.WriteElementString("NAME", item.NAME);
            writer.WriteElementString("AGE", item.AGE);
            writer.WriteElementString("DOB", item.DOB);
        }
    }
}

Upvotes: 0

nologo
nologo

Reputation: 6278

ok so what i did with the help from Nagg was to implement the IXmlSerializable interface.

[Serializable]
[XmlRoot("FAMILY")]
public class FamilyBlock : IXmlSerializable
{
   [XmlElement("NAME")]
   public List<string> NAME { get; set; }
   [XmlElement("AGE")]
   public List<int> AGE { get; set; }
   [XmlElement("DOB")]
   public List<DateTime?> DOB { get; set; }

   public FamilyBlock(string name, int age, DateTime? dob)
   {
      NAME = name;
      AGE = age;
      DOB = dob;
   }

   public void WriteXml(XmlWriter writer)
   {
      for (int i = 0; i < this.NAME.Count; i++)
      {
          writer.WriteElementString("NAME ", this.NAME[i]);
          writer.WriteElementString("AGE", this.AGE[i]);
          writer.WriteElementString("DOB", this.DOB[i]);
      }

   }
}

if its not the correct approach, please feel free to update but this works for me. Please note, the following to methods are also inherited:

public XmlSchema GetSchema()
{
    throw new NotImplementedException();
}
public void ReadXml(XmlReader reader)
{
    throw new NotImplementedException();
}

RESULT:

<FAMILY>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
 <NAME>
 <AGE>
 <DATEOFBIRTH>
</FAMILY>

Upvotes: 0

EgorBo
EgorBo

Reputation: 6142

[XmlRoot("Family")]
public class FamilyBlock
{
    [XmlElement("NAME")]
    public string[] NAME { get; set; }
    [XmlElement("AGE")]
    public int[] AGE { get; set; }
    [XmlElement("DOB")]
    public DateTime?[] DOB { get; set; }
}

After xml serialization looks like:

<?xml version="1.0"?>
<Family xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <NAME>a</NAME>
  <NAME>s</NAME>
  <AGE>1</AGE>
  <AGE>3</AGE>
  <DOB>2011-07-04T13:51:20.6757286+03:00</DOB>
  <DOB xsi:nil="true" />
</Family>

Upvotes: 1

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