Returning List of Custom Types

Question

I am working on xml files using serializations and deserializations.

I need a method for deserializations.

It will take two parameters(ReturnType Type ,string  XmlSourcePath) and return the list of Type.

Is it possible to do with a single method?

 

Thnx,

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please, mark this as answer if it is THE answer

Answer 1

http://msdn.microsoft.com/en-us/library/system.xml.serialization.xmlserializer.deserialize(v=vs.71).aspx

http://msdn.microsoft.com/en-us/library/ms731073.aspx

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One word frees us of all the weight and pain of life: that word is love.

Answer 2

use following method and cast your object instance!

void main()
{
    object obj = methodName("XmlSourcePath");
    
    //in example
    YourType objInstance = (YourType)obj;

}

public object methodName(string xmlSourcePath)
{
    //deserialize your object!
    return yourDeserializedObjectInstance;
}

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Any fool can know. The point is to understand.(Albert Einstein)

Answer 3

Hi,

I had a similar scenario:

My XML File is in a format:

<DocumentList>
  <Document>
    <DocumentTypeCode>UAP11</DocumentTypeCode>
    <DocumentDescription>Application Form</DocumentDescription>
  </Document>
  <Document>
    <DocumentTypeCode>UAP14</DocumentTypeCode>
    <DocumentDescription>Buy to Let App Form</DocumentDescription>
   </Document>
</DocumentList>

And the following Classes:

[Serializable]
    [XmlRoot("DocumentList")]
    public class DocumentList
    {
        public DocumentList()
        {
         ListElement=new List<Document>();   
        }
        [XmlElement("Document")]
        public List<Document> ListElement { get; set; }
    }

and

public partial class Document
    {
        public Document()
        {
        }

   [XMLElement]
public string DocumentTypeCode{get;set;}
   [XMLElement]
public string DocumentTypeDescription{get;set;}
}

Make sure your XML files and Classes are in synch.

Now you can use the following code to deserialize the xml's to Generic List:

public static T DesirializeMyXML<T>(string fileName)
        {
            XmlSerializer serializer = new XmlSerializer(typeof(T));
            FileStream fs = new FileStream(fileName, FileMode.Open);
            XmlReader reader = new XmlTextReader(fs);
            var returnList = (T)serializer.Deserialize(reader);
            return returnList;
        }

If this colves your problem, mark this as answer, or helpfull if this helped.

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Warm Regards, Deepak Pandit