Save List to XML file

Question

I want to save records fetched from database in an XML file,
take x number of records from XML file into a custom collection List<T>
process them and save updated items back into XML file.

'T' is a simple object with value type properties, something like -

public class T
{
   public int Id {get; set;}
   public string property1 {get; set;}
   public string property2 {get; set;}
}

Please guide me how can I save custom collection List<T> to XML file and vice-versa?

Also, because I am not sending out this XML file, will it make sense to go for XmlSerializer as suggested in some of the replies?

Answer 1

Here are two methods that we use to accomplish this using the XMLSerializer:

 public static T FromXML<T>(string xml)
 {
     using (StringReader stringReader = new StringReader(xml))
     {
        XmlSerializer serializer = new XmlSerializer(typeof(T));
        return (T)serializer.Deserialize(stringReader);
     }
 }

 public string ToXML<T>(T obj)
 {
    using (StringWriter stringWriter = new StringWriter(new StringBuilder()))
    {
        XmlSerializer xmlSerializer = new XmlSerializer(typeof(T));
        xmlSerializer.Serialize(stringWriter, obj);
        return stringWriter.ToString();
    }
 }

Answer 2

While you could use a serializer - and many times this is the right answer - I personally would use Linq to XML which would allow you to be more flexible on how your XML should look like, i.e. to create the following XML from a collection foos based on your class:

<Foos>
  <foo Id="1" property1="someprop1" property2="someprop2" />
  <foo Id="1" property1="another" property2="third" />
</Foos>

You could use:

var xml = new XElement("Foos", foos.Select( x=> new XElement("foo", 
                                                new XAttribute("Id", x.Id), 
                                                new XAttribute("property1", x.property1), 
                                                new XAttribute("property2", x.property2))));

Answer 3

Using the code below (Class T Taken from your code snippet) you will be able to serialize into an XML file with ease, and without the hassle of implementing ISerializable

[Serializable()]
public class T
{
    public int Id {get; set;}
    public string property1 {get; set;}
    public string property2 {get; set;}
}

...

List<T> data = new List<T>()

... // populate the list

//create the serialiser to create the xml
XmlSerializer serialiser = new XmlSerializer(typeof(List<T>));

// Create the TextWriter for the serialiser to use
TextWriter filestream = new StreamWriter(@"C:\output.xml");

//write to the file
serialiser.Serialize(filestream , data);

// Close the file
filestream.Close();

Answer 4

Use the XmlSerializer class. Scroll down about 1/3 of the way for the examples.

Answer 5

List<BI_QA_ChoiceEntity> choiceSet = new List<BI_QA_ChoiceEntity>();
        choiceSet = biEntityObj.ChoiceSet;

        XmlDocument ChoiceXML = new XmlDocument();
        ChoiceXML.AppendChild(ChoiceXML.CreateElement("CHOICESET"));
        foreach (var item in choiceSet)
        {
            XmlElement element = ChoiceXML.CreateElement("CHOICE");
           // element.AppendChild(ChoiceXML.CreateElement("CHOICE_ID")).InnerText = Convert.ToString(item.ChoiceID);
            element.AppendChild(ChoiceXML.CreateElement("CHOICE_TEXT")).InnerText = Convert.ToString(item.ChoiceText);
            element.AppendChild(ChoiceXML.CreateElement("SEQUENCE")).InnerText = Convert.ToString(item.Sequence);
            element.AppendChild(ChoiceXML.CreateElement("ISCORRECT")).InnerText = Convert.ToString(item.IsCorrect);
            ChoiceXML.DocumentElement.AppendChild(element);
        }

Pass ChoiceXML to Stored procedure then SQL Server Do like as below

@Choice_XML VARCHAR(MAX)=NULL

 IF(@Choice_XML<>'')  
                BEGIN  
                    SET @intDocHandle =0
                    --Create an internal representation of the XML document.  
                    EXEC sp_xml_preparedocument @intDocHandle OUTPUT, 

@Choice_XML  

                --SET @ChoiceID = (SELECT  max([choice_id])+1 AS 'ChoiceID'  from BI_QUESTION_CHOICE)

                --Insert 
                INSERT BI_QUESTION_CHOICE
                (
                    [choice_id],
                    [choice_descr],
                    [sequence],
                    [question_id],
                    [is_correct],
                    [created_by],
                    [created_dt],
                    [modified_by],
                    [modified_dt]
                )
                SELECT (SELECT  max([choice_id])+1 AS 'ChoiceID'  from BI_QUESTION_CHOICE),  
                    CASE WHEN CHOICE_TEXT='' THEN NULL ELSE CHOICE_TEXT END, 
                    CASE WHEN SEQUENCE='' THEN NULL ELSE SEQUENCE END,
                    QuestionID, 
                    CASE WHEN ISCORRECT='' THEN NULL ELSE ISCORRECT END,
                    'mbathini',  
                     GETDATE(),  
                    'mbathini', 
                    GETDATE()  
                FROM OPENXML(@intDocHandle,'/CHOICESET/CHOICE', 3)  
                WITH  
                (CHOICE_TEXT VARCHAR(500), 
                 SEQUENCE VARCHAR(50),
                 QuestionID INT,
                 ISCORRECT bit)

END

Answer 6

You Can save Your List<T> into a DataTable then WriteXml

    T t0 = new T();
       t0.id=1;
       t0.property1="John";
       t0.property2="Doe";
    List<T> Tlist = new List<T>();
       Tlist.Add(t0);

    DataTable dt = new DataTable();
       dt.TableName = "People";
       dt.Columns.Add("ID");
       dt.Columns.Add("Name");
       dt.Columns.Add("Lastname");


        foreach(var item in tlist)
        {
            dt.Rows.Add();
            dt.Rows[dt.Rows.Count-1]["ID"] = item.name;
            dt.Rows[dt.Rows.Count - 1]["Name"] = item.id_cod;
            dt.Rows[dt.Rows.Count - 1]["Lastname"] = item.id_cod;


        }
        dt.WriteXml("test.Xml");