C# Object-Serialization and -Deserialization in XML

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Hey,

I have the problem, that I have to serialize multiple objects of the same type to a xml-file with the XmlSerializer-Class. But I'm only able to create xml-files like this one here:

XML

<Projects>
  <Name>A</Name>
  <TimeTotal>0</TimeTotal>
  <TimeToday>0</TimeToday>
  <Comment>a</Comment>
</Projects>
<Projects>
  <Name>B</Name>
  <TimeTotal>0</TimeTotal>
  <TimeToday>0</TimeToday>
  <Comment>b</Comment>
</Projects>

But thats invalid xml. So i want that it looks like this:

XML

<TimeProjects>
  <Projects>
    <Name>A</Name>
    <TimeTotal>0</TimeTotal>
    <TimeToday>0</TimeToday>
    <Comment>a</Comment>
  </Projects>
  <Projects>
    <Name>B</Name>
    <TimeTotal>0</TimeTotal>
    <TimeToday>0</TimeToday>
    <Comment>b</Comment>
  </Projects>
</TimeProjects>

So the problem is that I don't know how to add ONLY ONE TIME a root element and then multiple sub-elements.

Thats my code how I try to add objects to a xml-file:

C#

XmlSerializer xmlSerializer = new XmlSerializer(typeof(TimeProjects));
ProjectWriter = new StreamWriter(filePath, true);
        
TimeProjects project = new TimeProjects { Name = projectName, Comment = comments, TimeToday = timeToday, TimeTotal = timeTotal };

xmlSerializer.Serialize(ProjectWriter, project);

ProjectWriter.Close();

I hope you can help me. If you need more informations with that, just ask me please!!

Posted 26-Nov-14 21:23pm

ToShX

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3 solutions

Solution 1

How about not dealing with XML manually at all? The best XML serialization is implemented using the Data Contract:
http://msdn.microsoft.com/en-us/library/ms733127%28v=vs.110%29.aspx[^].

Please see my past answers where I advocate this approach:
How can I utilize XML File streamwriter and reader in my form application?[^],
Creating property files...[^].

—SA

Posted 26-Nov-14 21:48pm

Sergey Alexandrovich Kryukov

Comments

ToShX 27-Nov-14 4:06am

Thanks for your answer. had a lot to read :P
But for my situation _Canny_Brisk_'s solution works fine for me. But I guess in time I have to go back to your solution.

Sergey Alexandrovich Kryukov 27-Nov-14 4:19am

I would highly recommend that...
—SA

ToShX 27-Nov-14 4:22am

yeah, please know that I just started my education as a software developer. So i have to learn these things step by step. Yet I don't know whats going on with all these metadatas everyone suggests, so I have to take the easiest answer I can get, if they're working for me :)

Solution 3

Hi Please try this Code

C#

public class MakeSer
    {

        public MakeSer()
        {
            TimeProject tm = new TimeProject();
            Projects[] proj = new Projects[2];
            proj[0] = new Projects { Comment = "X", name = "Y", TimeToday = "40", TimeTotal = "10" };
            proj[1] = new Projects { Comment = "XX", name = "CY", TimeToday = "50", TimeTotal = "30" };


            tm.TimeProjects = proj;
            System.Xml.Serialization.XmlSerializer xml = new System.Xml.Serialization.XmlSerializer(tm.GetType());
            StringBuilder builder = new StringBuilder();
            using (StringWriter writer = new StringWriter(builder))
            {
                xml.Serialize(writer, tm);
            }
            string value_ = builder.ToString();

        }
    }

    [Serializable]
    public class Projects
    {
        public string name { get; set; }
        public string TimeTotal { get; set; }
        public string TimeToday { get; set; }
        public string Comment { get; set; }
    }
    [Serializable]
    public class TimeProject
    {
      
        public Projects[] TimeProjects { get; set; }
    }

Posted 26-Nov-14 22:17pm

Er. Dinesh Sharma

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**Joezer BH** · Accepted Answer · 2014-11-26T21:52:00

Should be fairly simple.
Try creating the XmlSerializer with the constructor that also takes XmlRootAttribute

C#

// Fix this line of code, use the costructor that follows : XmlSerializer(Type, XmlRootAttribute)
XmlSerializer xmlSerializer = new XmlSerializer(typeof(TimeProjects));

// try using: XmlSerializer xmlSerializer = new XmlSerializer(typeof(TimeProjects), new XmlRootAttribute("TimeProjects"));

Cheers,
C