How do I serialize a list of objects as XML attributes

Question

0.00/5 (No votes)

See more:

Hello there,
I'm currently working on a project of WPF application which allows me to configure a XML file. This application has to load a XML file and permit the user to dynamically change the attributes values.

I try to use the XmlSerializer object, but its native behavior doesn't match what I need to do.

The idea is to obtain such an XML file, with a non known number of attributes (the names and the number of attribute comes from an other XML file) :

XML

<Parameters>
<elementOne attributeOne="ValueOne" attributeTwo="ValueTwo" ...attributeN="ValueN">
</elementOne>
</Parameters>

In order to be able to change the number of attributes, I created such a Class, and the idea is to implement IXmlSerializable, but I don't know how, to be able to create a dynamic list of attributes in my Xml file :

C#

public class element : IXmlSerializable
{
    List<attribute> AttributeList;

    public void WriteXml(XmlWriter writer)
    {
          ???
    }
}

Could you tell me, please :
1)If what I want is possible.
2)If I'm on the good way

Thank you.

Posted 14-Dec-15 4:51am

DimitriStPi

Add a Solution

Comments

Sinisa Hajnal 14-Dec-15 11:01am

Is there any special reason why you have to have attributes instead of nodes?

DimitriStPi 14-Dec-15 11:05am

Hi,
Yes, the reason is that my tool, the configurator, has to be added to a big existing solution all linked to this XML. So the XML I have to be able to configure can't be changed from his original scheme...

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Rajdeep Debnath · Accepted Answer · 2015-12-14T05:11:00

Solution 1

Simply use XMlElement class and xmlElement.setattribute() in loop to take care of this....

Posted 14-Dec-15 5:11am

Rajdeep Debnath

Comments

DimitriStPi 15-Dec-15 3:13am

Hi,
I really wanted to use the XmlSerializer which could have been usefull to make both the serialization/deserialization. BUT after all searches I did, I think having a fixed scheme is a big condition to use XmlSerializer

So I'll take your solution, and to deserialize I think I can use the classes you mentionned. I obtain all the XElements and attributes, and I just have to compare it with the list I contructed with my other Xml which gaves me the available Elements/Attributes.

Thank you !

Dimitri.SP

DimitriStPi · Accepted Answer · 2015-12-14T23:57:00

So... There is a solution !

What we have to do is rather simple..

If our class Attribute looks like the following one :

C#

{
    public class Attribute
    {
        public Attribute(string name, int value)
        {
            this.Name = name;
            this.Value = value;
        }

        public string Name;
        public int Value;
    }
}

The solution could be :

C#

public class element : IXmlSerializable
{
    List<Attribute> AttributeList;
 
    public void WriteXml(XmlWriter writer)
    {
          foreach(var attribute in AttributeList)
          {
            writer.WriteAttributeString(attribute.Name,attribute.Value.ToString());
          }
    }
}

And the result is what i was looking for.