Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 629707
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 13, 20262026-05-13T19:46:33+00:00 2026-05-13T19:46:33+00:00

I’ve got a very basic object object model that is being serialized by the

  • 0

I’ve got a very basic object object model that is being serialized by the System.Xml.XmlSerialization stuff. I need to use the XmlAttributeOverrides functionality to set the xml element names for a collection of child elements.

public class Foo{
  public List Bars {get; set; }
}

public class Bar {
  public string Widget {get; set; }
}

using the standard xml serializer, this would come out as

 <Foo>
  <Bars>
    <Bar>...</Bar>
  </Bars>
 </Foo>

I need to use the XmlOverrideAttributes to make this say

 <Foo>
  <Bars>
    <SomethingElse>...</SomethingElse>
  </Bars>
 </Foo>

but I can’t seem to get it to rename the child elements in the collection… i can rename the collection itself… i can rename the root… not sure what i’m doing wrong.

here’s the code I have right now:

XmlAttributeOverrides xOver = new XmlAttributeOverrides();

var bars = new XmlElementAttribute("SomethingElse", typeof(Bar));
var elementNames = new XmlAttributes();
elementNames.XmlElements.Add(bars);
xOver.Add(typeof(List), "Bars", elementNames);

StringBuilder stringBuilder = new StringBuilder();
StringWriter writer = new StringWriter(stringBuilder);
XmlSerializer serializer = new XmlSerializer(typeof(Foo), xOver);
serializer.Serialize(writer, someFooInstance);

string xml = stringBuilder.ToString();

but this doesn’t change the name of the element at all… what am I doing wrong?

thanks

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-13T19:46:34+00:00Added an answer on May 13, 2026 at 7:46 pm

    To do that you want [XmlArray] and [XmlArrayItem] (ideally both of to make it explicit):

    using System.Collections.Generic;
    using System.IO;
    using System.Xml.Serialization;
    public class Foo {
        public List<Bar> Bars { get; set; }
    }  
    public class Bar {
        public string Widget { get; set; }
    }
    static class Program {
        static void Main() {
            XmlAttributeOverrides xOver = new XmlAttributeOverrides();
            xOver.Add(typeof(Foo), "Bars", new XmlAttributes {
                XmlArray = new XmlArrayAttribute("Bars"),
                XmlArrayItems = {
                    new XmlArrayItemAttribute("SomethingElse", typeof(Bar))
                }
            });
            XmlSerializer serializer = new XmlSerializer(typeof(Foo), xOver);
            using (var writer = new StringWriter()) {
                Foo foo = new Foo { Bars = new List<Bar> {
                    new Bar { Widget = "widget"}
                }};
                serializer.Serialize(writer, foo);
                string xml = writer.ToString();
            }            
        }
    }
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Ask A Question

Stats

  • Questions 379k
  • Answers 379k
  • Best Answers 0
  • User 1
  • Popular
  • Answers
  • Editorial Team

    How to approach applying for a job at a company ...

    • 7 Answers
  • Editorial Team

    What is a programmer’s life like?

    • 5 Answers
  • Editorial Team

    How to handle personal stress caused by utterly incompetent and ...

    • 5 Answers
  • Editorial Team
    Editorial Team added an answer It's perfectly valid. You wouldn't want to put great big… May 14, 2026 at 9:35 pm
  • Editorial Team
    Editorial Team added an answer Lines 4-5 seem to be doing you no good. Also,… May 14, 2026 at 9:35 pm
  • Editorial Team
    Editorial Team added an answer Check out http://docs.heroku.com/config-vars, the first example on the page is… May 14, 2026 at 9:35 pm

Trending Tags

analytics british company computer developers django employee employer english facebook french google interview javascript language life php programmer programs salary

Top Members

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.