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 556599
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 13, 20262026-05-13T11:56:13+00:00 2026-05-13T11:56:13+00:00

My xml file has something like this: … <Keyword name = if /> <Keyword

  • 0

My xml file has something like this:
...
<Keyword name = "if" />
<Keyword name = "else" />
<Keyword name = "is" />
...

So how can I recursively get all of the values of the name attribute and add them to a List<string> or string[]. Maybe a foreach loop?


I followed codemeit’s and I keep getting an error:Data at the root level is invalid. Line 1, position 1. My xml file is
<KeyWords>
...
<KeyWord name = "if" />
...
</KeyWord>


New problem The '\' character, hexadecimal value 0x5C, cannot be included in a name. but the same file.

  • 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
    2026-05-13T11:56:13+00:00Added an answer on May 13, 2026 at 11:56 am

    Assume let variable testXml is equal to the follow xml string

    <Keywords>
     <Keyword name = "if" />
     <Keyword name = "else" />
     <Keyword name = "is" />
    </Keywords>
    

    Use XElement and LINQ to extract the name attribute values

    var myXml = XElement.Parse(testXml );
    var myArray = myXml.Elements().Where(n => n.Name.LocalName.Equals("Keyword"))
                        .Select(n => n.Attribute("name").Value)
                        .ToArray();
    

    myArray will contain {“if”, “else”, “is”}

    UPDATE

    Thanks to @SLaks comment, we could actually just do

    var myArray = myXml.Elements("Keyword").Attributes("name").Select(n => n.Value);
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Ask A Question

Stats

  • Questions 368k
  • Answers 368k
  • Best Answers 0
  • User 1
  • Popular
  • Answers
  • Editorial Team

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

    • 7 Answers
  • Editorial Team

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

    • 5 Answers
  • Editorial Team

    What is a programmer’s life like?

    • 5 Answers
  • Editorial Team
    Editorial Team added an answer Type genericSuperclass = x.getClass().getGenericSuperclass(); if (genericSuperclass instanceof ParameterizedType) { Type[]… May 14, 2026 at 5:07 pm
  • Editorial Team
    Editorial Team added an answer You'll need to query twice. I dont have my environment… May 14, 2026 at 5:07 pm
  • Editorial Team
    Editorial Team added an answer SELECT a.id, a.name, AVG(r.rating) AS average FROM albums a LEFT… May 14, 2026 at 5:07 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.