C#, LINQ and XML - Get Value

Hi experts,

I have this XML, which I am not allowed to change. My question is how to use LINQ and select the name,phone,key2 and val2 from this XML?

<Root>
 <OrganisationList>
  <Organization>
    <Name>xxx</Name>
    <Phone>yyy</Phone>
    <Features>
     <Feature>
      <Key>key1</Key>
      <Val>val1</Val>
     </Feature>
     <Feature>
      <Key>key2</Key>
      <Val>val2</val>
     </Feature>
     ...
    <Features>
  </Organization>

  <Organization>
     ...
  </Organization>
 </OrganisationList>
</Root>

just to be specific on the query, What I need is a query which returns the organization with the name,phone,key2 and val2 where the key2 value of the organization is val2 .

I hope someone can help!
LVL 21
masterpassAsked:
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

käµfm³d 👽Commented:
I don't fully understand the criteria you mention in your last paragraph, but see if this is what you are after:

using System;
using System.Linq;
using System.Xml.Linq;

...

static void Main(string[] args)
{
    XDocument xdoc = XDocument.Load("input.xml");

    var query = from org in xdoc.Descendants("Organization")
                from feature in org.Element("Features").Elements()
                where feature.Element("Val") != null && feature.Element("Val").Value == "val2"
                select new
                {
                    Name = org.Element("Name").Value,
                    Phone = org.Element("Phone").Value,
                    Key2 = feature.Element("Key").Value,
                    Val2 = feature.Element("Val").Value
                };

    foreach (var item in query)
    {
        Console.WriteLine(item.Name);
        Console.WriteLine(item.Phone);
        Console.WriteLine(item.Key2);
        Console.WriteLine(item.Val2);
        Console.WriteLine();
    }

    Console.ReadKey();
}

Open in new window

0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
Rahul AgarwalTeam LeaderCommented:
static XDocument GetStarbuzzData()
{
     /*
         * You can use an XDocument to create an XML file, and that includes XML
         * files you can read and write using DataContractSerializer.
         *
         * An XMLDocument object represents an XML document. It's part of the
         * System.Xml.Linq namespace.
         *
         * Use XElement objects to create elements under the XML tree.
         */
 
     XDocument doc = new XDocument(
         new XDeclaration("1.0", "utf-8", "yes"),
         new XComment("Starbuzz Customer Loyalty Data"),
         new XElement("starbuzzData",
             new XAttribute("storeName", "Park Slope"),
             new XAttribute("location", "Brooklyn, NY"),
             new XElement("person",
                 new XElement("personalInfo",
                     new XElement("name", "Janet Venutian"),
                     new XElement("zip", 11215)),
                 new XElement("favoriteDrink", "Choco Macchiato"),
                 new XElement("moneySpent", 255),
                 new XElement("visits", 50)),
             new XElement("person",
                 new XElement("personalInfo",
                     new XElement("name", "Liz Nelson"),
                     new XElement("zip", 11238)),
                 new XElement("favoriteDrink", "Double Cappuccino"),
                 new XElement("moneySpent", 150),
                 new XElement("visits", 35)),
             new XElement("person",
                 new XElement("personalInfo",
                     new XElement("name", "Matt Franks"),
                     new XElement("zip", 11217)),
                 new XElement("favoriteDrink", "Zesty Lemon Chai"),
                 new XElement("moneySpent", 75),
                 new XElement("visits", 15)),
             new XElement("person",
                 new XElement("personalInfo",
                     new XElement("name", "Joe Ng"),
                     new XElement("zip", 11217)),
                 new XElement("favoriteDrink", "Banana Split in a Cup"),
                 new XElement("moneySpent", 60),
                 new XElement("visits", 10)),
             new XElement("person",
                 new XElement("personalInfo",
                     new XElement("name", "Sarah Kalter"),
                     new XElement("zip", 11215)),
                 new XElement("favoriteDrink", "Boring Coffee"),
                 new XElement("moneySpent", 110),
                 new XElement("visits", 15))));
     return doc;
}

please use the following link for reference:
http://broadcast.oreilly.com/2010/10/understanding-c-simple-linq-to.html
http://stackoverflow.com/questions/3763999/extract-data-from-xml-into-c-sharp-object-using-linq-to-xml-enums
http://dotnet.dzone.com/articles/using-linq-xml-query-xml-data
0
masterpassAuthor Commented:
well, that is exactly what I needed. just a small change in condition. Thanks Kaufmed :) you are a champ!

var query = from org in xdoc.Descendants("Organization")
                        from feature in org.Element("Features").Elements()
                        where feature.Element("[b]Key[/b]") != null && feature.Element("Key").Value == "[b]key2[/b]"
                        select new
                        {
                            Name = org.Element("Name").Value,
                            Phone = org.Element("Phone").Value,
                            Key2 = feature.Element("Key").Value,
                            Val2 = feature.Element("Val").Value
                        };

Open in new window

0
masterpassAuthor Commented:
thanks :)
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
C#

From novice to tech pro — start learning today.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.