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c#linq-to-xml
Jitesh Dani
Jitesh Dani

Reputation: 385

XML with C# LINQ to XML

Following is the example of an XML document.

<People>
   <Person>
       <Name>ABC </Name>
       <SSN>111111</SSN>
       <Address>asdfg</Address>
   </Person>
</People>

I need to get the tag names but not the values between the tag names. That is, under the person tag, I should grab the Name, SSN, and Address nodes and not the ABC, 111111, and asdfg values.

I need to use LINQ to XML and query it in C#.

How can I do it?

Upvotes: 0

Views: 3838

Answers (4)

Brian ONeil
Brian ONeil

Reputation: 4339

Simple LINQ syntax to get the names is listed below. It assumes you have the XML loaded in a XDocument variable named doc.

var nodeNames = from node in doc.Descendants("Person").First().Descendants()
                select node.Name.LocalName;

It only looks at the first person. If you have more than one in the XML document, the list of names is probably not what you would want (no reason to repeat all the names of the nodes over and over). This way, you get a list of just the node names for the first person, but it does assume that the first one would have a complete list of names. If they vary, you would need to build a distinct list from all the nodes.

Upvotes: 0

Omar
Omar

Reputation: 40202

Create a class

public class Person
{
     public string Name {get; set;}
     public int SSN {get; set;}
     public string Address {get; set;}
}

And create a new person this way;

List<Person> NewPersons = new List<Person>();
XDocument doc = XDocument.Load(xmlFile);     
foreach(XElement xElem in doc.Descendants("Person"))
{
    NewPersons.Add(new Person
            {
                Name = xElem. Element("Name").Value,
                SSN = xElem.Element("SSN").Value,
                Address = xElem.Element("Address").Value,
            });
}

Upvotes: 1

Ahmad Mageed
Ahmad Mageed

Reputation: 96557

This returns the names as a list of strings:

var doc = XDocument.Parse(@"<People>
   <Person>
       <Name>ABC </Name>
       <SSN>111111</SSN>
       <Address>asdfg</Address>
   </Person>
</People>"
);

var list = doc.Root.Element("Person").Descendants()
              .Select(node => node.Name.LocalName).ToList();

In case you're using an XElement instead of an XDocument, you can remove the .Root from the above code and get the correct results.

Upvotes: 4

Michael La Voie
Michael La Voie

Reputation: 27926

You can get it this way...

string xml = "<People> <Person> <Name>ABC </Name> <SSN>111111</SSN> <Address>asdfg</Address> </Person> <Person> <Name>ABC </Name> <SSN>111111</SSN> <Address>asdfg</Address> </Person> </People>";

XElement xmlData = XElement.Parse(xml);

foreach(XElement xmlPerson in xmlData.Elements("Person"))
{
    List<string> TagsForThisPerson = new List<string>();
    foreach(XElement xmlTag in xmlPerson.Elements())
    {
        TagsForThisPerson.Add(xmlTag.Name.ToString());
    }
    TagsForThisPerson.Dump();
}

Upvotes: 0

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