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Best way to read through xml

Asked 2011-03-04T15:28:47.217
20

HI I have a xml document like this:

<Students>
<student name="A" class="1"/>
<student name="B"class="2"/>
<student name="c" class="3"/>
</Students>

I want to use XmlReader to read through this xml and return a list of students as List<student>. I know this can be achieved as follows:

 List<Student> students = new List<Student>();
    XmlReader reader = XmlReader.Create("AppManifest.xml");
    while (reader.Read())
    {
       if (reader.NodeType == XmlNodeType.Element && reader.Name == "student")
       {
            students.Add(new Student()
            {
                 Name = reader.GetAttribute("name"),
                 Class = reader.GetAttribute("Class")
             });
        }
     }

I just want to know if there is any better solution for this?

I am using silverlight 4. The xml structure is static, ie. it will have only one Students node and all the student node with above said attributes will only be there.

c# xml silverlight silverlight-4.0
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2 Answers

54

Absolutely - use LINQ to XML. It's so much simpler:

XDocument doc = XDocument.Load("AppManifest.xml");
var students = doc.Root
                  .Elements("student")
                  .Select(x => new Student {
                              Name = (string) x.Attribute("name"),
                              Class = (string) x.Attribute("class")
                          })
                  .ToList();

XmlReader is a relatively low-level type - I would avoid it unless you really can't afford to slurp the whole of the XML into memory at a time. Even then there are ways of using LINQ to XML in conjunction with XmlReader if you just want subtrees of the document.

answered 2011-03-04T15:31:17.387
8

It's alot easier if we're using Linq xml:

var xDoc = XDocument.Load("AppManifest.xml");

var students = 
    xDoc.Root.Elements("student")
    .Select(n =>
        new Student
        {
            Name = (string)n.Attribute("name"),
            Class = (string)n.Attribute("class"),
        })
    .ToList();
answered 2011-03-04T15:33:33.177

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