在 C# 中使用 LINQ 解析 XML

Question

In this xml file (http://www.studiovincent.net/list.xml):

<list version="1.0">
  <meta>
    <type>resource-list</type>
  </meta>
  <resources start="0" count="4">
    <resource classname="Quote">
      <field name="name">Vincent</field>
      <field name="username">Hill</field>
      <field name="age">31</field>
      <field name="hair">black</field>
    </resource>
    <resource classname="Quote">
      <field name="name">John</field>
      <field name="username">Tedelon</field>
      <field name="age">27</field>
      <field name="hair">brown</field>
    </resource>
    <resource classname="Quote">
      <field name="name">Michael</field>
      <field name="username">Lopez</field>
      <field name="age">20</field>
      <field name="hair">red</field>
    </resource>
    <resource classname="Quote">
      <field name="name">Frank</field>
      <field name="username">Lopez</field>
      <field name="age">25</field>
      <field name="hair">black</field>
    </resource>
  </resources>
</list>

using this code:

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

XmlReader reader = XmlReader.Create("http://www.studiovincent.net/list.xml");
XElement el = XElement.Load(reader);
reader.Close();

var items = el.Elements("resources").Elements("resource").Descendants().DescendantNodes();

var items = from item in el.Elements("resources").Elements("resource").Descendants() 
            where item.Attribute("name").Value == "name" select item.FirstNode;

foreach (XNode node in items)
{
    Console.WriteLine(node.ToString());
}

I have this OUTPUT:

Vincent
John
Michael
Frank

Code working very good, but I need get value 31 which corresponds field name="age" where field name="name" is Vincent. How Can I get this result?

Answer 1

I recommend you do as you would when reading XML naturally.

In your code, try to find all the fields with the name attribute set to "name" .

This process cannot be used to associate a name with an age. It is more natural to read the XML and check all resource elements. Then add to this element some information described in the field elements.

// Using LINQ to SQL
XDocument document = XDocument.Load("http://www.studiovincent.net/list.xml");  // Loads the XML document.
XElement resourcesElement = document.Root.Element("resources");  // Gets the "resources" element that is in the root "list" of the document.

XElement resourceElementVincent = (from resourceElement in resourcesElement.Elements("resource")// Gets all the "resource" elements in the "resources" element
                                    let fieldNameElement = resourceElement.Elements("field").Single(fieldElement => fieldElement.Attribute("name").Value == "name")  // Gets the field that contains the name (there one and only one "name" field in the "resource" element -> use of Single())
                                    where fieldNameElement.Value == "Vincent"  // To get only resources called "Vincent"
                                    select resourceElement).Single();  // We suppose there is one and only 1 resource called "Vincent" -> Use of Single()
XElement fieldAgeElement = resourceElementVincent.Elements("field").Single(fieldElement => fieldElement.Attribute("name").Value == "age");  // Gets the corresponding "age" field
int age = int.Parse(fieldAgeElement.Value, CultureInfo.InvariantCulture);  // Gets the age by Parse it as an integer
Console.WriteLine(age);

Does it do what you want?

Answer 2

Consider this below XML is there as one of the SQL table's column.

<Root>
  <Name>Dinesh</Name>
  <Id>2</Id>
</Root>

The objective of the query is to fetch the Name from the XML. In this example we will fetch the 'Dinesh' as the value.

var Query = (from t in dbContext.Employee.AsEnumerable()
where t.active == true 
select new Employee
{
    Id = t.AtpEventId,  
    Name = XDocument.Parse(t.Content).Descendants("Root").Descendants("Name").ToList().  
    Select(node => node.Value.ToString()).FirstOrDefault()  
});

Note the following:

1. t.active == true is just an example to make some condition if needed.

2. Please note, in the above LINQ query, always use the AsEnumerable, as I did in the first line.

3. Descendants("Root").Descendants("Name") - here "Root" should be the element matching with the XML, and under Root we have Name element.