Using LINQ to select a random XML node

I'm new to LINQ and am having a problem. I have a file that looks something like this:

开发者_如何学JAVA<?xml version="1.0" encoding="utf-8" ?>
<Galleries>
   <Gallery ID="10C31804CEDB42693AADD760C854ABD" Title="Test1">
      <Description>The first test gallery.  Picture of a cat and Wilford Brimley.  Can you tell the difference?</Description>
      <Images>
         <Image Title="t1Image1" FileName="tcats.jpg" />
         <Image Title="t1Image2" FileName="twb.jpg" />
      </Images>
   </Gallery>
   <Gallery ID="0420EC15405B488E1E0F157AC823A6" Title="Test2">
      <Description>The second test gallery.  A large image of Wilford Brimley and various cats.  The cats will be on the right.</Description>
      <Images>
         <Image Title="t2Image1" FileName="wilfordbrimley.jpg" />
      </Images>
   </Gallery>
</Galleries> 

Anyway, I know the ID of the Gallery I want, but I want to choose one of the images at random. Is there a LINQ statement that can do this?

Can you order the Images in the gallery by Random.Next() then select out the first element.

I dont know much about linq2xml but here's what i came up with

static void Main(string[] args)
{
    Random rnd = new Random();
    XDocument galleries = XDocument.Load(@"C:\Users\John Boker\Documents\Visual Studio 2008\Projects\ConsoleApplication1\ConsoleApplication1\Galleries.xml");
    var image = (from g in galleries.Descendants("Gallery")
                 where g.Attribute("ID").Value == "10C31804CEDB42693AADD760C854ABD"
                 select g.Descendants("Images").Descendants("Image").OrderBy(r=>rnd.Next()).First()).First();
    Console.WriteLine(image);
    Console.ReadLine();
}

I'm sure the selecting could be done a lot differently, but that's what i did to make it work with the random.next thing.

Here are a couple solutions that rely on computing the number of Image nodes; not terribly efficient, but I don't think you can do better since many Linq collection types are exposed as IEnumerable.

XElement GetRandomImage(XElement images)
{
    Random rng = new Random();
    int numberOfImages = images.Elements("Image").Count();

    return images.Elements("Image").Skip(rng.Next(0, numberOfImages)).FirstOrDefault();
}

XElement GetRandomImage(XElement images)
{
    Random rng = new Random();
    IList<XElement> images = images.Elements("Image").ToList();

    return images.Count == 0 :
        null ?
        images[rng.Next(0, images.Count - 1)];
}

I do not recommend using the selected answer as it uses a sort which is O(n log n) where n is the number of images in the selected gallery. You can select a random item from a list in O(1) time. Thus, I would use the following:

using(StreamReader sr = new StreamReader(File.Open(path, FileMode.Open))) {
    XDocument galleries = XDocument.Load(sr);
    string id = "10C31804CEDB42693AADD760C854ABD";
    var query = (from gallery in galleries.Descendants("Galleries")
                                          .Descendants("Gallery")
                 where (string)gallery.Attribute("ID") == id
                 select gallery.Descendants("Images")
                               .Descendants("Image")
                ).SingleOrDefault();
    Random rg = new Random();
    var image = query.ToList().RandomItem(rg);
    Console.WriteLine(image.Attribute("Title"));
}

Here I am using:

static class ListExtensions {
    public static T RandomItem<T>(this List<T> list, Random rg) {
        if(list == null) {
            throw new ArgumentNullException("list");
        }
        if(rg == null) {
            throw new ArgumentNullException("rg");
        }
        int index = rg.Next(list.Count);
        return list[index];
    }
}