CODEWITHSUNDEEP
c#xmllinqvar
LazyZebra
LazyZebra

Reputation: 1113

select only one random node from linq to xml

I have an XML file and I want to select only one random node. It seems like I'm almost there but the foreach with the var is looping. How do I select only one node and return it?

XML:

<human_check>
  <qa>
    <q>2 + 2</q>
    <a>4</a>
  </qa>
  <qa>
    <q>1 + 2</q>
    <a>3</a>
  </qa>
  <qa>
    <q>6 + 3</q>
    <a>9</a>
  </qa>
  <qa>
    <q>3 + 5</q>
    <a>7</a>
  </qa>
</human_check>

C#

public class human_check
{

    public static string get_q()
    {
        try
        {
            string h = string.Empty;
            Random rnd = new Random();
            XDocument questions = XDocument.Load(@"C:\Users\PETERS\Desktop\human_check.xml");
            var random_q = from q in questions.Descendants("qa")
                           select new
                           {
                               question = q.Descendants("q").OrderBy(r => rnd.Next()).First().Value
                           };

            foreach (var rq in random_q)
            {
                h = rq.question.ToString();
            }

            return h;

        }
        catch (Exception ex)
        {
            throw ex;
        }
    }

}

Thanks in advance,

EP

Upvotes: 2

Views: 1300

Answers (2)

Adam S
Adam S

Reputation: 3125

Instead of setting up an ordering you could just pick a random element.

var qas = questions.Descendants("qa");
int qaCount = qas.Count();
h = qas.ElementAt(rnd.Next(0, qaCount - 1)).Element("q").Value;

Upvotes: 4

Kittoes0124
Kittoes0124

Reputation: 5080

var random_q = (from q in questions.Descendants("qa")
                select q).OrderBy(r => rnd.Next()).First();

h = random_q.Descendants("q").SingleOrDefault().Value.ToString();

Upvotes: 3

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