Question

我有一个XML文件，我想只选择一个随机节点。好像我差不多了，但是var的foreach正在循环。如何只选择一个节点并将其返回？

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;
        }
    }

}

提前致谢，

EP

Answer 1

您可以选择随机元素，而不是设置排序。

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

Answer 2

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

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

从linq到xml只选择一个随机节点

2 个答案: