创建一个随机实例:

Random rnd = new Random();

获取一个随机字符串:

string s = arraylist[rnd.Next(arraylist.Count)];

记住，如果你经常这样做，你应该重用随机对象。将它作为类中的静态字段，这样它只初始化一次，然后访问它。

为什么不:

public static T GetRandom<T>(this IEnumerable<T> list)
{
   return list.ElementAt(new Random(DateTime.Now.Millisecond).Next(list.Count()));
}

为什么不[2]:

public static T GetRandom<T>(this List<T> list)
{
     return list[(int)(DateTime.Now.Ticks%list.Count)];
}

我将建议不同的方法，如果列表中项目的顺序在提取时并不重要(并且每个项目应该只选择一次)，那么您可以使用ConcurrentBag，而不是list，它是线程安全的，无序的对象集合:

var bag = new ConcurrentBag<string>();
bag.Add("Foo");
bag.Add("Boo");
bag.Add("Zoo");

事件:

string result;
if (bag.TryTake(out result))
{
    MessageBox.Show(result);
}

TryTake将尝试从无序集合中提取一个“随机”对象。

Create an instance of Random class somewhere. Note that it's pretty important not to create a new instance each time you need a random number. You should reuse the old instance to achieve uniformity in the generated numbers. You can have a static field somewhere (be careful about thread safety issues): static Random rnd = new Random(); Ask the Random instance to give you a random number with the maximum of the number of items in the ArrayList: int r = rnd.Next(list.Count); Display the string: MessageBox.Show((string)list[r]);

你可以:

list.OrderBy(x => Guid.NewGuid()).FirstOrDefault()