-2

我有两种方法。在第一个中，我将项目添加到名为 Persons 的列表中。在第二个中，我需要从列表（Persons）中选择一个随机项目并从列表中返回该人的姓名。

我不确定如何做到这一点。我试过生成一个随机数和一个随机字母，但是我不明白如何在列表中使用它们。

任何帮助将不胜感激！先感谢您

static void PopulatePersons()
        {
            Person Bill = new Person("Bill", "no", "brown", 'm');
            Person Eric = new Person("Eric", "yes", "brown", 'm');
            Person Robert = new Person("Robert", "no", "blue", 'm');
            Person George = new Person("George", "yes", "brown", 'm');
            Person Herman = new Person("Herman", "no", "green", 'm');
            Person Anita = new Person("Anita", "no", "blue", 'f');
            Person Maria = new Person("Maria", "yes", "green", 'f');
            Person Susan = new Person("Susan", "no", "brown", 'f');
            Person Claire = new Person("Claire", "yes", "brown", 'f');
            Person Anne = new Person("Anne", "no", "brown", 'f');

            Persons = new List<Person>() 
            { Bill, Eric, Robert, George, Herman, Anita, Maria, Susan, Claire, Anne };            
        }

        static Person GetRandomPerson()
        {
            PopulatePersons();
        }
4

2 回答 2

2

您可以使用 Random 类生成列表大小范围内的随机整数，这是一个示例：

public class Program
{
    public static void Main()
    {
        var persons = PopulatePersons();
        var random = new Random();
        var randomPeople = persons.ElementAt(random.Next(0, persons.Count));
        Console.WriteLine(randomPeople.Name);
    }
    
    public static List<Person> PopulatePersons()
        {
            Person Bill = new Person("Bill");
            Person Eric = new Person("Eric");
            Person Robert = new Person("Robert");
            Person George = new Person("George");
            Person Herman = new Person("Herman");
            Person Anita = new Person("Anita");
            return new List<Person>() { Bill, Eric, Robert, George, Herman, Anita };            
        }
    
    public class Person 
    {
        public Person(string name)
        {
            Name = name;
        }
        
        public string Name {get;set;}
    }
}
于 2020-12-29T00:40:23.467 回答
2

您正在寻找的是：

 var person = PopulatePersons.Persons[yourRandomNumber];

所以你的方法应该是这样的：

 static Person GetRandomPerson()
    {
      Random rndPerson= new Random();
      int rndNumber = rnd.Next(0, PopulatePersons.Pesrons.Count);
       return PopulatePersons.Persons[rndNumber];
    }
于 2020-12-29T00:44:40.747 回答