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) 中随机选择一个项目并从列表中返回此人的姓名。
我不确定如何执行此操作。我试过生成一个随机数和一个随机字母,但是我不明白如何将它们与列表一起使用。
任何帮助将不胜感激!提前致谢
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();
}
答案 0 :(得分:2)
您可以使用 Random 类在您的列表大小范围内生成一个随机整数,这里是一个例子:
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;}
}
}
答案 1 :(得分: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];
}