如何在c#console中读取用户输入

aid*_*n87 1 c# console console-application

我想这对你们来说应该很简单,但对我来说很难,因为我是c#的新手.

我有一个简单的"pacient"课程.

public class Pacient {

public Pacient(string _name, string _lastName, DateTime _date, string _phone, string _email)
{
    name = _name;
    lastname = _lastName
    dateOfBirth = _date;
    phone_num = _phone;
    email = _email;
}


private string name;
public string Name {
    get {
        return name;
    }
    set {
        name = value;
    }
}
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等等...

现在我想在控制台中读取输入用户类型...

我怎么做?它适用于预先输入的名称,如下图所示.

 Pacient John = new Pacient("John", " Doe ", new DateTime(1992,12,12) , " 045-999-333", "  example@example.com");
        John.Email = "example@example.com";
        John.Name ="JOHN ";
        John.LastName=" DOE ";*/
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总结 当控制台打开时,它应该询问名称.当用户键入名称时,控制台应将名称存储到"名称"中,然后再显示它.

感谢你们!

Mar*_*cus 6

name如果要将其拆分为示例中提供的名字和姓氏,则一个名称变量是不够的.

Console.Write("First name:");
var firstName = Console.ReadLine();
Console.Write("Last name:");
var lastName = Console.ReadLine();

Pacient John = new Pacient(firstName, lastName, new DateTime(1992,12,12) , " 045-999-333", "  example@example.com");
John.Email = "example@example.com";
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要打印它:

Console.WriteLine("Name: {0} {1}",firstName,lastName);
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PS霸牛逼 ient是英文拼写有T.