c#property get,设置不同的类型

Question

我有这样的枚举和财产.

        public enum Type
        {
            Hourly = 1,
            Salary = 2,
            None = 3
        };


        public string EmployeeType
        {
            get
            {
                string type;
                switch (employeeType)
                {
                    case Type.Hourly:
                        type = "Hourly Employee";
                        break;
                    case Type.Salary:
                        type = "Salary Employee";
                        break;
                    default:
                        type = "None";
                        break;
                }
                return type;
            }

            // **EDIT:**
            // Now I am trying to parse the string as enum Type.
            // But Constructor still waits a string to set EmployeeType.
            set
            {
                employeeType = (Type)Enum.Parse(typeof(Type), value);
            }
        }

这是我的班级:

public class Employee
{
     private Type employeeType;
}

我想创建这样一个构造函数:

Employee(Employee.Type type) 
{
      EmployeeType = type;
}

编辑:

无法将类型'Payroll.Employee.Type'隐式转换为'string'

我该如何编写属性的set访问器？

更新:

我希望get访问器返回字符串并设置访问器以获取参数类型Employee.Type.我了解到根据C#规范,不可能在属性中执行此操作.我必须编写单独的getter和setter方法.

Answer 1

请改用DescriptionAttribute.

public enum Type
{
    [Description("Hourly Employee")]
    Hourly = 1,
    [Description("Salary Employee")]
    Salary = 2,
    [Description("None")]
    None = 3
};

然后你会有一个

public Type EmployeeType {get; set;}

属性.如果有人想写出来,他们可以得到描述.我也称之为,Type而不是EmployeeType,因为电话myEmployee.EmployeeType听起来多余.您的另一个选择可能是展开该属性并有两种方法

public string GetEmployeeType() { //your switch statement }
public void SetEmployeeType(EmployeeType type)
{
    _type = type;
}

不像房产那么优雅,但很快就完成了工作.还记得IL中的属性只是方法.