Question

如何简化下面的代码以避免在泛型方法上传递类型推断的对象？

using System;
using System.Linq.Expressions;

namespace lambda
{
    class Program
    {
        static void Main(string[] args)
        {

            var area = new Area { Name = "New Area" };
            var propertyName = area.GetPropertyName(area, a => a.Name); // propertyName is COMPILE time checked

            Console.WriteLine(propertyName);

        }
    }

    public class Area
    {
        public int Id;
        public string Name { get; set; }
    }

    public static class Extension
    {
        public static string GetPropertyName<T>(this Area entity, T e, Expression<Func<T, object>> path) // T e for type inference
        {
            var member = path.Body as MemberExpression;
            if (member == null) throw new ArgumentException();
            return member.Member.Name;
        }
    }
}

我的意思是代替使用area.GetPropertyName（area，a =＆gt; a.Name）调用扩展方法只是像这个区域一样调用.GetPropertyName（a =＆gt; a.Name），避免仅为类型推断传递区域对象

我想除非我将方法的签名重构为GetPropertyName（此IEntity实体，Expression＆gt; path），否则我无法做到但是在那种情况下，我希望在编写代码时不太想要，因为我需要在每次调用时指定类型我的意思是area.GetPropertyName（a =＆gt; a.Name）似乎对我来说不像编写区域那么清晰.GetPropertyName（area，a =＆gt; a.Name）

Answer 1

下面的示例代码适用于要求的要求，在调用扩展方法时无需传递对象本身以进行类型推断我使用了一个基类和一个接口，对于我的所有域类都可以正常工作。见下面的代码

namespace UnitTestProject
{
    using Microsoft.VisualStudio.TestTools.UnitTesting;
    using System;
    using System.Linq.Expressions;

    public interface IEntity<T>
    {
    }

    public abstract class Entity<T> : IEntity<T> where T : class
    {
    }

    public class Area : Entity<Area>
    {
        public int Id { get; set; }
        public string Name { get; set; }
        public DateTime Created { get; set; }
        public bool Closed { get; set; }
        public string Description { get; set; }
    }

    public static class EntityExtension
    {
        public static string GetPropertyName<T>(this IEntity<T> entity, Expression<Func<T, object>> expr) // T e for type inference
        {
            var unary = expr.Body as UnaryExpression;
            var member = expr.Body as MemberExpression;
            return member != null ? member.Member.Name : (unary != null ? ((MemberExpression)unary.Operand).Member.Name : String.Empty);
        }
    }


    [TestClass]
    public class UnitTest
    {
        [TestMethod]
        public void GetPropertyName_Tests()
        {
            var area = new Area();

            var x = area.GetPropertyName(a => a.Id);
            var y = area.GetPropertyName(a => a.Name);
            var v = area.GetPropertyName(a => a.Created);
            var w = area.GetPropertyName(a => a.Closed);
            var z = area.GetPropertyName(a => a.Description);

            Assert.AreEqual(x, "Id");
            Assert.AreEqual(y, "Name");
            Assert.AreEqual(v, "Created");
            Assert.AreEqual(w, "Closed");
            Assert.AreEqual(z, "Description");

        }
    }
}

如何重新访问强类型的给定类型的属性名称？

1 个答案: