获取通用抽象类的属性名称 | 获取通用抽象类的属性名称

本文介绍了获取通用抽象类的属性名称的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

考虑以下通用抽象类的实现:

Considering the following implementation of a generic, abstract class:

public abstract class BaseRequest<TGeneric> : BaseResponse where TRequest : IRequestFromResponse
{
    public TGeneric Request { get; set; }
}

是否有机会获得属性Request的名称而又没有从其继承实例?

Is there any chance to get the name of the property Request without having an instance inherited from it?

我需要Request作为字符串"Request"，以避免使用硬编码的字符串.有什么想法可以通过反思做到这一点吗?

I need Request as string "Request" to avoid using hardcoded strings. Any ideas how to make this through reflection?

推荐答案

从C#6开始，您应该能够使用nameof运算符:

Starting from C# 6, you should be able to use the nameof operator:

string propertyName = nameof(BaseRequest<ISomeInterface>.Request);

用于BaseRequest<T>的通用类型参数是无关紧要的(只要它符合类型约束)，因为您没有从该类型实例化任何对象.

The generic type parameter used for BaseRequest<T> is irrelevant (as long as it meets the type constraints), since you are not instantiating any object from the type.

对于C#5和更早版本的C#，您可以使用 Cameron MacFarland的答案来从lambda表达式中检索属性信息.下面给出了一个大大简化的改编(无错误检查):

For C# 5 and older, you can use Cameron MacFarland's answer for retrieving property information from lambda expressions. A heavily-simplified adaptation is given below (no error-checking):

public static string GetPropertyName<TSource, TProperty>(
    Expression<Func<TSource, TProperty>> propertyLambda)
{
    var member = (MemberExpression)propertyLambda.Body;
    return member.Member.Name;
}

然后您可以像这样消费它:

You can then consume it like so:

string propertyName = GetPropertyName((BaseRequest<ISomeInterface> r) => r.Request);
// or //
string propertyName = GetPropertyName<BaseRequest<ISomeInterface>, ISomeInterface>(r => r.Request);

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