好吧，这是我最终创建的(基于我选择的答案和他引用的问题):

// <summary>
// Get the name of a static or instance property from a property access lambda.
// </summary>
// <typeparam name="T">Type of the property</typeparam>
// <param name="propertyLambda">lambda expression of the form: '() => Class.Property' or '() => object.Property'</param>
// <returns>The name of the property</returns>

public string GetPropertyName<T>(Expression<Func<T>> propertyLambda)
{
    var me = propertyLambda.Body as MemberExpression;

    if (me == null)
    {
        throw new ArgumentException("You must pass a lambda of the form: '() => Class.Property' or '() => object.Property'");
    }

    return me.Member.Name;
 }

用法:

// Static Property
string name = GetPropertyName(() => SomeClass.SomeProperty);

// Instance Property
string name = GetPropertyName(() => someObject.SomeProperty);

我修改了您的解决方案，以链接多个属性:

public static string GetPropertyName<T>(Expression<Func<T>> propertyLambda)
{
    MemberExpression me = propertyLambda.Body as MemberExpression;
    if (me == null)
    {
        throw new ArgumentException("You must pass a lambda of the form: '() => Class.Property' or '() => object.Property'");
    }

    string result = string.Empty;
    do
    {
        result = me.Member.Name + "." + result;
        me = me.Expression as MemberExpression;
    } while (me != null);

    result = result.Remove(result.Length - 1); // remove the trailing "."
    return result;
}

用法:

string name = GetPropertyName(() => someObject.SomeProperty.SomeOtherProperty);
// returns "SomeProperty.SomeOtherProperty"

这就是我实现它的方式，背后的原因是如果你想从它的成员中获取名称的类不是静态的，那么你需要创建它的一个实例，然后获取成员的名称。所以这里的一般性就有用了

public static string GetName<TClass>(Expression<Func<TClass, object>> exp)
{
    MemberExpression body = exp.Body as MemberExpression;

    if (body == null)
    {
         UnaryExpression ubody = (UnaryExpression)exp.Body;
         body = ubody.Operand as MemberExpression;
    }

     return body.Member.Name;
}

用法是这样的

var label = ClassExtension.GetName<SomeClass>(x => x.Label); //x is refering to 'SomeClass'

您可以使用Reflection来获取属性的实际名称。

http://www.csharp-examples.net/reflection-property-names/

如果你需要一种方法来分配一个“字符串名称”的属性，为什么不写一个属性，你可以反映得到字符串名称?

[StringName("MyStringName")]
private string MyProperty
{
    get { ... }
}

您可以使用StackTrace类来获取当前函数的名称(或者如果您将代码放在函数中，则降低一级并获取调用函数)。

见http://msdn.microsoft.com/en-us/library/system.diagnostics.stacktrace (VS.71) . aspx

有一个著名的黑客从lambda表达式中提取它(这是来自PropertyObserver类，由Josh Smith在他的MVVM基金会):

    private static string GetPropertyName<TPropertySource>
        (Expression<Func<TPropertySource, object>> expression)
    {
        var lambda = expression as LambdaExpression;
        MemberExpression memberExpression;
        if (lambda.Body is UnaryExpression)
        {
            var unaryExpression = lambda.Body as UnaryExpression;
            memberExpression = unaryExpression.Operand as MemberExpression;
        }
        else
        {
            memberExpression = lambda.Body as MemberExpression;
        }

        Debug.Assert(memberExpression != null, 
           "Please provide a lambda expression like 'n => n.PropertyName'");

        if (memberExpression != null)
        {
            var propertyInfo = memberExpression.Member as PropertyInfo;

            return propertyInfo.Name;
        }

        return null;
    }

抱歉，这里缺少一些上下文。这是一个更大的类的一部分，其中TPropertySource是包含属性的类。您可以将TPropertySource中的函数设为泛型，以便从类中提取它。我建议查看来自MVVM基金会的完整代码。