c# - 使用 lambda 表达式获取属性或类型名称

Question

当 lambda 表达式引用实际实例本身时，我如何调整下面的方法来工作？

例如而不是

x => x.Name

表达式是

x => x

所以如果我有一些类“Car”，我可以返回字符串“Car”，而不是只能对其属性进行操作(例如 Car.Colour)

方法:

public static string GetMemberName(Expression expression)
    {
        if (expression is LambdaExpression)
            expression = ((LambdaExpression)expression).Body;

        if (expression is MemberExpression)
        {
            var memberExpression = (MemberExpression)expression;
            if (memberExpression.Expression.NodeType ==
                ExpressionType.MemberAccess)
            {
                return GetMemberName(memberExpression.Expression)
                       + "."
                       + memberExpression.Member.Name;
            }
            return memberExpression.Member.Name;
        }


        if (expression is UnaryExpression)
        {
            var unaryExpression = (UnaryExpression)expression;

            if (unaryExpression.NodeType != ExpressionType.Convert)
                throw new Exception(string.Format(
                    "Cannot interpret member from {0}",
                    expression));
            return GetMemberName(unaryExpression.Operand);
        }
        throw new Exception(string.Format(
            "Could not determine member from {0}",
            expression));
    }

即我想要这样的东西:

if (expression is SomeExpressionThatReturnsAnInstance)
{
    return (name of type of instance);
}

Answer 1

我可能误解了，但是直接的 x => x 将是一个 ParameterExpression。只需在您现有的 is MemberExpression 测试下添加一个额外的测试:

if (expression is MemberExpression)
{
    // As-is
}
// New condition
if (expression is ParameterExpression)
{
    return expression.Type.Name;
}

使用这段代码:

class Car { public string Color { get; set; }}

Expression<Func<Car, string>> expr1 = x => x.Color;
Expression<Func<Car, Car>> expr2 = x => x;

Console.WriteLine(GetMemberName(expr1));

> Color

Console.WriteLine(GetMemberName(expr2));

> Car