当通过lambda表达式传入时,是否有更好的方法来获得属性名? 这是我目前拥有的。

eg.

GetSortingInfo<User>(u => u.UserId);

它只在属性为字符串时才将其转换为成员表达式。因为不是所有的属性都是字符串,我必须使用object,但它会为那些返回一个unaryexpression。

public static RouteValueDictionary GetInfo<T>(this HtmlHelper html, 
    Expression<Func<T, object>> action) where T : class
{
    var expression = GetMemberInfo(action);
    string name = expression.Member.Name;

    return GetInfo(html, name);
}

private static MemberExpression GetMemberInfo(Expression method)
{
    LambdaExpression lambda = method as LambdaExpression;
    if (lambda == null)
        throw new ArgumentNullException("method");

    MemberExpression memberExpr = null;

    if (lambda.Body.NodeType == ExpressionType.Convert)
    {
        memberExpr = 
            ((UnaryExpression)lambda.Body).Operand as MemberExpression;
    }
    else if (lambda.Body.NodeType == ExpressionType.MemberAccess)
    {
        memberExpr = lambda.Body as MemberExpression;
    }

    if (memberExpr == null)
        throw new ArgumentException("method");

    return memberExpr;
}

当前回答

现在在c# 6中,你可以简单地使用这样的nameof(User.UserId)

这有很多好处,其中之一是这是在编译时完成的,而不是在运行时。

https://msdn.microsoft.com/en-us/magazine/dn802602.aspx

其他回答

好吧,没有必要调用. name . tostring(),但大体上就是这样,是的。你可能需要考虑的唯一问题是x.f o.Bar是否应该返回“Foo”,“Bar”,或者一个异常——也就是说,你是否需要迭代。

(re comment)关于灵活排序的更多信息,请看这里。

使用c# 7模式匹配:

public static string GetMemberName<T>(this Expression<T> expression)
{
    switch (expression.Body)
    {
        case MemberExpression m:
            return m.Member.Name;
        case UnaryExpression u when u.Operand is MemberExpression m:
            return m.Member.Name;
        default:
            throw new NotImplementedException(expression.GetType().ToString());
    }
}

例子:

public static RouteValueDictionary GetInfo<T>(this HtmlHelper html, 
    Expression<Func<T, object>> action) where T : class
{
    var name = action.GetMemberName();
    return GetInfo(html, name);
}

[更新]c# 8模式匹配:

public static string GetMemberName<T>(this Expression<T> expression) => expression.Body switch
{
    MemberExpression m => m.Member.Name,
    UnaryExpression u when u.Operand is MemberExpression m => m.Member.Name,
    _ => throw new NotImplementedException(expression.GetType().ToString())
};

现在在c# 6中,你可以简单地使用这样的nameof(User.UserId)

这有很多好处,其中之一是这是在编译时完成的,而不是在运行时。

https://msdn.microsoft.com/en-us/magazine/dn802602.aspx

public string GetName<TSource, TField>(Expression<Func<TSource, TField>> Field)
{
    return (Field.Body as MemberExpression ?? ((UnaryExpression)Field.Body).Operand as MemberExpression).Member.Name;
}

这个函数处理成员表达式和一元表达式。区别在于,如果你的表达式表示值类型,你将得到一个UnaryExpression,而如果你的表达式表示引用类型,你将得到一个MemberExpression。所有内容都可以转换为对象,但值类型必须被装箱。这就是UnaryExpression存在的原因。参考。

出于可读性考虑(@Jowen),这里有一个扩展的等效内容:

public string GetName<TSource, TField>(Expression<Func<TSource, TField>> Field)
{
    if (object.Equals(Field, null))
    {
        throw new NullReferenceException("Field is required");
    }

    MemberExpression expr = null;

    if (Field.Body is MemberExpression)
    {
        expr = (MemberExpression)Field.Body;
    }
    else if (Field.Body is UnaryExpression)
    {
        expr = (MemberExpression)((UnaryExpression)Field.Body).Operand;
    }
    else
    {
        const string Format = "Expression '{0}' not supported.";
        string message = string.Format(Format, Field);

        throw new ArgumentException(message, "Field");
    }

    return expr.Member.Name;
}

下面是基于这个答案获取PropertyInfo的另一种方法。它消除了对对象实例的需要。

/// <summary>
/// Get metadata of property referenced by expression. Type constrained.
/// </summary>
public static PropertyInfo GetPropertyInfo<TSource, TProperty>(Expression<Func<TSource, TProperty>> propertyLambda)
{
    return GetPropertyInfo((LambdaExpression) propertyLambda);
}

/// <summary>
/// Get metadata of property referenced by expression.
/// </summary>
public static PropertyInfo GetPropertyInfo(LambdaExpression propertyLambda)
{
    // https://stackoverflow.com/questions/671968/retrieving-property-name-from-lambda-expression
    MemberExpression member = propertyLambda.Body as MemberExpression;
    if (member == null)
        throw new ArgumentException(string.Format(
            "Expression '{0}' refers to a method, not a property.",
            propertyLambda.ToString()));

    PropertyInfo propInfo = member.Member as PropertyInfo;
    if (propInfo == null)
        throw new ArgumentException(string.Format(
            "Expression '{0}' refers to a field, not a property.",
            propertyLambda.ToString()));

    if(propertyLambda.Parameters.Count() == 0)
        throw new ArgumentException(String.Format(
            "Expression '{0}' does not have any parameters. A property expression needs to have at least 1 parameter.",
            propertyLambda.ToString()));

    var type = propertyLambda.Parameters[0].Type;
    if (type != propInfo.ReflectedType &&
        !type.IsSubclassOf(propInfo.ReflectedType))
        throw new ArgumentException(String.Format(
            "Expression '{0}' refers to a property that is not from type {1}.",
            propertyLambda.ToString(),
            type));
    return propInfo;
}

它可以这样调用:

var propertyInfo = GetPropertyInfo((User u) => u.UserID);