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

我在ObjectStateEntry上创建了一个扩展方法,以便能够以类型安全的方式标记(实体框架POCO类的)属性,因为默认方法只接受字符串。这是我从属性中获取名称的方式:

public static void SetModifiedProperty<T>(this System.Data.Objects.ObjectStateEntry state, Expression<Func<T>> action)
{
    var body = (MemberExpression)action.Body;
    string propertyName = body.Member.Name;

    state.SetModifiedProperty(propertyName);
}

其他回答

我也在玩同样的东西,然后做了这个。它还没有完全测试过,但似乎处理了值类型的问题(你遇到的unaryexpression问题)

public static string GetName(Expression<Func<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;
}

我发现了另一种方法,就是让源和属性具有强类型,并显式地推断lambda的输入。不确定这是否是正确的术语,但这是结果。

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

    return GetInfo(html, name);
}

然后像这样叫它。

GetInfo((User u) => u.UserId);

瞧,它起作用了。

如果你想要获得多个字段,我就保留这个函数:

/// <summary>
    /// Get properties separated by , (Ex: to invoke 'd => new { d.FirstName, d.LastName }')
    /// </summary>
    /// <typeparam name="T"></typeparam>
    /// <param name="exp"></param>
    /// <returns></returns>
    public static string GetFields<T>(Expression<Func<T, object>> exp)
    {
        MemberExpression body = exp.Body as MemberExpression;
        var fields = new List<string>();
        if (body == null)
        {
            NewExpression ubody = exp.Body as NewExpression;
            if (ubody != null)
                foreach (var arg in ubody.Arguments)
                {
                    fields.Add((arg as MemberExpression).Member.Name);
                }
        }

        return string.Join(",", fields);
    }

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())
};

我已经更新了@Cameron的回答,包括一些针对转换类型lambda表达式的安全检查:

PropertyInfo GetPropertyName<TSource, TProperty>(
Expression<Func<TSource, TProperty>> propertyLambda)
{
  var body = propertyLambda.Body;
  if (!(body is MemberExpression member)
    && !(body is UnaryExpression unary
      && (member = unary.Operand as MemberExpression) != null))
    throw new ArgumentException($"Expression '{propertyLambda}' " +
      "does not refer to a property.");

  if (!(member.Member is PropertyInfo propInfo))
    throw new ArgumentException($"Expression '{propertyLambda}' " +
      "refers to a field, not a property.");

  var type = typeof(TSource);
  if (!propInfo.DeclaringType.GetTypeInfo().IsAssignableFrom(type.GetTypeInfo()))
    throw new ArgumentException($"Expresion '{propertyLambda}' " + 
      "refers to a property that is not from type '{type}'.");

  return propInfo;
}