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

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;
}

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;
}

getpropertyaccess()是可用的,如果你可以引用efcore。

using Microsoft.EntityFrameworkCore.Infrastructure;

var propertyInfo = lambda.GetPropetyAccess(); //PropertyInfo
var propertyName = propertyInfo.Name;

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

这是卡梅隆提出的方法的更新。第一个参数不是必需的。

public PropertyInfo GetPropertyInfo<TSource, TProperty>(
    Expression<Func<TSource, TProperty>> propertyLambda)
{
    Type type = typeof(TSource);

    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 (type != propInfo.ReflectedType &&
        !type.IsSubclassOf(propInfo.ReflectedType))
        throw new ArgumentException(string.Format(
            "Expresion '{0}' refers to a property that is not from type {1}.",
            propertyLambda.ToString(),
            type));

    return propInfo;
}

您可以执行以下操作:

var propertyInfo = GetPropertyInfo<SomeType>(u => u.UserID);
var propertyInfo = GetPropertyInfo((SomeType u) => u.UserID);

扩展方法:

public static PropertyInfo GetPropertyInfo<TSource, TProperty>(this TSource source,
    Expression<Func<TSource, TProperty>> propertyLambda) where TSource : class
{
    return GetPropertyInfo(propertyLambda);
}

public static string NameOfProperty<TSource, TProperty>(this TSource source,
    Expression<Func<TSource, TProperty>> propertyLambda) where TSource : class
{
    PropertyInfo prodInfo = GetPropertyInfo(propertyLambda);
    return prodInfo.Name;
}

您可以:

SomeType someInstance = null;
string propName = someInstance.NameOfProperty(i => i.Length);
PropertyInfo propInfo = someInstance.GetPropertyInfo(i => i.Length);

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);
    }