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

假设(TModel作为类)

Expression<Func<TModel, TValue>> expression

检索属性的名称

expression.GetPropertyInfo().Name;

扩展函数:

public static PropertyInfo GetPropertyInfo<TType, TReturn>(this Expression<Func<TType, TReturn>> property)
{
  LambdaExpression lambda = property;
  var memberExpression = lambda.Body is UnaryExpression expression
      ? (MemberExpression)expression.Operand
      : (MemberExpression)lambda.Body;

  return (PropertyInfo)memberExpression.Member;
}

其他回答

假设(TModel作为类)

Expression<Func<TModel, TValue>> expression

检索属性的名称

expression.GetPropertyInfo().Name;

扩展函数:

public static PropertyInfo GetPropertyInfo<TType, TReturn>(this Expression<Func<TType, TReturn>> property)
{
  LambdaExpression lambda = property;
  var memberExpression = lambda.Body is UnaryExpression expression
      ? (MemberExpression)expression.Operand
      : (MemberExpression)lambda.Body;

  return (PropertyInfo)memberExpression.Member;
}

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。

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。

从。net 4.0开始,你可以使用ExpressionVisitor来查找属性:

class ExprVisitor : ExpressionVisitor {
    public bool IsFound { get; private set; }
    public string MemberName { get; private set; }
    public Type MemberType { get; private set; }
    protected override Expression VisitMember(MemberExpression node) {
        if (!IsFound && node.Member.MemberType == MemberTypes.Property) {
            IsFound = true;
            MemberName = node.Member.Name;
            MemberType = node.Type;
        }
        return base.VisitMember(node);
    }
}

下面是如何使用这个访问者:

var visitor = new ExprVisitor();
visitor.Visit(expr);
if (visitor.IsFound) {
    Console.WriteLine("First property in the expression tree: Name={0}, Type={1}", visitor.MemberName, visitor.MemberType.FullName);
} else {
    Console.WriteLine("No properties found.");
}

这是一个通用的实现,用于获取struct/class/interface/delegate/array的字段/属性/索引器/方法/扩展方法/委托的字符串名称。我已经测试了静态/实例和非泛型/泛型变体的组合。

//involves recursion
public static string GetMemberName(this LambdaExpression memberSelector)
{
    Func<Expression, string> nameSelector = null;  //recursive func
    nameSelector = e => //or move the entire thing to a separate recursive method
    {
        switch (e.NodeType)
        {
            case ExpressionType.Parameter:
                return ((ParameterExpression)e).Name;
            case ExpressionType.MemberAccess:
                return ((MemberExpression)e).Member.Name;
            case ExpressionType.Call:
                return ((MethodCallExpression)e).Method.Name;
            case ExpressionType.Convert:
            case ExpressionType.ConvertChecked:
                return nameSelector(((UnaryExpression)e).Operand);
            case ExpressionType.Invoke:
                return nameSelector(((InvocationExpression)e).Expression);
            case ExpressionType.ArrayLength:
                return "Length";
            default:
                throw new Exception("not a proper member selector");
        }
    };

    return nameSelector(memberSelector.Body);
}

这个东西也可以写在一个简单的while循环中:

//iteration based
public static string GetMemberName(this LambdaExpression memberSelector)
{
    var currentExpression = memberSelector.Body;

    while (true)
    {
        switch (currentExpression.NodeType)
        {
            case ExpressionType.Parameter:
                return ((ParameterExpression)currentExpression).Name;
            case ExpressionType.MemberAccess:
                return ((MemberExpression)currentExpression).Member.Name;
            case ExpressionType.Call:
                return ((MethodCallExpression)currentExpression).Method.Name;
            case ExpressionType.Convert:
            case ExpressionType.ConvertChecked:
                currentExpression = ((UnaryExpression)currentExpression).Operand;
                break;
            case ExpressionType.Invoke:
                currentExpression = ((InvocationExpression)currentExpression).Expression;
                break;
            case ExpressionType.ArrayLength:
                return "Length";
            default:
                throw new Exception("not a proper member selector");
        }
    }
}

我喜欢递归方法,尽管第二种方法可能更容易阅读。我们可以这样称呼它:

someExpr = x => x.Property.ExtensionMethod()[0]; //or
someExpr = x => Static.Method().Field; //or
someExpr = x => VoidMethod(); //or
someExpr = () => localVariable; //or
someExpr = x => x; //or
someExpr = x => (Type)x; //or
someExpr = () => Array[0].Delegate(null); //etc

string name = someExpr.GetMemberName();

打印最后一个成员。

注意:

对于像a.b.c.这样的链式表达式,将返回“C”。 这并不适用于const,数组索引器或枚举(不可能涵盖所有情况)。