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;
}
我发现一些建议的答案钻到MemberExpression/UnaryExpression不捕获嵌套/子属性。
o =>。Thing2返回Thing1而不是Thing1.Thing2。
如果您试图使用EntityFramework DbSet.Include(…)，这种区别就很重要。
我发现只要解析Expression.ToString()就可以了，而且速度相对较快。我将它与UnaryExpression版本进行了比较，甚至从成员/UnaryExpression中获得ToString，以查看是否更快，但差异可以忽略不计。如果这是个糟糕的主意，请纠正我。
可拓法
/// <summary>
/// Given an expression, extract the listed property name; similar to reflection but with familiar LINQ+lambdas. Technique @via https://stackoverflow.com/a/16647343/1037948
/// </summary>
/// <remarks>Cheats and uses the tostring output -- Should consult performance differences</remarks>
/// <typeparam name="TModel">the model type to extract property names</typeparam>
/// <typeparam name="TValue">the value type of the expected property</typeparam>
/// <param name="propertySelector">expression that just selects a model property to be turned into a string</param>
/// <param name="delimiter">Expression toString delimiter to split from lambda param</param>
/// <param name="endTrim">Sometimes the Expression toString contains a method call, something like "Convert(x)", so we need to strip the closing part from the end</param>
/// <returns>indicated property name</returns>
public static string GetPropertyName<TModel, TValue>(this Expression<Func<TModel, TValue>> propertySelector, char delimiter = '.', char endTrim = ')') {
var asString = propertySelector.ToString(); // gives you: "o => o.Whatever"
var firstDelim = asString.IndexOf(delimiter); // make sure there is a beginning property indicator; the "." in "o.Whatever" -- this may not be necessary?
return firstDelim < 0
? asString
: asString.Substring(firstDelim+1).TrimEnd(endTrim);
}//-- fn GetPropertyNameExtended
(检查分隔符甚至可能是多余的)
演示 （LinqPad）
演示+比较代码—https://gist.github.com/zaus/6992590
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。
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;
}
