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。
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
