当lambda表达式引用实际实例本身时,如何使以下方法起作用?
例如代替
x => x.Name
表达式是
x => x
因此,如果我有某个类“ Car”,我可以返回字符串“ Car”,而不是只能对其属性进行操作(例如Car.Colour)
方法:
public static string GetMemberName(Expression expression)
{
if (expression is LambdaExpression)
expression = ((LambdaExpression)expression).Body;
if (expression is MemberExpression)
{
var memberExpression = (MemberExpression)expression;
if (memberExpression.Expression.NodeType ==
ExpressionType.MemberAccess)
{
return GetMemberName(memberExpression.Expression)
+ "."
+ memberExpression.Member.Name;
}
return memberExpression.Member.Name;
}
if (expression is UnaryExpression)
{
var unaryExpression = (UnaryExpression)expression;
if (unaryExpression.NodeType != ExpressionType.Convert)
throw new Exception(string.Format(
"Cannot interpret member from {0}",
expression));
return GetMemberName(unaryExpression.Operand);
}
throw new Exception(string.Format(
"Could not determine member from {0}",
expression));
}
即我想要类似的东西:
if (expression is SomeExpressionThatReturnsAnInstance)
{
return (name of type of instance);
}
我将其追溯到表达式的构造。它实际上不包含任何实例信息,也没有获取类型名称的方法。
static Expression<Func<object, object>> thisObject = x => x;
因此,无法从甚至没有类型(对象除外)的表达式派生类型名称。
用于生成返回属性名称的表达式的方法:
LambdaExpression BuildExpression(Type rootType, string propertyName)
{
try
{
var properties = propertyName.Split('.');
ParameterExpression arg = Expression.Parameter(rootType, "x");
Expression expr = arg;
foreach (string property in properties)
{
PropertyInfo propertyInfo = rootType.GetProperty(property);
if (propertyInfo == null)
return null;
expr = Expression.Property(expr, propertyInfo);
rootType = propertyInfo.PropertyType;
}
return Expression.Lambda(expr, arg);
}
catch (System.Exception ex)
{
return null;
}
}
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