我有一个问题一直困扰着我一段时间,我找不到答案。
我需要获取Lambda表达式中引用的属性的名称。我会将lambda表达式提供给一个返回字符串的方法。例如,如果我有:
x => x.WeirdPropertyName
然后该方法将返回:
"WeirdPropertyName"
我已经读过它可以用表达式树来完成,但答案已经没有了。
感谢您的帮助
答案 0 :(得分:28)
你走了:
string GetPropertyName<T>(Expression<Func<T>> property)
{
var propertyInfo = (property.Body as MemberExpression).Member as PropertyInfo;
if (propertyInfo == null)
{
throw new ArgumentException("The lambda expression 'property' should point to a valid Property");
}
return propertyInfo.Name;
}
答案 1 :(得分:6)
我有一个非常全面的答案here。
除了处理x => x.WeirdPropertyName
之类的表达式外,它还可以处理“扩展”表达式,例如x => x.WeirdMember.WeirdPropertyName
。
以下是该答案的代码:
// code adjusted to prevent horizontal overflow
static string GetFullPropertyName<T, TProperty>
(Expression<Func<T, TProperty>> exp)
{
MemberExpression memberExp;
if (!TryFindMemberExpression(exp.Body, out memberExp))
return string.Empty;
var memberNames = new Stack<string>();
do
{
memberNames.Push(memberExp.Member.Name);
}
while (TryFindMemberExpression(memberExp.Expression, out memberExp));
return string.Join(".", memberNames.ToArray());
}
// code adjusted to prevent horizontal overflow
private static bool TryFindMemberExpression
(Expression exp, out MemberExpression memberExp)
{
memberExp = exp as MemberExpression;
if (memberExp != null)
{
// heyo! that was easy enough
return true;
}
// if the compiler created an automatic conversion,
// it'll look something like...
// obj => Convert(obj.Property) [e.g., int -> object]
// OR:
// obj => ConvertChecked(obj.Property) [e.g., int -> long]
// ...which are the cases checked in IsConversion
if (IsConversion(exp) && exp is UnaryExpression)
{
memberExp = ((UnaryExpression)exp).Operand as MemberExpression;
if (memberExp != null)
{
return true;
}
}
return false;
}
private static bool IsConversion(Expression exp)
{
return (
exp.NodeType == ExpressionType.Convert ||
exp.NodeType == ExpressionType.ConvertChecked
);
}
答案 2 :(得分:-4)
我知道获取属性的字符串名称的唯一方法是通过反射。