Question

我有以下代码：

List<MultiServiceRequestMember> _memberList = new List<MultiServiceRequestMember>();
var type = Type.GetType(svc.NotificationClassName); <- this is a string of the class name.
MultiServiceRequestMember newMember = (MultiServiceRequestMember)Activator.CreateInstance(type);

_memberList.add(newMember);

MultServiceRequestMember是基本类型，我想为特定于type的属性赋值。我的问题是：如何转换newMember以输入和访问其属性？

Answer 1

如何投射newMember以输入并访问它的属性？

你不能强制它，因为你在编译时不知道具体的类型。如果你这样做，你首先不需要反思！

您还必须通过反射设置属性：

// TODO: Checking that you managed to get the property, that's it's writable etc.
var property = type.GetProperty("PropertyName");
property.SetValue(newMember, "new value", null);

Answer 2

您必须将代码更改为如下所示：

List<MultiServiceRequestMember> _memberList = new List<MultiServiceRequestMember>();
var type = Type.GetType(svc.NotificationClassName);
MultiServiceRequestMember newMember = null;
if (type == typeof(MultiServiceRequestMemberA))
{
    newMember = new MultiServiceRequestMemberA();
    //set specific properties
}
else if (type == typeof(MultiServiceRequestMemberB)) //etc.
{
    //...
}
else
{
    //throw or some default
}

_memberList.add(newMember);

然而，它看起来像代码味道。我想你是在尝试基于其他一些对象初始化一个对象（让我们称之为NotificationInfo）。然后代替看起来像这样的代码：

if (type == typeof(MultiServiceRequestMemberA))
{
    newMember = new MultiServiceRequestMemberA();
    newMember.A = notificationInfo.A;
}

也许应该考虑以下设计：

class MultiServiceRequestMember
{
    public virtual void Initialize(NotificationInfo notificationInfo) //or abstract if you wish
    {
    }
}

class MultiServiceRequestMemberA : MultiServiceRequestMember
{
    public override void Initialize(NotificationInfo notificationInfo)
    {
        base.Initialize(notificationInfo);
        this.A = notificationInfo.A;
    }
}

然后您就可以保留以前的代码，只需调用Initialize。

访问由反射创建的类型的属性

2 个答案: