如何以其ClassName作为参数打开Silverlight ChildWindow?

问题描述

嗨
我需要基于定义子窗口名称或类的参数创建一种方法来打开子窗口.
此参数存储在SQL的表中，我不知道该打开哪个女巫窗口，因为它取决于用户选择.

我以为我会使用通用类ChildWindow并创建类似以下内容的内容:

Hi
I need to create way to open a child window based on a parameter that would define the child window name or class.
This parameter is stored in a table in SQL and I don''t know in advance witch child window is supposed to be open for it would depend on a user choice.

I thought I''d use generic class ChildWindow and create something like:

ChildWindow cw = new ChildWindow();
//How to direct cw to myCwClass?
cw.Show();

我有很多子窗口，使用switch()命令或嵌套if只是不切实际.

Google并没有很大的帮助！

有谁知道实现此目标的方法或任何可以在其中看到示例的Web链接?

感谢

I have a lot of child windows and it''s just not practical to use the switch() command or nested if''s.

Google hasn''t been of great help!

Does anyone know a way to achieve this or knows any web link where I can see an example?

Thanks

推荐答案

Type cwType = Type.GetType("myNamespace.myCwClass");
object cw = Activator.CreateInstance(cwType);
((ChildWindow)cw).Show();

我的5！

My 5!

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