我想列出所有打开Windows资源管理器窗口，在一个列表框的活动路径。该应用程序应该刷新列表框如果在浏览器窗口中导航到不同的路径。

I want to list all open Windows Explorer windows with their active path in a ListBox. The app should refresh the ListBox if the explorer window is navigated to a different path.

有关如两块Explorer窗口是打开的。一个是导航到 C：\ WINDOWS ，另一种是导航到 D：\东西。当应用程序运行它，它增加了 C：\ WINDOWS 和 D：\东西到列表框。然后，用户导航到不同的文件夹中的打开资源管理器窗口，比如一C：\ WINDOWS \ SYSTEM32 。那么应用程序应刷新列表框和列表 C：\ WINDOWS \ SYSTEM32 和 D： \东西代替。

For e.g. two explorer windows are open. One is navigated to C:\Windows and the other is navigated to D:\Stuff. When the app it run, it adds C:\Windows and D:\Stuff to the ListBox. Then, the user navigates to a different folder in one of the open explorer windows like C:\Windows\system32. The app should then refresh the ListBox and list C:\Windows\system32 and D:\Stuff instead.

我没有对如何做到这一点的任何想法。任何指针将AP preciated。

I don't have any ideas on how to do this. Any pointers would be appreciated.

推荐答案

在这里你可以找到一个例子如何访问 WindowsExplorer 的路径和 InternetExplorer的： HTTP：//omega$c$cr.com/？ P = 63

Here you can find an example how to access the paths in WindowsExplorer and InternetExplorer : http://omegacoder.com/?p=63

怎么样才能够收到关于该用户导航到不同的路径的事实的通知，也没有办法，我所知道的，真的。

What about to be able to receive a notification about the fact that user navigated to different path, there is no way that I'm aware of, honestly.

让我脑海的第一个解决方案，就是使用定时器并检查每一个勾号。

So the first solution that comes to my mind, is use a Timer and check on every tick.