获取打开目录的路径 [英] Get the path of opened directories

问题描述

我需要获取用户打开的所有目录的路径，以便获取用户可能拥有的所有文件和子目录，以打开并扫描它们.有人可以帮忙吗?

非常感谢.

Hi,

I need to get the path of all directories which are opened by the user so that I can get all the files and subdirectories which the user may wnat to open and scan them. Can anybody help?

Thanks alot.

推荐答案

http://www.cplusplus.com/ [ ]

根据您要执行的操作，shell扩展 [我的答案 [ ^ ]到几天前的类似问题. Windows Explorer和Internet Explorer在很多方面都非常相似，此代码将找到所有Windows Explorer和Internet Explorer窗口的URL.如果URL以file://开头，则它是Windows资源管理器窗口.

如我所说，也许有更好的方法，但是由于没有人回答，我想我会给你的.

Depending on what you are wanting to do, a shell extension[^] may be the best thing. This will integrate your code directly into Windows Explorer, and allow you to do things such as add a toolbar or a column to the file view.

If you just want to know what they are looking at, without changing the Windows Explorer experience, then this would not be a good way.

This may or may not be the best way, but check out my answer[^] to a similar question a few days ago. Windows Explorer and Internet Explorer are very similar in a lot of respects, this code will find the URL of all Windows Explorer and Internet Explorer windows. If the URL starts with file:// then it is a Windows Explorer window.

As I said, there may be a better way, but since no one has answered yet, I thought I''d give you this.

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