C# OpenFileDialog 存储路径
在我的应用程序中，我有两个地方需要用户选择文件。在这两种情况下，文件位于不同的目录中，但它们在运行之间通常是相同的。
OpenFileDialog 似乎默认为最后使用的目录，但这对我来说毫无用处，因为它几乎总是错误的文件夹，并且我最终在两个文件夹之间交替。我想要的是以某种方式让第一个对话框记住上次打开时使用的路径，第二个对话框也记住它自己的路径。
示例：路径 A 是 C:\foo\bar\something\x.dll，路径 B 是 C:\foo\baz\whatever\y.xml
对话框 a 打开，我选择 A，然后打开对话框 b（默认为 A ）并且我必须向后导航到 B。 当我再次打开应用程序时，对话框 a 打开（默认为 B），我必须再次导航回 A。
我想通过单独记住路径来避免所有这些额外的导航。有没有好的方法可以做到这一点？
In my application I have two places where the user needs to select a file. In both cases, the files are in different directories, but they are generally the same between runs.
The OpenFileDialog appears to be defaulting to the last directory used, but this is useless to me since it is almost always the wrong folder, and I end up alternating between the two folders. What I would like is to somehow have the first dialog remember the path that was used the last time it was opened, and the second to remember its own path as well.
Example: Path A is C:\foo\bar\something\x.dll, Path B is C:\foo\baz\whatever\y.xml
Dialog a opens and I select A, then later dialog b opens(defaulted to A) and I have to navigate back and up to B.
When I open the app again dialog a opens (defaulted to B) and I have to navigate back up to A again.
I would like to avoid all of this extra navigation by remembering the paths separately. Is there a good way to do that?
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打开每个对话框时，只需设置对话框的 InitialDirectory 属性到该对话框上次使用的文件夹。
当然，这需要保存每个对话框的目录，但它将提供您想要实现的行为。
When you open each dialog, just set the dialog's InitialDirectory property to the folder that was last used for that dialog.
Granted, this will require saving the directory for each dialog, but it will provide the behavior you are looking to achieve.