How to retrieve the openfiledialog opening path? C#

Question

I have a openfiledialog in my application. I do not set the initial directory and user can navigate to open file and import it to the application (for example imports excel file from C:\ Test location). Next time, when user wants to open another file, system (by default, since initial directory is not set) remember the previous location and openfiledialog opens in that directory (C:\ Test). Is there anyway to retrieve the path that openfiledialog opens? I am sorry if I couldn't explain my problem well and thanks for the help.

possible duplicate of [OpenFileDialog default path](http://stackoverflow.com/questions/9980262/openfiledialog-default-path) — Hans Passant, Sep 29 '12 at 21:56

score 2 · Accepted Answer · answered Sep 29 '12 at 22:33

2

Registry is hard to handle, I would suggest that you use Application Properties instead.

answered Sep 29 '12 at 22:33

Aleksandar Toplek

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How to retrieve the openfiledialog opening path? C#

1 Answers1