Windows 7 forum


Define the default opening application depending on folders

by sunseeker2k / April 24, 2013 4:55 PM PDT


is there any way, how I can create different settings for the default program to open a certain file type, depending on which directory the file is located?
For example, let's say there's a folder "\boooks ", where I have files I want to open using SumatraPDF and then there's another folder "\papers", which I I'd like to open using Adobe Acrobat Reader.
I know that it's not difficult to do that using the context menu, but still there are stiuations where you can save a lot of time. Especially because it often happens that one instincitvely double clicks on a file for opening it. My example was just an illustration, but imagine situations where complex programs will open, which means you have to wait until it's loaded, close it, and then choose the right program to open using the context menu.

I would really appreciate it, if anyone could tell me a solution for that.

Thanks in advance!

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All Answers

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Re: default opening depending on folders
by Kees_B Forum moderator / April 24, 2013 5:06 PM PDT

That's not a feature of Windows. If you need it you have to write a program or script yourself. That seems quite easy.


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by sunseeker2k / April 24, 2013 5:16 PM PDT

Do you know what kind of scripting language would be perfect to achieve this?

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Re: scripting
by Kees_B Forum moderator / April 24, 2013 5:36 PM PDT
In reply to: Thanks!

In Windows we have VBS (Visual Basic Scripting) and Powershell. For both, you need to have some programming experience. But any first year IT student should be able to do it for you.


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by sunseeker2k / April 24, 2013 6:11 PM PDT
In reply to: Re: scripting

I used to program using C and Visual C++ a few years ago and Commodore Basic was the first programming I taught to myself, so I guess that shouldn't be the problem. Anyway, I heard about Powershell before and now that you mention it, I think it would be the easiest way to do it maybe.
Anyway thanks!

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