Use The Application.StartUp Property To Connect To The Database?

Jul 9, 2009

I added an Access database to my project and properties are: Build Action: Content Copy to Output Directory:Copy Always Full Path = directory where program code is.Within the program, I use the Application.StartUp property to connect to the database.My problem is that the program opens with the database in the root directory, but all changes are only made to the database in the bin/Debug folder. So the next time I run the program (through Visual Studio), it again opens up the database that didn't have the change smade to it.

I believe I want the database in the bin/debug folder because that's where the application is.I'm not an ADO.net maven. Is there anyone who can explain what i'm doing wrong and how I could correct it. If I run the program directly from the bin/debug folder, it works correctly.

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[Code]...

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[code].....

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PHP
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[Code]...

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[Code]....

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