Command button to open EXCEL path dab1477 12/8/2008 11:30:02 PM How do I code a command button to open a specific EXCEL sheet or file located in a specific location each time I click the command button? Is this possible? I understand that I can output to EXCEL, but in this case, I only wish to open then print an EXCEL file. This file acts as a data collection form. It was too cumbersome to create in Access. I'm hoping to be able to print the from from the Access database switchboard that is accepting the manual data input from the file. Thank in advance for the assist.

Re: Command button to open EXCEL path "Ken Snell \(MVP\)" <kthsneisllis9[ at ]ncoomcastt.renaetl> 12/9/2008 3:34:57 AM Private Sub CommandButton_Click() Application.FollowHyperlink "C:\MyFolder\MyFile.xls" End Sub Above example will open the MyFile.xls file located in the C:\MyFolder\ folder. You can replace the string with a variable or constant that contains the full path and file name. -- Ken Snell <MS ACCESS MVP> http://www.accessmvp.com/KDSnell/ "dab1477" <dab1477[ at ]discussions.microsoft.com> wrote in message news:DB7E0388-9950-42FB-923A-8417947C0276[ at ]microsoft.com... [Quoted Text] > How do I code a command button to open a specific EXCEL sheet or file > located > in a specific location each time I click the command button? Is this > possible? I understand that I can output to EXCEL, but in this case, I > only > wish to open then print an EXCEL file. This file acts as a data > collection > form. It was too cumbersome to create in Access. I'm hoping to be able to > print the from from the Access database switchboard that is accepting the > manual data input from the file. Thank in advance for the assist. >