Set version of Access a database opens with

[hobbes]

Hi,

We are going through a computer/software upgrade in our office and I am in charge of making sure all of the databases function properly. Due to processing requirements, Access 97 and Access 2003 will be installed on the computers affected by these databases. How do I program a specific database to only open (or at least default open) with Access 97? (Access 2003 will be the primary .mdb program.)

Thank you very much for your help and input!

As a sidenote: I will be converting all of the databases eventually, but can't do it during the conversion.

Hobbes

 

[Crilen007]

How do I program a specific database to only open (or at least default open) with Access 97? (Access 2003 will be the primary .mdb program.)


So, the default is 97, but 2003 is the primary?

Shouldnt the default be 2003 and only sometimes it open with the 97?

Just make shortcuts or run commands from VB that run the database with access.

"C:\Access2003\MSACCESS.exe" file2k3.mdb

"C:\Access97\MSACCESS.exe" file97.mdb


Map these to the proper location of course.
I think that will work just fine.

 

[ghudson]

The users PC [registry] will remember which version of Access was used last and that version will be the default the next time a db is opened from Windows Explorer.

You can create shortcuts for each Access file but that could be a large task depending on how many *.mdb files the users have. Advanced warning of how their computer remembers the last version of Access used will also help.

Here is how the Target: field of a shortcut should look to open Access 2003 for a specific .mdb [non secured] file.

"C:\Program Files\Microsoft Office\Office11\MSACCESS.EXE" "X:\DB1.mdb"

Here is how the Target: field of a shortcut should look to open Access 97 for a specific .mdb [non secured] file.

"C:\Program Files\Microsoft Office\Office8\MSACCESS.EXE" "X:\DB2.mdb"