VBA Code for Open another Access Application with command Button

[hnhpak]

I have following code to open another access database with a command button:



Application.FollowHyperlink "D:\Database\database.accdb"

I want to save the path in a table and want to execute the command with taking path and file name from table "PathReferences"
Say for example have table with fields

ID:
ApplicationName
ApplicationPath

According to above the ApplicationName is database.accdb
and the ApplicationPath is "D:\Database\"

I want to make a code something like this:

Application.FollowHyperlink " [PathReference].[ApplicationPath]"&"[PathReference].[ApplicationName]"

Do any one can help to make the code for this purpose?

Further to that:


I have a Navigation form in an Access database.
When I click on a button I want to open a particular form in a (different) particular Access Database.
Is this possible with some VBA code?

Thanks in advance

Haseeb

[June7]

Review https://www.accessforums.net/forms/h...ing-52443.html

Do not put variables within quote marks. Cannot reference tables directly.

Have the form in other db set as the form to open by default when db opens.

[hnhpak]

thanks June the matter solved.