G
Guest
Can someone tell me how to write a vba code to open an Access application
(MyApp2_rev_n.mdb) from the same folder as my current project (MyApp1.mdb) as
shown in the example below. MyApp2_rev_n.mdb is rev controlled and the name
is stored on zDbName table in MyApp1.mdb. Thanks for the help.
current project path
c:\folder1\folder2\myProjYYYYMMDD\MyApp1.mdb
database I want to open
c:\folder1\folder2\myProjYYYYMMDD\MyApp2_rev_n.mdb
(MyApp2_rev_n.mdb) from the same folder as my current project (MyApp1.mdb) as
shown in the example below. MyApp2_rev_n.mdb is rev controlled and the name
is stored on zDbName table in MyApp1.mdb. Thanks for the help.
current project path
c:\folder1\folder2\myProjYYYYMMDD\MyApp1.mdb
database I want to open
c:\folder1\folder2\myProjYYYYMMDD\MyApp2_rev_n.mdb