Command-Line Startup

[Guest]

I was looking for the syntax, of the command-line, to startup Access. This
particular db is Access97.

Thanks, in advance.

Sharkbyte

 

[aaron.kempf]

command line?

c:\program files\microsoft office\msaccess.exe "c:\myAccessApp.mdb" /X
myMacroName

 

[Guest]

Hi.

If you just want to start Access, then the path and file name of the
executable located on your computer is appropriate:

"C:\Program Files\Microsoft Office\OFFICE\MSACCESS.EXE"

If there are any illegal characters contained in the path or file name, then
ensure that this is enclosed in double quotes.

However, if you want to open a specific database file, with or without
command-line switches, then the syntax requires the paths of the executable
and the database file, and would be like the following (watch out for word
wrap, as this is all one line):

"C:\Program Files\Microsoft Office\OFFICE\MSACCESS.EXE"
"C:\MyDir\MyDB.mdb" /excl

.. . . which opens the MyDB.mdb file exclusively. For a list of all of the
command-line switches for Access 97, please see the following Web page:

http://support.microsoft.com/kb/105128/

HTH.
Gunny

See http://www.QBuilt.com for all your database needs.
See http://www.Access.QBuilt.com for Microsoft Access tips and tutorials.
http://www.Access.QBuilt.com/html/expert_contributors2.html for contact info.

 

[Guest]

I have a question on the next step to this, which is to be able to open a
particular object (in my case a particular table) in the database from the
command line.

What I'm really trying to do is open a database and table from C# as a new
process (i.e., launch MSACCESS and have it open the desired table in the
database) , but I'm figuring if I know that command line structure I'll have
my answer.

TIA,

DanK