app.path i dunno whats wrong HELP newbie here?

AM USING THIS for
now----------------------------------------------------------------------------------------
Public Sub OpenDBConnection()
Set conn = New ADODB.Connection
Set rec = New ADODB.Recordset
strconek = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=C:\Program
Files\balik sa dati\data\MFMS.mdb;Persist Security Info=False"
conn.CursorLocation = adUseClient
conn.Open strconek
End Sub
----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
ICANT USE THIS
Public Sub OpenDBConnection()
Set conn = New ADODB.Connection
Set rec = New ADODB.Recordset
strconek = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" &
app.path & "\data\MFMS.mdb;Persist Security Info=False"
conn.CursorLocation = adUseClient
conn.Open strconek
End Sub
-----------------------------------------------------------------------------------------
the error is:

If you're trying to use this from within Access, there is no App object in
Access, so App.Path will (naturally) fail.

Assuming you're using Access 2000 or newer, use CurrentApplication.Path
instead.

BTW, note that both App.Path and CurrentApplication.Path have an annoying
little "feature". If the file is in the root of a drive, what's returned
will have a terminating slash: E:\, not simply E:. That means that something
like

strconek = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" &
App.Path & "\data\MFMS.mdb;Persist Security Info=False"

or

strconek = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" &
CurrentApplication.Path & "\data\MFMS.mdb;Persist Security Info=False"

will fail, as you'll end up with two slashes in a row.

To get around this, try:

Dim strPath As String

strPath = CurrentApplication.Path
If Right$(strPath, 1) "\" Then
strPath = strPath & "\"
End If
strconek = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" &
strPath & "\data\MFMS.mdb;Persist Security Info=False"





--
Doug Steele, Microsoft Access MVP
http://I.Am/DougSteele
(no e-mails, please!)


wrote in message
ups.com...
AM USING THIS for
now----------------------------------------------------------------------------------------
Public Sub OpenDBConnection()
Set conn = New ADODB.Connection
Set rec = New ADODB.Recordset
strconek = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=C:\Program
Files\balik sa dati\data\MFMS.mdb;Persist Security Info=False"
conn.CursorLocation = adUseClient
conn.Open strconek
End Sub
----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
ICANT USE THIS
Public Sub OpenDBConnection()
Set conn = New ADODB.Connection
Set rec = New ADODB.Recordset
strconek = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" &
app.path & "\data\MFMS.mdb;Persist Security Info=False"
conn.CursorLocation = adUseClient
conn.Open strconek
End Sub
-----------------------------------------------------------------------------------------
the error is:

"Douglas J. Steele" wrote in
message

Assuming you're using Access 2000 or newer, use
CurrentApplication.Path instead.

CurrentProject.Path, I think Doug means.

--
Dirk Goldgar, MS Access MVP
www.datagnostics.com

(please reply to the newsgroup)

"Dirk Goldgar" wrote in message
...
"Douglas J. Steele" wrote in
message

Assuming you're using Access 2000 or newer, use
CurrentApplication.Path instead.

CurrentProject.Path, I think Doug means.

Thanks, Dirk. I've only got Access 97 installed on this machine, so I was
going from memory.

For the record, in Access 97 or earlier, you'd use

Left$(CurrentDb.Name, Len(CurrentDb.Name) - Len(Dir$(CurrentDb.Name)))

--
Doug Steele, Microsoft Access MVP
http://I.Am/DougSteele
(no e-mails, please!)