open a Access database with VBA
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open a Access database with VBA

 
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Alexander Heide
Guest
Posted: Tue May 18, 2004 9:38 pm    Post subject: open a Access database with VBA Reply with quote
Hi Group!

If i run following code i receive message

"Error '53'"
"File not found"

But file "House3.mdb" is there. What is wrong?

My code:

Sub OpenDatabaseFile()
Dim strFile as String
Dim strDataPath as String
Dim strdefault As String
Dim RetVal

strDataPath = Application.ActiveDesignFile.Path
strdefault = InStrRev(strDataPath, "Projects")
If strdefault > 0 Then
strDataPath = Left$(strDataPath, strdefault - 1)& "Data\House3.mdb"
End If

strFile = "strDataPath"
RetVal = Shell(strFile, 1)
End Sub



Thank You!
Alexander

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AHA
Guest
Posted: Tue May 18, 2004 10:53 pm    Post subject: Re: open a Access database with VBA Reply with quote
Just an idea. Print the contents of the resulting filename ("strDataPath") to
screen, so you can verify it is pointing to where you think it should be.

Alex

"Alexander Heide" <AHeide@yandex.ru> schreef in bericht
news:29fbeeec.0405180938.7ef4e4f3@posting.google.com...
| Hi Group!
|
| If i run following code i receive message
|
| "Error '53'"
| "File not found"
|
| But file "House3.mdb" is there. What is wrong?
|
| My code:
|
| Sub OpenDatabaseFile()
| Dim strFile as String
| Dim strDataPath as String
| Dim strdefault As String
| Dim RetVal
|
| strDataPath = Application.ActiveDesignFile.Path
| strdefault = InStrRev(strDataPath, "Projects")
| If strdefault > 0 Then
| strDataPath = Left$(strDataPath, strdefault - 1)&
"Data\House3.mdb"
| End If
|
| strFile = "strDataPath"
| RetVal = Shell(strFile, 1)
| End Sub
|
|
|
| Thank You!
| Alexander
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Jon Summers
Guest
Posted: Wed May 19, 2004 7:44 am    Post subject: Re: open a Access database with VBA Reply with quote
Alexander ...

Quote:
strDataPath = Application.ActiveDesignFile.Path
strdefault = InStrRev(strDataPath, "Projects")
If strdefault > 0 Then
strDataPath = Left$(strDataPath, strdefault - 1) & "Data\House3.mdb"
End If

strFile = "strDataPath"
RetVal = Shell(strFile, 1)

1 The Shell() function attempt to run an executable
2 A .mdb file is a data file

You can't execute a data file. VBA does not perform magic: it can't run a
data file. If you want to understand the limits of Shell(), open a
command window and enter the line you want to execute. In this case:
C:\Program Files\Bentley\Program\MicroStation\Data\House3.mdb

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Alexander Heide
Guest
Posted: Wed May 19, 2004 12:57 pm    Post subject: Re: open a Access database with VBA Reply with quote
Hi "AHA"!

Thanks for answer!

Debug.Print delivers the right path!!!
That is so curiouse!
And the file is also in the right folder.

Greetings
AHeide

"AHA" <alexWANTSNOSPAMhaerens@hotmail.com> wrote in message news:<c8dlps$94e$1@reader13.wxs.nl>...
Quote:
Just an idea. Print the contents of the resulting filename ("strDataPath") to
screen, so you can verify it is pointing to where you think it should be.

Alex

"Alexander Heide" <AHeide@yandex.ru> schreef in bericht
news:29fbeeec.0405180938.7ef4e4f3@posting.google.com...
| Hi Group!
|
| If i run following code i receive message
|
| "Error '53'"
| "File not found"
|
| But file "House3.mdb" is there. What is wrong?
|
| My code:
|
| Sub OpenDatabaseFile()
| Dim strFile as String
| Dim strDataPath as String
| Dim strdefault As String
| Dim RetVal
|
| strDataPath = Application.ActiveDesignFile.Path
| strdefault = InStrRev(strDataPath, "Projects")
| If strdefault > 0 Then
| strDataPath = Left$(strDataPath, strdefault - 1)&
"Data\House3.mdb"
| End If
|
| strFile = "strDataPath"
| RetVal = Shell(strFile, 1)
| End Sub
|
|
|
| Thank You!
| Alexander
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GeoffH
Guest
Posted: Thu May 20, 2004 9:47 am    Post subject: Re: open a Access database with VBA Reply with quote
"Jon Summers" <j.summers-no@spam-computer.org> wrote in
news:opr78k3utpav4ic8@news.btconnect.com:

Quote:
Alexander ...

strDataPath = Application.ActiveDesignFile.Path
strdefault = InStrRev(strDataPath, "Projects")
If strdefault > 0 Then
strDataPath = Left$(strDataPath, strdefault - 1) &
"Data\House3.mdb"
End If

strFile = "strDataPath"
RetVal = Shell(strFile, 1)

1 The Shell() function attempt to run an executable
2 A .mdb file is a data file

You can't execute a data file. VBA does not perform magic: it can't
run a data file. If you want to understand the limits of Shell(),
open a command window and enter the line you want to execute. In
this case: C:\Program
Files\Bentley\Program\MicroStation\Data\House3.mdb

However, if there is an association between an MDB file and MS Access, the
above command you suggest *should* start up MS Access and open the
House3.mdb database, as desired.

GeoffH
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Tim Arheit
Guest
Posted: Thu May 20, 2004 8:44 pm    Post subject: Re: open a Access database with VBA Reply with quote
On Thu, 20 May 2004 05:47:36 GMT, GeoffH <hksys@c_o_m_c_a_s_t.net>
wrote:

Quote:
"Jon Summers" <j.summers-no@spam-computer.org> wrote in
news:opr78k3utpav4ic8@news.btconnect.com:

Alexander ...

strDataPath = Application.ActiveDesignFile.Path
strdefault = InStrRev(strDataPath, "Projects")
If strdefault > 0 Then
strDataPath = Left$(strDataPath, strdefault - 1) &
"Data\House3.mdb"
End If

strFile = "strDataPath"
RetVal = Shell(strFile, 1)

1 The Shell() function attempt to run an executable
2 A .mdb file is a data file

You can't execute a data file. VBA does not perform magic: it can't
run a data file. If you want to understand the limits of Shell(),
open a command window and enter the line you want to execute. In
this case: C:\Program
Files\Bentley\Program\MicroStation\Data\House3.mdb

However, if there is an association between an MDB file and MS Access, the
above command you suggest *should* start up MS Access and open the
House3.mdb database, as desired.


Yes and no. Shell() ignores any associations. Your options are to
either use the ShellExecute API (which does use associations) or to
use the command line. ie.

RetVal = Shell("cmd /c " & strFile,1)

or (under windows 95/98):
RetVal = Shell("command /c " & strFile,1)


-Tim
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GeoffH
Guest
Posted: Fri May 21, 2004 8:27 pm    Post subject: Re: open a Access database with VBA Reply with quote
tarheit@wcoil.com (Tim Arheit) wrote in news:c8in8i$1pj$0@65.17.150.25:

Quote:
On Thu, 20 May 2004 05:47:36 GMT, GeoffH <hksys@c_o_m_c_a_s_t.net
wrote:

"Jon Summers" <j.summers-no@spam-computer.org> wrote in
news:opr78k3utpav4ic8@news.btconnect.com:

Alexander ...

strDataPath = Application.ActiveDesignFile.Path
strdefault = InStrRev(strDataPath, "Projects")
If strdefault > 0 Then
strDataPath = Left$(strDataPath, strdefault - 1) &
"Data\House3.mdb"
End If

strFile = "strDataPath"
RetVal = Shell(strFile, 1)

1 The Shell() function attempt to run an executable
2 A .mdb file is a data file

You can't execute a data file. VBA does not perform magic: it can't
run a data file. If you want to understand the limits of Shell(),
open a command window and enter the line you want to execute. In
this case: C:\Program
Files\Bentley\Program\MicroStation\Data\House3.mdb

However, if there is an association between an MDB file and MS Access,
the above command you suggest *should* start up MS Access and open the
House3.mdb database, as desired.


Yes and no. Shell() ignores any associations. Your options are to
either use the ShellExecute API (which does use associations) or to
use the command line. ie.

I was referring to the "open a command window" Jon was suggesting, because
it *will* start up MS Access.

GeoffH
http://home.comcast.net/~hksys/

Quote:

RetVal = Shell("cmd /c " & strFile,1)

or (under windows 95/98):
RetVal = Shell("command /c " & strFile,1)


-Tim
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Jon Summers
Guest
Posted: Sun May 23, 2004 10:26 am    Post subject: Re: open a Access database with VBA Reply with quote
Geoff ...

Quote:
If you want to understand the limits of Shell(),
I was referring to the "open a command window" Jon was suggesting,
because it *will* start up MS Access.

Yes, it was a bad example because the Windows Explorer file association
will perform the magic. I was hoping to illustrate the difference between
running an executable and attempting to run a data file. Next time, I'll
be more careful 8-(
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