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From: Daniel Pineault on 4 Jun 2010 16:44
Try

username = Me!Text23 & vbNullString

Have you setup your form to have a control name Text23 which has the user's
name?!
--
Hope this helps,

Daniel Pineault
http://www.cardaconsultants.com/
For Access Tips and Examples: http://www.devhut.net
Please rate this post using the vote buttons if it was helpful.



"Bob H" wrote:

> Thanks for the coding but now I get runtime error 2465 at the line:
> username = Me!Text23& vbNullString.
>
> Thanks
>
> On 04/06/2010 20:21, Douglas J. Steele wrote:
> > There is no App object in Access.
> >
> > Private Sub CreateAcc_Click()
> > Dim intFile As Integer
> > Dim username As String
> >
> > username = Me!Text23& vbNullString
> > If Len(username) = 0 Then
> > MsgBox "Enter a Username !"
> > Else
> > If Len(Me!Text25& vbNullString) = 0 Then
> > MsgBox "Enter a Password !"
> > Else
> > intFile = FreeFile()
> > Open CurrentProject.Path& "\Accounts\Users\"& username& ".txt" For
> > Output As #intFile
> > Print #intFile, Me!Text23
> > Print #intFile, Me!Text25
> > Close #intFile
> > End If
> > End If
> >
> > End Sub
> >
> >
>
> .
>
From: Bob H on 4 Jun 2010 18:17
On 04/06/2010 21:44, Daniel Pineault wrote:
> Try
>
> username = Me!Text23& vbNullString
>
> Have you setup your form to have a control name Text23 which has the user's
> name?!

If you mean something like Text23.Username, then no I haven't.
But as I am attempting to create an account, the username could be anything.
Or should I have pre defined usernames and passwords.

Thanks
From: Bob H on 4 Jun 2010 18:23
On 04/06/2010 20:50, Daniel Pineault wrote:
> To pull the path of your currently open database you'd use:
>
> Application.CurrentProject.Path

Ok thanks, but where does this line fit in with the other lines.
Do I have to actually create the path first then put it in the line.

Thanks
From: Bob H on 5 Jun 2010 11:27
On 04/06/2010 23:23, Bob H wrote:
> On 04/06/2010 20:50, Daniel Pineault wrote:
>> To pull the path of your currently open database you'd use:
>>
>> Application.CurrentProject.Path
>
> Ok thanks, but where does this line fit in with the other lines.
> Do I have to actually create the path first then put it in the line.
>
> Thanks

Update:
I have used this line to create a text file with given information:

Open "C:\Accounts\Users\" + username + ".txt" For Output As #1
Print #1, Text23
Print #1, Text25

But now on another cmd button I want access to read that said
information in that text file, and grant access.
With this line:

Open "C:\Accounts\Users\" + username + ".txt" For Output As #2
Input #2, openfile
If username.Text = openfile Then
Input #2, datafile

Access throws up a runtime error 54 'Bad File mode' at Input #2 ,
openfile openfile

Also the information in the previously created text file has been deleted.

Thanks
From: Douglas J. Steele on 5 Jun 2010 15:38
"Bob H" <bob(a)despammer.com> wrote in message
news:edudnd0UkMVw8ZfRnZ2dnUVZ8kidnZ2d(a)giganews.com...
>
> Update:
> I have used this line to create a text file with given information:
>
> Open "C:\Accounts\Users\" + username + ".txt" For Output As #1
> Print #1, Text23
> Print #1, Text25

I guess I should have highlighted in my previous reply that you should never
refer to file handles by hard-coded numbers. While it may work fine for you,
if other applications are reading or writing to files at the same time, you
can run into problems. You should ALWAYS use the FreeFile function, even if
you're positive no other applications will ever be running concurrent with
yours.

Dim intFile As Integer

intFile = FreeFile()
Open "C:\Accounts\Users\" & username & ".txt" For Output As #intFile
Print #intFile, Text23
Print #intFile, Text25

You should also use & for concatenating text, not +.

> But now on another cmd button I want access to read that said information
> in that text file, and grant access.
> With this line:
>
> Open "C:\Accounts\Users\" + username + ".txt" For Output As #2
> Input #2, openfile
> If username.Text = openfile Then
> Input #2, datafile

Try using

Open "C:\Accounts\Users\" & username & ".txt" For Output Shared As #2

> Access throws up a runtime error 54 'Bad File mode' at Input #2 , openfile
> openfile
>
> Also the information in the previously created text file has been deleted.

As you've found, opening a file For Output deletes the previous file if it
exists. use For Append. That will create the file if it doesn't already
exist, and append to the file if it does.

--
Doug Steele, Microsoft Access MVP
http://www.AccessMVP.com/djsteele
Co-author: "Access 2010 Solutions", published by Wiley
(no e-mails, please!)




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