[AccessD] Record navigation - find versus new

So after the find, the values in the found record are displayed in the
form?   If not, it t sounds like you might have AllowAdditions set to Yes,
in which case, you are trying to create a duplicate record even if you don't
intend to.  DataEntry is only in effect if AllowAdditions is turned on
IIRRC.

Charlotte Foust

On Sat, Oct 15, 2011 at 1:43 PM, Bob Gajewski <rbgajewski at roadrunner.com>wrote:

> DataEntry = No
> AllowEdits = Yes
>
> -----Original Message-----
> From: accessd-bounces at databaseadvisors.com
> [mailto:accessd-bounces at databaseadvisors.com] On Behalf Of Charlotte Foust
> Sent: Saturday, October 15, 2011 12:57 PM
> To: Access Developers discussion and problem solving
> Subject: Re: [AccessD] Record navigation - find versus new
>
> What are the Data Entry and Allow Edits properties on the form set to?
>
> Charlotte Foust
> On Sat, Oct 15, 2011 at 8:58 AM, Bob Gajewski
> <rbgajewski at roadrunner.com>wrote:
>
> > Hi All
> >
> > I am having trouble with what is probably one of the most basic
> > functions, but I just can't figure it out and would really appreciate
> > if someone could point me in the right direction.
> >
> > I have a data entry form, with one of the key fields called
> > TransactionsEntryNumber (text, length=8, required, indexed, duplicates
> > not allowed). I am trying to write the code so that when a user types
> > in an entry number, the code looks to see if it exists ... if it does,
> > go to the existing record ... if not, continue to the next field as a new
> record.
> >
> >    <code>
> >    Private Sub TransactionsEntryNumber_Exit(Cancel As Integer)
> >    If IsNull(TransactionsEntryNumber) Or TransactionsEntryNumber = ""
> Then
> >        Dim Msg As String, Response As Integer
> >        Msg = "Do you want to have an entry number assigned?"
> >        Response = MsgBox(Msg, vbYesNo + vbQuestion + vbDefaultButton1,
> > "Entry Number Assignment")
> >        If Response = vbYes Then
> >            TransactionsEntryNumber =
> > fAssignEntryNumber(TransactionsFilerCodeID.Column(1),
> > DMax("TransactionsEntryNumber", "tblTransactions"), 1)
> >        Else
> >            Cancel = True
> >            Exit Sub
> >        End If
> >    Else
> >        DoCmd.FindRecord TransactionsEntryNumber, acEntire, True,
> > acSearchAll, True, acCurrent, True
> >    End If
> >    End Sub
> >    </code>
> >
> > The 'new' part works perfectly ... probably because I'm starting the
> > user in a new record:
> >
> >    Private Sub Form_Open(Cancel As Integer)
> >    DoCmd.GoToRecord , , acNewRec
> >    End Sub
> >
> > ... it's the 'find' that keeps failing (error: unable to save; would
> > create
> > duplicate)
> >
> >    <error>
> >    Run-time error '3022':
> >
> >    The changes you requested to the table were not successful because
> > they would
> >    create duplicate values in the index, primary key, or relationship.
> > Change the data
> >    in the field or fields that contain the duplicate data, remove the
> > index, or redefine
> >    the index to permit duplicate entries and try again.
> >    </error>
> >
> > Is it as simple as trapping the error? Or should I use two controls
> > and stack them?
> >
> > Thanks, as usual
> > Bob Gajewski
> >
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>
> >
> >
> >
> >
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