Error 2105 - when opening form via button

[akika]

hi,
Ive added below code in form load for 'FormEntry'.
however, the issue that am getting is that i have another continuous form 'FormSearchBy' which contains a button.
This button 'opens selected record', open the formEntry(condition is ID = ID).

But this i have below code in formEntry.. I'm getting error 2105, you cannot go to the specified record.

Private Sub Form_Load()
DoCmd.GoToRecord , , acNewRec


End Sub

pls help.

[Minty]

You can't specify the ID of a new record if that is the primary key.

Your code is saying
Open the form on "This ID"

Then your form code is saying
Open on a new record

You can't do both.

[akika]

ID is indeed the primary key.. wht would u suggest?
is it possible via the macro or any code to bypass this error?

in the Microsoft Visual Basic dialog, when click End button instead of debug.. the form is opened and record displayed.
pls advise.

[Minty]

What are you trying to achieve?
If it's to simply open the form on the record with the ID then remove the on load code from the form.

[Missinglinq]

If you want to open the secondary Form to a given Record (if it exists in the secondary Form) but open it to a New Record (if it doesn't exist in the secondary Form) and assign it the ID of the primary Form's Record...here's some boilerplate code I did ages ago for a poster to do just that:

In the Primary Form:

Private Sub Go2SecondaryForm_Click()

DoCmd.RunCommand acCmdSaveRecord

 If Nz(Me.RecordID,"") <> "" Then
  DoCmd.OpenForm "Secondary Form", , , , , , Me.RecordID
 Else
  MsgBox "A RecordID Must Be Entered First!"
 End If

End Sub

In the Secondary Form:

Private Sub Form_Load()

Dim rst As Recordset

If Nz(Me.OpenArgs,"") <> "" Then

 Set rst = Me.RecordsetClone
 
 rst.FindFirst "[SecondaryRecordID] = " & Me.OpenArgs 
 
   If Not rst.NoMatch Then
      Me.Bookmark = rst.Bookmark
   Else
    DoCmd.GoToRecord , , acNewRec
    Me.SecondaryRecordID = Me.OpenArgs
   End If

rst.Close

Set rst = Nothing

End If

End Sub

The above code assumes that RecordID is Numeric. If it is actually defined Text, replace the line

rst.FindFirst "[SecondaryRecordID] = " & Me.OpenArgs

with

rst.FindFirst "[SecondaryRecordID] = '" & Me.OpenArgs & "'"

You'll also need to replace Go2SecondaryForm with the actual name of your Command Button and SecondaryForm.

Also note that, as already suggested, the SecondaryRecordID, which acts as a Foreign Key, in this scenario, cannot be an AutoNumber Field, but must be defined as a Datatype that matches the Datatype of the RecordID, from the primary Form.

Linq ;0)>