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Old June 25th, 2009, 11:38 AM
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Default Open an existing record in access 2

Hello everyone,
I have a form (frmgood) and its subform (formparts2) . I designed another form (frmrecordnumber) that has a textbox (where the record number is entered by the user), and a findrecord button ( once is clicked it would open frmgood with the formparts2 at the existing record entered by the user). The record textbox on the frmgood is named IDA. The code i tried is shown below: ( i know is totally wrong)

Private Sub findrecord_Click()
Dim IntNum As Integer
Dim intID As Integer
Set intID = Forms!frmgood!frmParts2!IDA (Once run it keeps giving me 2450 error saying that the frmgood doesn't exist, eventhough it does)
Set IntNum = Me.text1
If intID = IntNum Then ( I dont know what to write to make it go to the existing record entered by the user)
MsgBox "The record does not exist"
End If
End Sub
I'm , i have been trying for the last 3 days and nothing. I would really appreciate any help. Thank you. I'm using access 2002
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Old June 26th, 2009, 12:15 PM
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First: Declare those integers as Long instead since that will eventually cause problems if those are autonumber fields.

Second: When you reference "Set intID = Forms!frmgood!frmParts2!IDA ", that form is not open yet. To Access, it doesn't exist. What it means is, it is not opened. You can only reference a control on a form that is open.

The other code I posted should resolve the other issues you have.

Look it up at: http://wrox.books24x7.com
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