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  • How to connect Crystal Report 8.5 to VB 6

    Hi everyone, Im really new at VB and Crystal Report and i know nothing about Crystal Report 8.5. My problem is, i have this application called CustomerManagement that will keep data in Microsoft Access called 'dbCustomer'
    In the form customer ('frmCustomer') i've created a command button 'cmdView' so that the user may view the data in Crystal Report.
    My problem is, i've created the link from the form to the CrystalReport1 but this error appear when i click the button 'View' (cmdView) on my frmCustomer:
    "This group section cannot be printed becase its condition field is nonexistent or invalid. Format the section to choose another condition field."          Please help me!!!

    Thank you in advance!!


    ****************************************************************************
    on the customer form ("frmCustomer"):
    Private Sub cmdView_Click()
    Call GenerateCrystalReport
    End Sub
    ****************************************************************************
    on module 1: (I put this on global)

    Public Function GenerateCrystalReport() As Long

    Dim recReport As ADODB.Recordset
    Dim strSQL As String

    ''Call connect
    Set recReport = New ADODB.Recordset
    strSQL = "select * from TableCustomer order by TransactionID, TranDateTime"
    recReport.Open strSQL, DBconnection, adOpenDynamic, adLockOptimistic, adCmdText
    Set m_crReport = New CrystalReport1

    m_crReport.Database.Tables(1).SetDataSource recReport

    frmViewReport.Show

    GenerateCrystalReport = 1

    End Function

    ****************************************************************************
    on the frmViewReport:

    Private Sub Form_Load()
    Screen.MousePointer = vbHourglass
    CRViewer1.ReportSource = m_crReport
    CRViewer1.ViewReport
    Screen.MousePointer = vbDefault

    End Sub

    Private Sub Form_Resize()
    CRViewer1.Top = 0
    CRViewer1.Left = 0
    CRViewer1.Height = ScaleHeight
    CRViewer1.Width = ScaleWidth

    End Sub

     

     

     

     

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