Creating a control on EXISTING form at runtime

[mhnaccess]

I am trying to create controls, EG a text box, at runtime. ALL examples I find are creating a new form at the same time. I want to create a control on an existing form.

The following does not work:
Dim frm As Form
Dim Box_1 As Control

Set Box_1 = CreateControl(f rm.Name, acRectangle)
With Box_1
.Width = 100
.Height = 100
.Top = 10
.Left = 10
End With

Thanks for any help

[ADezii]

I am trying to create controls, EG a text box, at runtime. ALL examples I find are creating a new form at the same time. I want to create a control on an existing form.

The following does not work:
Dim frm As Form
Dim Box_1 As Control

Set Box_1 = CreateControl(f rm.Name, acRectangle)
With Box_1
.Width = 100
.Height = 100
.Top = 10
.Left = 10
End With

Thanks for any help

To create a Control on an existing Form (frmTest), execute the following code outside of the context of frmTest. The following code will:

1. Open frmTest in Design Mode.
2. Create a Rectangle on frmTest with the following Properties:
   1. Width = 2.5 inches
   2. Height = 1 inch
   3. Top = 3 inches
   4. Left = 4 inches
   5. Special Effect = Sunken
   6. Border Style = Solid
   7. Back Style = Normal (color will show through)
   8. Backcolor = Blue
3. Open frmTest in order to display the fruits of your labor.
4. The following code has been tested and is fully operational.

Dim frm As Form
Dim Box_1 As Control
Const conInch As Integer = 1440     'Measurements are in Twips
                                    '1 inch = 1,440 Twips
 
DoCmd.OpenForm "frmTest", acDesign
 
Set frm = Forms!frmTest
 
Set Box_1 = CreateControl(frm.Name, acRectangle)
 
With Box_1
  .Width = 2.5 * conInch
  .Height = 1 * conInch
  .Top = 3 * conInch
  .Left = 4 * conInch
  .SpecialEffect = 2    'Sunken
  .BorderStyle = 1      'Solid
  .BackStyle = 1        'Normal
  .BackColor = vbBlue
End With
 
DoCmd.OpenForm "frmTest", acNormal, , , acFormEdit

[kaddoura1]

There is bug in MS Access that even if you follow the instructions from the official website of MS still you won't be able to "create" and "edit" controls at "run-time".
I found one way to go around this bug (witch is a combination of solutions I got from others).
Non working solution: Don't tray to define a variable to create the object:-

Sample 1:

        DoCmd.OpenForm "Form1", acDesign
        Dim t as TextBox
        set t = CreateControl("Form1", acTextBox)
        t.FontName = "Arial"  'Any sample as test.   
        .ForeColor = vbRead   'Any sample as test.   
        .DefaultValue = 10    'Any sample as test.   
        'etc...
 

This way will give you error (by the way it is what you get from MS official site).

If you use the "With" it will work, but....
Sample 2:

        DoCmd.OpenForm "Form1", acDesign
        With CreateControl("Form1", acTextBox)
            .FontName = "Arial"  'Any sample as test.   
            .ForeColor = vbRead  'Any sample as test.   
            .DefaultValue = 10   'Any sample as test.   
            'etc...     
        End With
 

This way Won't give you any error (bug free) But....
You won't be able to edit it later through the code; because we didn't use an object or variable to hold it or point to it (as in Sample 1).


My solution is: To use the name of the control to call it using "Controls() ", and that is after useing "With" to create the control.:-
Sample 3:

        DoCmd.OpenForm "Form1", acDesign
        Dim strName as string
        With CreateControl("Form1", acTextBox)
            .FontName = "Arial"  'Any sample as test.   
            .ForeColor = vbRead     'Any sample as test.   
            .DefaultValue = 10   'Any sample as test.   
            ' etc...
            strName = .Name   'Get the default name from the control (or you can give one).
        End With
        'And to be able to edit it later use the name you got in the previous lines.
        Me.Controls(strName).FontName = "Times New Roman"    'Any sample as test.   
        Me.Controls(strName).ForeColor = vbGreen            'Any sample as test.   
        Me.Controls(strName).DefaultValue = 20                    'Any sample as test.   
        ' etc...