Access: Load data from one form to another

[gheyman]

I have code that loads data that is selected on one form to another form (Someone here helped me with - thanks!!!!!)

The code runs great except for the last line that I tried to add. I want filed "_Upload" to equal "Y". Its a Combo box Formatted True/False

Private Sub LoadItemId_Click()
[Forms]![frm_VendorQuoteData].[ITEM_ID].Value = Me.ItemID.Value
[Forms]![frm_VendorQuoteData].[_ITEM_ID_DESC].Value = Me.Description.Value
[Forms]![frm_VendorQuoteData].[ITEM_RVSN_ID].Value = Me.Rev.Value
[Forms]![frm_VendorQuoteData].[_COMMD_CODE].Value = Me.Commodity.Value
[Forms]![frm_VendorQuoteData].[QT_UM_CD].Value = Me.UM.Value
[Forms]![frm_VendorQuoteData].[_Upload] = "Y"
End Sub

Also how do I ensure that form "frm_VendorQuoteData" is open when code is run?. I don't want to open the form, I want to make sure it was already open.

Thanks

 

[Joe4]

If it is a True/False field, it stored values are typically True/False, or numerically -1, 0 (and not the string value "Y").

So try:

[Forms]![frm_VendorQuoteData].[_Upload] = True

or

[Forms]![frm_VendorQuoteData].[_Upload] = -1

 

[gheyman]

Thanks Joe. Is there a way to make sure the form "frm_VendorQuoteData" is open before the code is run?

 

[Joe4]

How is this code being called? Is it being called from an event on that Form (if so, the Form would have to be open already)?
If not, you will need to add a "DoCmd.OpenForm" command at the beginning of your code.

Note, I see that you have a few question recently that involve using the various "DoCmd" commands.
Most of these are available under the "Macros" sections. A good way to get the exact code that you need with all the right arguments/values is to create a simple Macro to perform that action (i.e. Open a specific form). Then, convert that Macro to VBA code using the "Convert Macros to Visual Basic" action (found on Macro toolbar). Then, you can just copy/paste the needed lines of code that you need from the converted function to your VBA code.

 

[welshgasman]

https://social.msdn.microsoft.com/F...test-to-see-if-a-form-is-open?forum=accessdev

Found using https://www.google.com/search?ei=K5.....0.0..0.0.0.......0......gws-wiz.L9NbBf0X6h0

I have code that loads data that is selected on one form to another form (Someone here helped me with - thanks!!!!!)

The code runs great except for the last line that I tried to add. I want filed "_Upload" to equal "Y". Its a Combo box Formatted True/False

Private Sub LoadItemId_Click()
[Forms]![frm_VendorQuoteData].[ITEM_ID].Value = Me.ItemID.Value
[Forms]![frm_VendorQuoteData].[_ITEM_ID_DESC].Value = Me.Description.Value
[Forms]![frm_VendorQuoteData].[ITEM_RVSN_ID].Value = Me.Rev.Value
[Forms]![frm_VendorQuoteData].[_COMMD_CODE].Value = Me.Commodity.Value
[Forms]![frm_VendorQuoteData].[QT_UM_CD].Value = Me.UM.Value
[Forms]![frm_VendorQuoteData].[_Upload] = "Y"
End Sub

Also how do I ensure that form "frm_VendorQuoteData" is open when code is run?. I don't want to open the form, I want to make sure it was already open.

Thanks