Put the following code in the Click event of the command button:
DoCmd.OpenForm "FormB"
Forms!FormB!SupplierID = Me!SupplierID
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> Hi,
>
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>
> Nathan
MacDermott - 31 Aug 2004 16:01 GMT
Ummm....
If FormB is bound to a table, and its DataEntry property is not set to
Yes, this will change the the SupplierID for the first record to the
SupplierID which is being displayed on FormA. I doubt this is the behavior
you want.
If you want to start a new record in FormB every time you click the
button, you can set FormB's DataEntry property to Yes, and use the code
below.
If you want to look at existing records for this supplier, you'll need a
different approach.
HTH
- Turtle
> Put the following code in the Click event of the command button:
> DoCmd.OpenForm "FormB"
[quoted text clipped - 23 lines]
> >
> > Nathan