How to create a button to open a form on a specific record?

[hesca]

Hello experts.



I have two forms: Form1: tabular form showing the latest order date for every customer. Form2: A form showing the history of this customer's interactions with my CSreps.

I want to create a button on form 1 that opens form 2 AND shows the same customer´s details. This button should show on every row so it opens form2 on the record that shows the customer name from this row.

How do I do this?

I have tried to create a button on design view that opens form2 and then tried linking the button to form1 via "Customer name". It doesnt work. All it does is open form2 on the same record every time, no matter which button from which row do I press on the tabular form (form1)


Help!!

[pbaldy]

See if this helps:

BaldyWeb wherecondition

[ssanfu]

Without knowing your table structure and relationships (also PK fields and SQL of the form record sources", it is difficult to be specific. Look in HELP for the "OpenForm Method".


Something like:
DoCmd.OpenForm "Employees", , ,"MY_PK = " & Me.PK_Field

The syntax is:
DoCmd.OpenForm formname ,,, wherecondition