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Error 40036 when passing subform as a variable to procedure

Hi, I'm trying to pass a subform as a variable to a pocedure to be able to access this subform's properties (i.e refresh it later on)

If used the syntax described here but to no effect. I keep getting error 40036 "Method Item of object Forms failed" on the following line of code:

260   Set fForm = Forms!New_frmZamowienia_DaneRzeczywiste!subform_Notes.Form

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My main form is named: New_frmZamowienia_DaneRzeczywiste
The sub form is named: subform_Notes

The sub form is on a page control tab
Sub and main form are linked through a field.

I'm attaching the whole procedure.
Sub.txt
0
kmila
Asked:
kmila
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2 Solutions
 
Jim Dettman (Microsoft MVP/ EE MVE)PresidentCommented:
When you call your procedure, call it using

[form]

So procedure looks like this:

Public sub MySub( frm as form)

Now in the procedure.  Use the variable form just as you would with Me

Jim
0
 
Jim Dettman (Microsoft MVP/ EE MVE)PresidentCommented:
That should be frm not form.   Darn phone made a correction

Jim
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PatHartmanCommented:
If you are calling the procedure from the subform, then use Me.

Call SaveUpdate_OrderNote(Me, OtherParm)
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kmilaAuthor Commented:
Hi, substituting
Set fForm = Forms!New_frmZamowienia_DaneRzeczywiste!subform_Notes.Form

with a form object I've passed the Me variable earlier on to worked.

Set fForm = fCallingForm

Thanks for that.

Is there some inherent problem with the way I was trying to assign the form variable. I mean I can still think of a situation where I would need to point directly to a subform from the forms collection.
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Jim Dettman (Microsoft MVP/ EE MVE)PresidentCommented:
<<Is there some inherent problem with the way I was trying to assign the form variable.>>

The only tricky part to this:

Set fForm = Forms!New_frmZamowienia_DaneRzeczywiste!subform_Notes.Form

 is that you need to refer to the sub form *control* name on the main form, not the sub forms actual name form name as it appears in the database container.  They may not be one in the same.

Jim.
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