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Thanks, zeroaccess. It is done

So as not to post a new topic if I may ask something more. I have formA that have cboA with names of people and search button to filter the names (people) and their activities. So when I try to add new activite with comand button DoCmd.OpenForm "formB", , , , acFormAdd and save it with that code...

On the Main form I had timer. I removed it and now VBA Editor working great. Thanks all.

I have some problem with VBA Editor. When I try to use spacebar it does not allow me to move it forward and the beginning of the code turns red. What could be the problem

Тhank you for the advice. It is done.

I want to make 3 accde files on my database for 3 people. In my database I have a form with combobox with 3 names (name1, name2, name3) of that people to make their reports. So I want when the name1 open his accde file, in that form with combobox to see In that combobox only his name (name1) to...

In this case I get Run-time error 32538

The code finally works! TempVars("temVarLControl") = LControlsID.Value DoCmd.OpenForm "Form_hm", , , "[LControlsID]=" & Me.LControlsID Thanks, Gasman

Yes, I need the same result as in cmdMB and Form_mb When I add this code on my Access 2019: it shows my Run time error: 32538: TemVar can only stor data :banghead:

What should the code look like on click on the button cmdHM? I'm trying with this, but nothing Dim tvLControlsID As TempVars Me.LControlsID = TempVars!tvLControlsID DoCmd.OpenForm "Form_hm" On the form "Form_hm" in the field LControlsID I set Default Value =[TempVars]![tvLControlsID]

Is there a way when I press the button cmdHM to open the form "Form_hm" to the same LControlsID as on the form "subformNonStandard" and to add in the form "Form_hm"?

Oh, sorry. Here is the DB in Access 2002 - 2003 Database

The db is created on Access 2019. OK, not a problem. :) Thanks anyway for your time. Maybe someone else will help with my problem.

Would you try now? I replaced it

Here is my database. I put two buttons on the form"subformNonStandard" cmdHm and cmdMB. When I click on the cmdMB I have Macro to open form "Form_mb" to specific record , and it opens the form "Form_mb" correct. When I try to open the form "Form_hm" by click on the button cmdHm with this code...

It turned out to be quite confusing with all these explanations. I'll upload db

No, it is not a control. LaboratoryID on the second form is set by Form_Load() DoCmd.GoToRecord , , acNewRec.

I can't fix the problem the code: DoCmd.OpenForm "sub_Indicators", , , "[LaboratoryID] = " & Me.LaboratoryID In Immediate windows show me this error: Compile error: Variable not yet created in this context. Аny ideas for the reason?

No, it is not. Have I set to open as Add?

I have tried with DoCmd.OpenForm "sub_Indicators", , , "[LaboratoryID] = " & Me.LaboratoryID, but when the form is open LaboratoryID is set to "0". I wonder why in the loaded form "sub_Indicators" Property sheet - Filter [LaboratoryID]= 652 (this is the correct number in first form), but on...