Open Subform With Button

Jul 25, 2006

Hello all,

I know I have seen this before and I have searched the forums several times and cannot find it now.

Maybe I am searching the wrong thing. Can someone point me in the right direction?

I need a button on my form that opens a subform or subforms when clicked.

How do I set this up?

Thanks,
Di

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Forms :: Getting Subform To Open To A New Record When Button Is Clicked

Mar 3, 2015

I have a subform [ctrlLogDetail] on a parent form [incidentdetails] that is opened by the user when they click on a button on a navigation form. These forms are used for a variety of purposes. The problem I'm having is that the user needs to be able to select an incident number and go to the appropriate form (I accomplish this by using this code: DoCmd.OpenForm "IncidentDetails", acNormal, , "Activity_ID = " & Me.cboINum in the on click event of the button.) This works appropriately. The subform is also appropriately linked to the parent form.

I need an additional line of code to have the subform go to a new record when the form opens to an existing incident number. Since I use this form/subform when doing different tasks, having the Docmd.RunCommand acCmdRecordsGoToNew in the Form on open event isn't optimal.

I only want the LogDetail subform to open to a new record when the user wants to add an entry, but not when they need to edit a specific entry. What is the appropriate syntax to use either in the openargs event of the openform command or elsewhere in the procedure so that the gotonew function on the subform only occurs when this button is clicked? I'm having difficulty getting access to understand that I want the subform to open to a new record but not the parent form.

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Jun 1, 2006

Problem:
Visual Basic run-time error 7792: You can't open a subform when it is also open in Design view.

I have a subform with properties for SourceObject, LinkMasterFields, LinkChildFields changing according to objects and events in the master form.

The subform was bound to a query before which meant that the subform load event was happening as soon as the master form is loaded. I needed to restrict the subform loading until a certain point so I removed the SourceObject property for the masterform's subform. The subform on the master form is now Unbound.

(This is because I'm now running some code on the FormLoad event for the subform which needs to be restricted until the LinkMasterFields and LinkChild Fields properties have been assigned correctly otherwise it takes ages to load.)

Now I'm getting the above error. Obviously, i do not have the subform open anywhere in design view. No Visual Basic windows are open. I've closed the db, closed access, reopened it and clicked on nothing except the masterform. The error occurs when I raise the event in the master form which assigns the sourceobject property to the subform, i.e.

Me.sfmQryAllOV.SourceObject = "sfmQryAllOV"

Can anyone help me? I've googled this but finding no answers.

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Private Sub Notes_DblClick(Cancel As Integer)

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Hi people,

I have made a form. By giving an ID up firstly this form of the concerning person appears. Now I want to make a link to a report on this form. The intention is then that only those data are shown in the report of the concerning person (therefore with the ID that is reflected in the form).

Now I can open the report by means of a button... but then to get the correct data I have now used a query with a parameter value (the ID) (the report has been based on these query). But the situation is now therefore that if you click on the button that you must introduce the parameter value firstly. Is there also a solution where this is no longer necessary??... therefore that you have to click only on the button and then the concerning data of the report are shown. Therefore by giving in or an other manner the ID with. ??

Hopefully it is understandable, (my english is not that good...)

thanks!
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tables names
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________________
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