SubForm Problem

[DD]

I want to link a form to another form by a Cmd Button.
CustomerID = CustomerID
I want the form to open from the Cmd Button as if it was acting in the
same way as subform
What would be the code? as Access dose not show you.
DD

[MGFoster]

Re: SubForm Problem

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Actually, there are examples in the Help files (if you can find them).
Anyway...

Private Sub cmdOpen_Click()

Const FORM_SUB = "frmMyOtherForm "

DoCmd.OpenForm FORM_SUB, _
WhereCondition: ="CustomerID =" & Me!CustomerID

End Sub

Change the FORM_SUB constant to the name of the form you want to open.
You'll need a control named "CustomerID " on the main form (it can
have property Visible=False). This control should have a value before
the "Open" CommandButton is clicked.

See the Help files on the "OpenForm method."

HTH,

MGFoster:::mgf
Oakland, CA (USA)
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DD wrote:
[color=blue]
> I want to link a form to another form by a Cmd Button.
> CustomerID = CustomerID
> I want the form to open from the Cmd Button as if it was acting in the
> same way as subform
> What would be the code? as Access dose not show you.
> DD[/color]

[DD]

Re: SubForm Problem

I am still not getting it right.
MY ORIGINAL QUESTION
I want to link a form to another form by a Cmd Button.[color=blue][color=green]
> > CustomerID = CustomerID
> > I want the form to open from the Cmd Button as if it was acting in the
> > same way as subform
> > What would be the code? as Access dose not show you.
> > DD[/color][/color]



The form Opens ok but is not displaying that it is linked to the
BuilderID
(OPENS A FORM WITH NO LINK)

Here is what i have written in the CmdOpen Click

frmWorkSheet(ch ild) frmBuilder(Pare nt)

Const frmWorkSheet = "frmBuilder "

DoCmd.OpenForm "frmWorkShe et", , , Forms![frmWorkSheet] =
"SitesID=" & Me!BuilderID, acWindowNormal

WHERE AM I WRONG



MGFoster <me@privacy.com > wrote in message news:<%Wmub.468 9$sb4.4383@news read2.news.pas. earthlink.net>. ..[color=blue]
> -----BEGIN PGP SIGNED MESSAGE-----
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>
> Actually, there are examples in the Help files (if you can find them).
> Anyway...
>
> Private Sub cmdOpen_Click()
>
> Const FORM_SUB = "frmMyOtherForm "
>
> DoCmd.OpenForm FORM_SUB, _
> WhereCondition: ="CustomerID =" & Me!CustomerID
>
> End Sub
>
> Change the FORM_SUB constant to the name of the form you want to open.
> You'll need a control named "CustomerID " on the main form (it can
> have property Visible=False). This control should have a value before
> the "Open" CommandButton is clicked.
>
> See the Help files on the "OpenForm method."
>
> HTH,
>
> MGFoster:::mgf
> Oakland, CA (USA)
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> OfgcHQoCOJoAe4I BtXXHX9Fs
> =EqQB
> -----END PGP SIGNATURE-----
>
> DD wrote:
>[color=green]
> > I want to link a form to another form by a Cmd Button.
> > CustomerID = CustomerID
> > I want the form to open from the Cmd Button as if it was acting in the
> > same way as subform
> > What would be the code? as Access dose not show you.
> > DD[/color][/color]

[MGFoster]

Re: SubForm Problem

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You are using the Constant incorrectly. Do not put quotes around the
name of the Constant. You do not need to include the name of the
child form in the WhereCondition parameter. You do not need to
explicitly set the WindowMode parameter to acWindowNormal because that
is the default setting.

==

Since you want the child form snyched with the parent form you will
have to have the child form's RecordSource something like this (should
be one line):

SELECT * FROM <table/query name> WHERE SitesID =
Forms!frmBuilde r!SitesID

Put in the table/query name the child form uses.

There must be a control on the parent form named SitesID.

==

Put the child form Constant declaration in the Declarations section of
the parent form's module.

Const FORM_CHILD = "frmWorkShe et"
========

Private Sub cmdChild_Click( )
DoCmd.OpenForm FORM_CHILD
End Sub

When the parent form moves to another record/closes the child form
must be "notified." This can be accomplished by putting the
appropriate code in the parent form's OnCurrent and OnClose events.

== air code ==

Private Sub Form_Current()

On Error Resume Next

Forms(FORM_CHIL D).Requery

End Sub

Private Sub Form_Close()

On Error Resume Next

DoCmd.Close acForm, FORM_CHILD

End Sub

HTH,

MGFoster:::mgf
Oakland, CA (USA)

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DD wrote:[color=blue]
> I am still not getting it right.
> MY ORIGINAL QUESTION
> I want to link a form to another form by a Cmd Button.
>[color=green][color=darkred]
>>>CustomerID = CustomerID
>>>I want the form to open from the Cmd Button as if it was acting in the
>>>same way as subform
>>>What would be the code? as Access dose not show you.
>>>DD[/color][/color]
>
>
>
>
> The form Opens ok but is not displaying that it is linked to the
> BuilderID
> (OPENS A FORM WITH NO LINK)
>
> Here is what i have written in the CmdOpen Click
>
> frmWorkSheet(ch ild) frmBuilder(Pare nt)
>
> Const frmWorkSheet = "frmBuilder "
>
> DoCmd.OpenForm "frmWorkShe et", , , Forms![frmWorkSheet] =
> "SitesID=" & Me!BuilderID, acWindowNormal
>
> WHERE AM I WRONG
>
>
>
> MGFoster <me@privacy.com > wrote in message news:<%Wmub.468 9$sb4.4383@news read2.news.pas. earthlink.net>. ..
>[color=green]
>>-----BEGIN PGP SIGNED MESSAGE-----
>>Hash: SHA1
>>
>>Actually, there are examples in the Help files (if you can find them).
>> Anyway...
>>
>>Private Sub cmdOpen_Click()
>>
>> Const FORM_SUB = "frmMyOtherForm "
>>
>> DoCmd.OpenForm FORM_SUB, _
>> WhereCondition: ="CustomerID =" & Me!CustomerID
>>
>>End Sub
>>
>>Change the FORM_SUB constant to the name of the form you want to open.
>> You'll need a control named "CustomerID " on the main form (it can
>>have property Visible=False). This control should have a value before
>>the "Open" CommandButton is clicked.
>>
>>See the Help files on the "OpenForm method."
>>
>>HTH,
>>
>>MGFoster:::mg f
>>Oakland, CA (USA)
>>-----BEGIN PGP SIGNATURE-----
>>Version: PGP for Personal Privacy 5.0
>>Charset: noconv
>>
>>iQA/AwUBP7n7OIechKq OuFEgEQIIjwCdHs FTMN2JCRdOfTzP5 nQRR5PX8gEAoJjI
>>OfgcHQoCOJoAe 4IBtXXHX9Fs
>>=EqQB
>>-----END PGP SIGNATURE-----
>>
>>DD wrote:
>>
>>[color=darkred]
>>>I want to link a form to another form by a Cmd Button.
>>>CustomerID = CustomerID
>>>I want the form to open from the Cmd Button as if it was acting in the
>>>same way as subform
>>>What would be the code? as Access dose not show you.
>>>DD[/color][/color][/color]