Hi John,

The form does open using the command:
DoCmd.OpenForm "iraprovalformprint", acViewNormal
under a button ,but what I'd like to do is send the above form 
directly to print,then close?

Joe



--- In [email protected], "John Viescas" <[EMAIL PROTECTED]> wrote:
>
> Joe-
> 
> First, the query should be:
> 
> PARAMETERS [Forms]![irneedaproval]![iralist] int;
> SELECT [Incident Query].* FROM [Incident Query] 
> WHERE ((([Incident Query].ID)=[Forms]![irneedaproval]![iralist])); 
> 
> The second form should open just fine as long as the irneedaproval 
form is
> open and the iralist control has a value.
> 
> John Viescas, author
> "Building Microsoft Access Applications"
> "Microsoft Office Access 2003 Inside Out"
> "Running Microsoft Access 2000"
> "SQL Queries for Mere Mortals"
> http://www.viescas.com/
> 
> 
> -----Original Message-----
> From: [email protected] [mailto:[EMAIL PROTECTED] 
On Behalf
> Of eliotchs
> Sent: Tuesday, December 06, 2005 11:40 AM
> To: [email protected]
> Subject: [ms_access] ? on Print button on form to print another form
> 
> Hi,
> 
> Now that I have my listbox working(Thanks John!)
> I'm trying to setup a button on the form which sends a different 
form 
> to the printer.
> Form that opens = "IRneedaproval" it gets its record from
> SELECT [Incident Query].*, [Incident Query].ID FROM [Incident 
Query] 
> WHERE ((([Incident Query].ID)=Forms!irneedaproval!iralist));
> 
> the button is sent to open "irneenaprovalPrint", using the same 
record 
> source above.
> 
> The problem is it print blank.
> I think I need to pass the ID somewhere?
> 
> Thanks In Advance 
> 
> Joe
> 
>  
> 
> 
> 
> 
> 
> 
>  
> Yahoo! Groups Links
>







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