Open a Report with two different forms

I have a report (rptRepairRequest) that I have command button that
prints it out from a form based on the ID of that form.  This works
fine.  What I would like, and not sure if it is possible, is to have
another form where I can just select the ID from a drop-down and open
the same report.

My record source for the report is currently:
SELECT tblRepairList.* FROM tblRepairList WHERE
(((tblRepairList.RepairID)=Forms!frmSerReqEntry!txtRepairID));

I tried adding an OR statement at the end:
"OR (tblRepairList.RepairID)=Forms![frmRptRepairRequest]!
[cboRepairID]"
but it asks me for 'Forms!frmSerReqEntry!txtRepairID'

Are is there anyway to make this work, without making a new report?

0
Eric
10/9/2007 7:05:34 PM
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Eric wrote:

>I have a report (rptRepairRequest) that I have command button that
>prints it out from a form based on the ID of that form.  This works
>fine.  What I would like, and not sure if it is possible, is to have
>another form where I can just select the ID from a drop-down and open
>the same report.
>
>My record source for the report is currently:
>SELECT tblRepairList.* FROM tblRepairList WHERE
>(((tblRepairList.RepairID)=Forms!frmSerReqEntry!txtRepairID));
>
>I tried adding an OR statement at the end:
>"OR (tblRepairList.RepairID)=Forms![frmRptRepairRequest]!
>[cboRepairID]"
>but it asks me for 'Forms!frmSerReqEntry!txtRepairID'
>
>Are is there anyway to make this work, without making a new report?


Remove the criteria from the query.

Instead use the button code to use the OpenReport's
WhereCondition argument:

DoCmd.OpenReport "report name", _
		WhereCondition:= "RepairID=" & Me!txtRepairID

-- 
Marsh
MVP [MS Access]
0
Marshall
10/9/2007 8:08:49 PM
On Oct 9, 1:08 pm, Marshall Barton <marshbar...@wowway.com> wrote:
> Eric wrote:
> >I have a report (rptRepairRequest) that I have command button that
> >prints it out from a form based on the ID of that form.  This works
> >fine.  What I would like, and not sure if it is possible, is to have
> >another form where I can just select the ID from a drop-down and open
> >the same report.
>
> >My record source for the report is currently:
> >SELECT tblRepairList.* FROM tblRepairList WHERE
> >(((tblRepairList.RepairID)=Forms!frmSerReqEntry!txtRepairID));
>
> >I tried adding an OR statement at the end:
> >"OR (tblRepairList.RepairID)=Forms![frmRptRepairRequest]!
> >[cboRepairID]"
> >but it asks me for 'Forms!frmSerReqEntry!txtRepairID'
>
> >Are is there anyway to make this work, without making a new report?
>
> Remove the criteria from the query.
>
> Instead use the button code to use the OpenReport's
> WhereCondition argument:
>
> DoCmd.OpenReport "report name", _
>                 WhereCondition:= "RepairID=" & Me!txtRepairID
>
> --
> Marsh
> MVP [MS Access]

Oh fantastic!  That worked exactly like I wanted it.  Thank you!

0
Eric
10/9/2007 9:46:17 PM
Reply:

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