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Open Form pass record id

Posted on 2013-11-18
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Last Modified: 2013-11-18
Hello, I'm trying to open a form with a particular recordset via a record id passed in like this:

DoCmd.OpenForm "DMR Tracking_Edit", , , "[DMR#] =" & NewMDRNumber & ""

When the above executes I'm asked to provide the record id. The user will not know the record id as the record was just created in another process.

How can I pass the record id to open a specific record on a form?
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Question by:gogetsome
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3 Comments
 
LVL 61

Expert Comment

by:mbizup
ID: 39657343
How are you creating the new record (what code are you using)?
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Author Comment

by:gogetsome
ID: 39657403
This is all the code. I get the next record id, then fill the current forms record set with the record I'm duplicating. I then clone the record giving it the next record Id. Finely I want to open another form and display the cloned record.

The record is being duplicated and created.

If MIRROR.Value > "" Then

Dim NewMDRNumber As Integer

'get the selected mdr
Dim myDMR As Integer
myDMR = MIRROR.Value

'fill NewMDRNumber number with next mdr number
NewMDRNumber = DLookup("myDMR", "GetNextDMRID")


 Dim cn As ADODB.Connection
   Dim rs As ADODB.Recordset
         
   'Use the ADO connection that Access uses
   Set cn = CurrentProject.AccessConnection

   'Create an instance of the ADO Recordset class, and
   'set its properties
   Set rs = New ADODB.Recordset
   With rs
      Set .ActiveConnection = cn
      .Source = "Select * from [DMR Tracking] Where [DMR#] = " & myDMR & ""
      .LockType = adLockOptimistic
      .CursorType = adOpenKeyset
      .Open
   End With
   
   'Set the form's Recordset property to the ADO recordset
   Set Me.Recordset = rs

   Set rs = Nothing
   Set cn = Nothing


'Duplicate the record
With Me.RecordsetClone
.AddNew
![DMR#] = NewMDRNumber
![DATE Issued] = Me![DATE Issued]
![TYPE OF DMR] = Me![TYPE OF DMR]
![PART NUMBER] = Me![PART NUMBER]
![PART NAME] = Me![PART NAME]
![Program] = Me![Program]
![Customer] = Me![Customer]
'![SUPPLIER] = Me![SUPPLIER]
![Area] = Me![Area]
![LOT NUMBER] = Me![LOT NUMBER]
![QTY SUSPECT] = Me![QTY SUSPECT]
![PPM QUANTITY] = Me![PPM QUANTITY]
![WHERE LOCATED] = Me![WHERE LOCATED]
![WHO PREPARED] = Me![WHO PREPARED]
![QA Eng] = Me![QA Eng]
![JobNumber] = Me![JobNumber]
![PINNumber] = Me![PINNumber]
![PartCost] = Me![PartCost]
![REASON FOR REJECTION] = Me![REASON FOR REJECTION]
.Update

DoCmd.OpenForm "DMR Tracking_Edit", , , "[DMR#] =" & NewMDRNumber & ""

End With

Exit_Handler:
Exit Sub

Err_Handler:
MsgBox "Error " & Err.Number & " - " & Err.Description, , "cmdDupe_Click"
Resume Exit_Handler

Else

MsgBox "Select the MDR to duplicate."

End If

 

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LVL 61

Accepted Solution

by:
mbizup earned 2000 total points
ID: 39657418
Assuming the new ID is numeric, your syntax looks good.

Alternatively:

DoCmd.OpenForm "DMR Tracking_Edit", WhereCondition :=  "[DMR#] =" & NewMDRNumber 

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OR if the code is on the SAME FORM that you are trying to open:
Me.Requery
Me.Filter = "[DMR#] =" & NewMDRNumber
Me.FilterOn = true

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