Open form with specific record based on List box selection 1

[ScorpioX]

Hi All,
I have a little issue. I am trying open a form, showing a specific record based on the selection in a list box.
The form I am opening is "frmWork_preview", which is opened by double clicking on a list box, named "List58". The List box is populated by "CNumber" in the Main form. I have it working to some extent. It does open the right form, but will not show the correct record. Anyone have any ideas?

Thanks,
ScorpioX

 

[pbaldy]

This is still rough (any feedback would be appreciated), but see if this helps:

http://www.baldyweb.com/wherecondition.htm

Paul
MS Access MVP 2007/2008

 

[fneily]

An easy way is to create a macro. Under the Action Column in the first cell put Echo. At the bottom under Action Arguments where it says Echo On put No.
In the next Action cell, put OpenForm. At the bottom, put in your formname, in the next cell the word View, leave Filter Name Blank,Where Condition would look like this -
[CNumber]=[Forms]![FrmWork_Preview]![List58], leave Data Mode blank, Window is Normal.
Save the macro and place the name on the On Dbl Click event of the listbox's property.

Make sure your Listbox is bound to CNumber.

 

[fneily]

Oops. the Where condition is wrong. I put the name of the form you're opening as the form you're coming from. It should be:
[CNumber]=[Forms]![FormnameOfListbox]![List58]

 

[ScorpioX]

Hi All,
Thank you for your replies. I was able to get it working though, using part of pbaldy's code in mine. I have included the working code below.

On Error GoTo Err_List58_DblClick
    Dim stDocName As String
    Dim stLinkCriteria As String
    stDocName = "frmWork_preview"
    stLinkCriteria = "[Work_#]=" & Me![List58]
    DoCmd.OpenForm stDocName, , , stLinkCriteria
Exit_List58_DblClick:
    Exit Sub
Err_List58_DblClick:
    MsgBox Err.Description
    Resume Exit_List58_DblClick
End Sub

Thanks again,
ScorpioX