I am using a list box to select the record to display when opening another form in Access 2002(XP). When I run this code from a command button, I get an error message that says:

Run-time error '2450':
Microsoft Access can't find the form 'frmTest' referred to in a macro expression or Visual Basic code.

The form that won't open is in the same database and can be opened by this code:

DoCmd.OpenForm "frmTest"

Here is my code:

Dim rst As Recordset
Set rst = Forms!frmTest.RecordsetClone 'here is where the code stops
rst.Find "CustomerNumber = " & List0
Forms!frmTest.Bookmark = rst.Bookmark
DoCmd.Close acForm, "frmGoToRecordDialog"

Thanks for your help,

Bill Murrin
Nashville, TN

Nashville_Bill

__________________
Nashville_Bill

Not sure if this will fix yr problem or not - but try getting rid of the bang (!), as it's really just a remnant of the past - try using either
Forms.frmTest
or
Form_frmTest
HTH
Steven

I am a loud man with a very large hat. This means I am in charge

Thanks for the reply Steven. I tried you suggestion, but I am still having problems and think maybe I should just make it a VB application. It will be a lot more work, but VB seems to be more stable for me.

Nashville_Bill

It looks like that form must first be open, then run your code.

Once the form is opened, you can set the recordsource as follows:

Set rst = Me.RecordsetClone

But to open a form with a recordset based on the criteria selected in a list box or text box, your code would look something like the following:

Dim strCriteria as string

strCriteria = "[CustomerNumber]=" & Me![List0]
DoCmd.OpenForm "frmTest", , , strCriteria

HTH,

Beth Moffitt

Beth,

Thanks a lot for the help. I followed an example in a VB6 manual and it wouldn't work. I guess it shouldn't have worked. It seems really simple when you get the code right, huh?

Bill Murrin

Nashville_Bill