|
Subject:
|
Access can't find the form error
|
|
Posted By:
|
bmurrin
|
Post Date:
|
12/31/2003 1:55:09 PM
|
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
|
|
Reply By:
|
Steven
|
Reply Date:
|
12/31/2003 11:30:21 PM
|
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
|
|
Reply By:
|
bmurrin
|
Reply Date:
|
1/1/2004 3:31:13 PM
|
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
|
|
Reply By:
|
jemacc
|
Reply Date:
|
1/1/2004 9:35:34 PM
|
It looks like that form must first be open, then run your code.
|
|
Reply By:
|
BethMoffitt
|
Reply Date:
|
1/2/2004 9:12:23 AM
|
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
|
|
Reply By:
|
bmurrin
|
Reply Date:
|
1/3/2004 1:27:03 PM
|
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
|