Open a form with VBA coding

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G

Guest

I need to say if a particular value is entered, open this form.
I've tried "If [field] = [value] then
DoCmd.Open [form name]
End If
End Sub

This doesn't work. Help Please :)
 
Mandorallin said:
I need to say if a particular value is entered, open this form.
I've tried "If [field] = [value] then
DoCmd.Open [form name]
End If
End Sub
Click to expand...


It should be:
DoCmd.OpenForm "form name"

See VB Help - Index - OpenForm for details on how to use
this method.
 
Hi.

If the code is in a form or report module, and the value to be checked is a
text string, then try the following syntax:

If (Nz(Me!SomeField.Value, "") = someValue) Then
DoCmd.OpenForm formName

.... where formName is a string variable.

If the value to be checked is a numerical value, then try the following
syntax:

If (Nz(Me!SomeField.Value, 0) = someValue) Then
DoCmd.OpenForm formName

If the code is in a module, then try the following syntax:

If (Nz(recSet.Fields("SomeField").Value, 0) = someValue) Then
DoCmd.OpenForm formName

.... where recSet is a Recordset object.

.... or maybe even the following syntax for a text field in a form:

If (Nz(Forms!formName.SomeField.Value, "") = someValue) Then
DoCmd.OpenForm formName

HTH.

Gunny

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thankyou for your help but it still doesn't work. i have written:

If (Nz(Me![FIELDNAME].Value, "") = "[VALUE]") Then
DoCmd.OpenForm [FORMNAME]
Else
End If

and it still doesn't work. it highlights the DoCmd.OpenForm[FORMNAME] part
when i go to debug. i have entered the correct form to open and all other
details are correct :( am i still missing something?
 
Mandorallin said:
thankyou for your help but it still doesn't work. i have written:

If (Nz(Me![FIELDNAME].Value, "") = "[VALUE]") Then
DoCmd.OpenForm [FORMNAME]
Else
End If

and it still doesn't work. it highlights the DoCmd.OpenForm[FORMNAME] part
when i go to debug. i have entered the correct form to open and all other
details are correct :( am i still missing something?
Click to expand...

The square brackets are the remaining part of the problem.

Gunny and I both indicated that the form name is a STRING,
so it must be enclosed in quotes:
DoCmd.OpenForm "FORMNAME"

or you can use a string variable:
strFormName = "FORMNAME"
DoCmd.OpenForm strFormName
 
yes i understand that, but it still fails to work. i've found an alternative
way though, thankyou very much for your help.

Alternative: stDocName = "[FORM_NAME_TO_OPEN]"
If Me![FIELD_NAME] = "[STRING_VALUE]" Then
DoCmd.OpenForm stDocName, , , stLinkCriteria
Else

turns out your quotation bits are true there though so many thanks.
 
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