Switchboard open form based on function value

[silverspr]

I can't seem to get the syntax right to call a function in the macro used to open an existing form:



If [Command]=2 And (FindUserId())=True then

openform....etc

finduserid is a function that evaluates to true if the userid is stored in the urser id table.

[jzwp11]

I assume that you have to pass the current user's ID to the function so that it can evaluate it. You'll have to post the code for us to be sure.

If [Command]=2 And (FindUserId(userID))=True then...

So the issue then becomes, where can you get the current user's ID? From where are you executing the macro? Are you having the user's login to the database via a form? Do you leave the form open so that you can reference the control on the form that holds the user's ID?

[silverspr]

This IF statement runs on the click event of an option button on the switchboard form, no parameters are specified for the funtion:

If [Command]=2 And (FindUserId())=True then.....
the problem is the above line of code is not evaluating, which makes me think there is something wrong with the syntax used.

The userid function evaluates to true if the userid is found in the userid table.

Function FindUserId()as Boolean
Dim db As Database
Dim rs As Recordset
Dim lngCount As Integer
Dim strCriteria As String
Set db = CurrentDb
Set rs = CurrentDb.OpenRecordset("tblUserId", dbOpenDynaset)

x = GetNTUser
strCriteria = "[userid]= '" & x & "'"
rs.FindFirst strCriteria
If rs.NoMatch Then
FindUserId = False
Else
FindUserId = True
End If
End Function

[jzwp11]

I see, you are using a global variable instead of having to pass the ID to the function directly.

What I would do would be to put a debug.print statement in where shown in blue to make sure the userID is being populated correctly.

Function FindUserId()as Boolean
Dim db As Database
Dim rs As Recordset
Dim lngCount As Integer
Dim strCriteria As String
Set db = CurrentDb
Set rs = CurrentDb.OpenRecordset("tblUserId", dbOpenDynaset)

x = GetNTUser
strCriteria = "[userid]= '" & x & "'"
debug.print strCriteria
rs.FindFirst strCriteria
If rs.NoMatch Then
FindUserId = False
Else
FindUserId = True
End If
End Function

You would attempt to run the function and then check the immediate window in VBA to see what is returned by the debug.print statement

[silverspr]

Hi
the function works fine when called from the debug window, its only here in the "If [Command]=2 And (FindUserId())=True then....." that it doesn't evaluate correctly.

Am I calling the function correctly from within this IF statement? This IF statement is part of the embedded macro triggered on the onclick event of the option button of the switchboard form.

I want the function to evaluate the userid only when the [Command]=2 option button is clicked.

Not sure how else to explain this better.

[jzwp11]

The IF...THEN is structured properly, and if the function appears to be working properly then perhaps something is wrong with the [command]=2.

First I would try me.[command]=2. Also, I would guess that the word command is a reserved word in Access & VBA, so you might change the name of the option button to something else. Other than that I cannot guess what else might be the cause.