Solved

Access 2003 Type Mismatch Error

Posted on 2012-03-27
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376 Views
Last Modified: 2012-08-08
Can anyone tell me why I get this error, "The expression On Current you entered as the event property setting produced the following error: Type mismatch."  I get the error when the program is opened on a computer with only the Access 2003 runtime installed.  On computers with full version Access all works as expected.

This is the code I have in the On Current event.

If Me.txtD36 <= 3 Or Me.txtD36 >= 87 Then
    Me.txtD36.BackColor = 65280
    Me.txtD36.ForeColor = 0
Else
    Me.txtD36.BackColor = 2366701
    Me.txtD36.ForeColor = 16777215
End If

Any help appreciated.

Gary
0
Comment
Question by:garymcgowan
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7 Comments
 
LVL 34

Accepted Solution

by:
Norie earned 500 total points
ID: 37772118
Is txtD36 a text field?

If it is try this.
If Val(Me.txtD36) <= 3 Or (Me.txtD36) >= 87 Then

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0
 

Author Comment

by:garymcgowan
ID: 37773233
Its an unbound field

Format: Fixed
Decimal Places: 0

I tried your suggestion anyway but it didn't fix the problem.  Any other ideas?

Thanks,

Gary
0
 
LVL 34

Expert Comment

by:Norie
ID: 37773318
Gary

Does it have a value when this code runs?
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Author Comment

by:garymcgowan
ID: 37773442
Yes, a number.  It works fine in the full version of access but not the runtime.

Gary
0
 
LVL 34

Expert Comment

by:Norie
ID: 37773475
Gary

Is that all the code?

Is there no Sub/End Sub?
0
 

Author Comment

by:garymcgowan
ID: 37773588
Private Sub Form_Current()

If Me.txtD36 <= 3 Or Me.txtD36 >= 87 Then
    Me.txtD36.BackColor = 65280
    Me.txtD36.ForeColor = 0
Else
    Me.txtD36.BackColor = 2366701
    Me.txtD36.ForeColor = 16777215
End If



End Sub
0
 

Author Closing Comment

by:garymcgowan
ID: 38272775
The answer by imnorie did not work for me but I am giving him the points for at least trying to solve my problem.

Gary
0

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