can anyone help me for "#DIV/0!" Error
i am doing some caculations in excel cells, my result is in cell A1,
if the result comes "#DIV/0!" then While executing the following macro, VBA gives me Debug message. The value of A becomes "Error 2007". How can i solve this problem? Sub Postmacro() A = Cells(1, 1).Value B = 1.5 if(A<0) then C = B / A End if End Sub |
can anyone help me for "#DIV/0!" Error
Hi,
I don't understand the problem. The line if(A<0) then ensures you won't get a #DIV/0 error but you could get a type mismatch. I suggest you change that line to If a < 0 And IsNumeric(a) Then Mike "Arvind Mane" wrote: i am doing some caculations in excel cells, my result is in cell A1, if the result comes "#DIV/0!" then While executing the following macro, VBA gives me Debug message. The value of A becomes "Error 2007". How can i solve this problem? Sub Postmacro() A = Cells(1, 1).Value B = 1.5 if(A<0) then C = B / A End if End Sub |
can anyone help me for "#DIV/0!" Error
On Fri, 26 Sep 2008 00:07:01 -0700, Arvind Mane
wrote: i am doing some caculations in excel cells, my result is in cell A1, if the result comes "#DIV/0!" then While executing the following macro, VBA gives me Debug message. The value of A becomes "Error 2007". How can i solve this problem? Sub Postmacro() A = Cells(1, 1).Value B = 1.5 if(A<0) then C = B / A End if End Sub 1. It is good practice to explicitly DIM your variables. You need to test the contents of Cells(1,1) to see if there is an error value: ==================== Option Explicit Sub Postmacro() Dim A As Double Dim r As Range Dim B As Double Dim C As Double Set r = Cells(1, 1) If IsError(r) Then Exit Sub A = r.Value B = 1.5 If (A < 0) Then C = B / A End If End Sub =================== --ron |
can anyone help me for "#DIV/0!" Error
Thanks Ron Rosenfeld
"Ron Rosenfeld" wrote: On Fri, 26 Sep 2008 00:07:01 -0700, Arvind Mane wrote: i am doing some caculations in excel cells, my result is in cell A1, if the result comes "#DIV/0!" then While executing the following macro, VBA gives me Debug message. The value of A becomes "Error 2007". How can i solve this problem? Sub Postmacro() A = Cells(1, 1).Value B = 1.5 if(A<0) then C = B / A End if End Sub 1. It is good practice to explicitly DIM your variables. You need to test the contents of Cells(1,1) to see if there is an error value: ==================== Option Explicit Sub Postmacro() Dim A As Double Dim r As Range Dim B As Double Dim C As Double Set r = Cells(1, 1) If IsError(r) Then Exit Sub A = r.Value B = 1.5 If (A < 0) Then C = B / A End If End Sub =================== --ron |
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