.Formula Runtime error 1004 on some machines and not the other










1















I'll go straight to the point. Here is the line code that is geving me the error:



ThisWorkbook.Worksheets("Dados").Cells(16, 10).FormulaLocal = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"


The above code works on my home machine (where created) and does not work on my lab or work machine



ThisWorkbook.Worksheets("Dados").Cells(16, 10).Formula = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"


And the second one works on my lab and work machine, but does not in my home machine.
The only difference is Formula and FormulaLocal. I know FormulaLocal could give me errors because of different installations of excel might have different language.... but shouldn't Formula work everywhere?



Also, if I change cells to R16C9 instead of I16 it works in my home machine and bugs in others. I really want a code that can work everywhere here. Can someone help? Thanks



P.S.: Variables are all ok and well defined, but the issue is probably that they have decimals, thus the comma or dot issue



Debug.Print (With FormulaLocal) gives me:



=-(I16)+0,549840740436919+(0,0513418101762096*((604,958825198337 ^ 1,04086742681856)*(I16 ^ 1,04086742681856)/(49,44 ^ 1,04086742681856)))









share|improve this question



















  • 2





    Difficult to know what the problem might be without seeing the missing pieces - Debug.Print the formula and add the output to the question. Copy the output and try adding it to the cell via the formula bar - do you get an error? If Yes, what error? Formula should work if the string you're passing is a valid formula in a version of Excel with English as the language setting.

    – Tim Williams
    Nov 15 '18 at 5:09












  • As already mentioned the first thing to do is Debug.Print and inspect the formula. As is this means nothing. You have five variables going into that string whose value is unknown. It's quite possible the variable values are the issue

    – Nick.McDermaid
    Nov 15 '18 at 7:46











  • Completely forgot to mention that yes, variables are the issue because they have decimals.

    – Marco Faria
    Nov 15 '18 at 11:48















1















I'll go straight to the point. Here is the line code that is geving me the error:



ThisWorkbook.Worksheets("Dados").Cells(16, 10).FormulaLocal = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"


The above code works on my home machine (where created) and does not work on my lab or work machine



ThisWorkbook.Worksheets("Dados").Cells(16, 10).Formula = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"


And the second one works on my lab and work machine, but does not in my home machine.
The only difference is Formula and FormulaLocal. I know FormulaLocal could give me errors because of different installations of excel might have different language.... but shouldn't Formula work everywhere?



Also, if I change cells to R16C9 instead of I16 it works in my home machine and bugs in others. I really want a code that can work everywhere here. Can someone help? Thanks



P.S.: Variables are all ok and well defined, but the issue is probably that they have decimals, thus the comma or dot issue



Debug.Print (With FormulaLocal) gives me:



=-(I16)+0,549840740436919+(0,0513418101762096*((604,958825198337 ^ 1,04086742681856)*(I16 ^ 1,04086742681856)/(49,44 ^ 1,04086742681856)))









share|improve this question



















  • 2





    Difficult to know what the problem might be without seeing the missing pieces - Debug.Print the formula and add the output to the question. Copy the output and try adding it to the cell via the formula bar - do you get an error? If Yes, what error? Formula should work if the string you're passing is a valid formula in a version of Excel with English as the language setting.

    – Tim Williams
    Nov 15 '18 at 5:09












  • As already mentioned the first thing to do is Debug.Print and inspect the formula. As is this means nothing. You have five variables going into that string whose value is unknown. It's quite possible the variable values are the issue

    – Nick.McDermaid
    Nov 15 '18 at 7:46











  • Completely forgot to mention that yes, variables are the issue because they have decimals.

    – Marco Faria
    Nov 15 '18 at 11:48













1












1








1








I'll go straight to the point. Here is the line code that is geving me the error:



ThisWorkbook.Worksheets("Dados").Cells(16, 10).FormulaLocal = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"


The above code works on my home machine (where created) and does not work on my lab or work machine



ThisWorkbook.Worksheets("Dados").Cells(16, 10).Formula = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"


And the second one works on my lab and work machine, but does not in my home machine.
The only difference is Formula and FormulaLocal. I know FormulaLocal could give me errors because of different installations of excel might have different language.... but shouldn't Formula work everywhere?



Also, if I change cells to R16C9 instead of I16 it works in my home machine and bugs in others. I really want a code that can work everywhere here. Can someone help? Thanks



P.S.: Variables are all ok and well defined, but the issue is probably that they have decimals, thus the comma or dot issue



Debug.Print (With FormulaLocal) gives me:



=-(I16)+0,549840740436919+(0,0513418101762096*((604,958825198337 ^ 1,04086742681856)*(I16 ^ 1,04086742681856)/(49,44 ^ 1,04086742681856)))









share|improve this question
















I'll go straight to the point. Here is the line code that is geving me the error:



ThisWorkbook.Worksheets("Dados").Cells(16, 10).FormulaLocal = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"


The above code works on my home machine (where created) and does not work on my lab or work machine



ThisWorkbook.Worksheets("Dados").Cells(16, 10).Formula = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"


And the second one works on my lab and work machine, but does not in my home machine.
The only difference is Formula and FormulaLocal. I know FormulaLocal could give me errors because of different installations of excel might have different language.... but shouldn't Formula work everywhere?



Also, if I change cells to R16C9 instead of I16 it works in my home machine and bugs in others. I really want a code that can work everywhere here. Can someone help? Thanks



P.S.: Variables are all ok and well defined, but the issue is probably that they have decimals, thus the comma or dot issue



Debug.Print (With FormulaLocal) gives me:



=-(I16)+0,549840740436919+(0,0513418101762096*((604,958825198337 ^ 1,04086742681856)*(I16 ^ 1,04086742681856)/(49,44 ^ 1,04086742681856)))






excel vba excel-vba excel-formula






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edited Nov 15 '18 at 12:48







Marco Faria

















asked Nov 15 '18 at 3:26









Marco FariaMarco Faria

62




62







  • 2





    Difficult to know what the problem might be without seeing the missing pieces - Debug.Print the formula and add the output to the question. Copy the output and try adding it to the cell via the formula bar - do you get an error? If Yes, what error? Formula should work if the string you're passing is a valid formula in a version of Excel with English as the language setting.

    – Tim Williams
    Nov 15 '18 at 5:09












  • As already mentioned the first thing to do is Debug.Print and inspect the formula. As is this means nothing. You have five variables going into that string whose value is unknown. It's quite possible the variable values are the issue

    – Nick.McDermaid
    Nov 15 '18 at 7:46











  • Completely forgot to mention that yes, variables are the issue because they have decimals.

    – Marco Faria
    Nov 15 '18 at 11:48












  • 2





    Difficult to know what the problem might be without seeing the missing pieces - Debug.Print the formula and add the output to the question. Copy the output and try adding it to the cell via the formula bar - do you get an error? If Yes, what error? Formula should work if the string you're passing is a valid formula in a version of Excel with English as the language setting.

    – Tim Williams
    Nov 15 '18 at 5:09












  • As already mentioned the first thing to do is Debug.Print and inspect the formula. As is this means nothing. You have five variables going into that string whose value is unknown. It's quite possible the variable values are the issue

    – Nick.McDermaid
    Nov 15 '18 at 7:46











  • Completely forgot to mention that yes, variables are the issue because they have decimals.

    – Marco Faria
    Nov 15 '18 at 11:48







2




2





Difficult to know what the problem might be without seeing the missing pieces - Debug.Print the formula and add the output to the question. Copy the output and try adding it to the cell via the formula bar - do you get an error? If Yes, what error? Formula should work if the string you're passing is a valid formula in a version of Excel with English as the language setting.

– Tim Williams
Nov 15 '18 at 5:09






Difficult to know what the problem might be without seeing the missing pieces - Debug.Print the formula and add the output to the question. Copy the output and try adding it to the cell via the formula bar - do you get an error? If Yes, what error? Formula should work if the string you're passing is a valid formula in a version of Excel with English as the language setting.

– Tim Williams
Nov 15 '18 at 5:09














As already mentioned the first thing to do is Debug.Print and inspect the formula. As is this means nothing. You have five variables going into that string whose value is unknown. It's quite possible the variable values are the issue

– Nick.McDermaid
Nov 15 '18 at 7:46





As already mentioned the first thing to do is Debug.Print and inspect the formula. As is this means nothing. You have five variables going into that string whose value is unknown. It's quite possible the variable values are the issue

– Nick.McDermaid
Nov 15 '18 at 7:46













Completely forgot to mention that yes, variables are the issue because they have decimals.

– Marco Faria
Nov 15 '18 at 11:48





Completely forgot to mention that yes, variables are the issue because they have decimals.

– Marco Faria
Nov 15 '18 at 11:48












1 Answer
1






active

oldest

votes


















0














There are probably better solutions, mainly on the dumb way i used that if, but this now works on every machine i have. Thanks everyone.



...
bloqueio = 5
On Error GoTo alternativa
ThisWorkbook.Worksheets("Dados").Cells(16, 10).Formula = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"
If bloqueio = 0 Then
alternativa:
On Error GoTo -1
On Error GoTo Erro
ThisWorkbook.Worksheets("Dados").Cells(16, 10).FormulaLocal = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"
End If
On Error GoTo -1
...





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    1 Answer
    1






    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    0














    There are probably better solutions, mainly on the dumb way i used that if, but this now works on every machine i have. Thanks everyone.



    ...
    bloqueio = 5
    On Error GoTo alternativa
    ThisWorkbook.Worksheets("Dados").Cells(16, 10).Formula = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"
    If bloqueio = 0 Then
    alternativa:
    On Error GoTo -1
    On Error GoTo Erro
    ThisWorkbook.Worksheets("Dados").Cells(16, 10).FormulaLocal = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"
    End If
    On Error GoTo -1
    ...





    share|improve this answer



























      0














      There are probably better solutions, mainly on the dumb way i used that if, but this now works on every machine i have. Thanks everyone.



      ...
      bloqueio = 5
      On Error GoTo alternativa
      ThisWorkbook.Worksheets("Dados").Cells(16, 10).Formula = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"
      If bloqueio = 0 Then
      alternativa:
      On Error GoTo -1
      On Error GoTo Erro
      ThisWorkbook.Worksheets("Dados").Cells(16, 10).FormulaLocal = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"
      End If
      On Error GoTo -1
      ...





      share|improve this answer

























        0












        0








        0







        There are probably better solutions, mainly on the dumb way i used that if, but this now works on every machine i have. Thanks everyone.



        ...
        bloqueio = 5
        On Error GoTo alternativa
        ThisWorkbook.Worksheets("Dados").Cells(16, 10).Formula = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"
        If bloqueio = 0 Then
        alternativa:
        On Error GoTo -1
        On Error GoTo Erro
        ThisWorkbook.Worksheets("Dados").Cells(16, 10).FormulaLocal = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"
        End If
        On Error GoTo -1
        ...





        share|improve this answer













        There are probably better solutions, mainly on the dumb way i used that if, but this now works on every machine i have. Thanks everyone.



        ...
        bloqueio = 5
        On Error GoTo alternativa
        ThisWorkbook.Worksheets("Dados").Cells(16, 10).Formula = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"
        If bloqueio = 0 Then
        alternativa:
        On Error GoTo -1
        On Error GoTo Erro
        ThisWorkbook.Worksheets("Dados").Cells(16, 10).FormulaLocal = "=-(I16)+" & a & "+(" & c & "*((" & PiZi & " ^ " & d & ")*(I16 ^ " & d & ")/(" & Pc & " ^ " & d & ")))"
        End If
        On Error GoTo -1
        ...






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 15 '18 at 16:23









        Marco FariaMarco Faria

        62




        62





























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