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From Явор Маринов <ymari...@neterra.net>
Subject Re: Problem with a formula
Date Mon, 03 Jun 2013 11:17:47 GMT
```Thanks a lot for the suggestion it worked out!

BR

On 06/03/2013 01:13 PM, Luis Iglesias wrote:
> Hi.
> The problem is not with fractions but with the values you expect in
> your conditional. Your formula only applies when A23 < B9 or when A23
> = (1,2,3,4,5,6,7,8). You can fix in your last IF:
> ...IF(A23=8;\$F\$16;1)...
> or maybe you're looking for something like:
> =IF(A23<\$B\$9;\$F\$8;IF(A23>8;\$F\$16;IF(A23>7;\$F\$15;IF(A23>6;\$F\$14;IF(A23>5;\$F\$13;IF(A23>4;\$F\$12;IF(A23>3;\$F\$11;IF(A23>2;\$F\$10;IF(A23>1;\$F\$9;1)))))))))*A23*D23
>
>
> ------------------------------------------------------------------------
> *De: *"Явор Маринов" <ymarinov@neterra.net>
> *Para: *users@openoffice.apache.org
> *Enviados: *Lunes, 3 de Junio 2013 11:18:35
> *Asunto: *Problem with a formula
>
> Hello,
>
> I'm having a problem with a formula that i have
>
> =IF(A23<\$B\$9;\$F\$8;IF(A23=1;\$F\$9;IF(A23=2;\$F\$10;IF(A23=3;\$F\$11;IF(A23=4;\$F\$12;IF(A23=5;\$F\$13;IF(A23=6;\$F\$14;IF(A23=7;\$F\$15;IF(A23=8;\$F\$16)))))))))*A23*D23
>
> If the content of A23 is integer the formula works fine, but if the
> content of A23 is а fraction the result is 0.
>
> Can anyone point where can be the mistake?
>
>
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>
> --
> */*/Luis Iglesias Rejas