#### SUM cells in every other column IF adjacent cell equals a criteria???

```I'm setting up a spreadsheet that will be added to daily. Each day has
two columns... column A is a drop down list with limited choices (lets
say 2). Column B will be a number. Day two will be column C and D... C
will be the drop down list and D will be a number. Day three will be
column E and F and so on. Now the tricky part is for me to SUM the
numbers from each day for the matching drop down choice.

Example:
I want to SUM B1,D1,F1... but only if A1,C1,E1... equals the criteria.

Trying to be as clear as possible... what I should have in the end is
2 (the drop down choices) different cells with the SUM of all the days
entries. If day one I selected X and the adjacent cells number was 10
and day two I selected X and the number was 3 and day three I select Y
and the number was 5, then the total for the "X" selection would be 13
and the total for the "Y" selection would be 5.

Thanks for any help.
```
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9/17/2008 12:29:45 AM
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```Check your other post.

christopherphartley@gmail.com wrote:
>
> I'm setting up a spreadsheet that will be added to daily. Each day has
> two columns... column A is a drop down list with limited choices (lets
> say 2). Column B will be a number. Day two will be column C and D... C
> will be the drop down list and D will be a number. Day three will be
> column E and F and so on. Now the tricky part is for me to SUM the
> numbers from each day for the matching drop down choice.
>
> Example:
> I want to SUM B1,D1,F1... but only if A1,C1,E1... equals the criteria.
>
> Trying to be as clear as possible... what I should have in the end is
> 2 (the drop down choices) different cells with the SUM of all the days
> entries. If day one I selected X and the adjacent cells number was 10
> and day two I selected X and the number was 3 and day three I select Y
> and the number was 5, then the total for the "X" selection would be 13
> and the total for the "Y" selection would be 5.
>
> Thanks for any help.

--

Dave Peterson
```
 0
petersod (12005)
9/17/2008 1:19:18 AM

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