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# Re: [libreoffice-users] Add column values according to part of date value

```At 08:59 24/08/2018 +0200, Hylton Conacher wrote:
```
I have data in 4 columns titled Date:Text:A:B. A typical row example is: 20180822:PNP:blank:500.00 OR 20180822:PNP:500.00:blank
```
```
A row will only have a value in column A or B, never both on the same row. There may be multiple rows with the same date but different Text and A or B value.
```
Example Data:
20180803:Cattle Baron:430.00:
20180805:Dischem::1233.39
20180805:Checkers::606.71
20180901:PNP:2000.00:
20181001:WW::150.00

1) How to SUM the values in column B ...
```
```
Er, that's column D, then?!

```
```... for September i.e. 201809
```
```
That's easy: there are no such values, so the answer is zero!

```
```Pointers and solutions GREATLY appreciated.
```
```
```
I'm having to guess that your date values are actually plain eight-digit numbers; if they are date values formatted similarly, you will need to modify my suggestion.
```
Try =SUMPRODUCT(INT(A2:A99/100)=201809;D2:D99)

Explanation:
o A2:A99 is the array of date values (as integers).
```
o Dividing those by 100 and taking the integral part of the result - INT(A2:A99/100) - gives the year and month values only. o The expression INT(A2:A99/100)=201809 then gives an array of boolean values, TRUE for September 2018 and FALSE otherwise. o The SUMPRODUCT function then sums the numerical products of members of the array of boolean values and corresponding members in column D (your "B"). In this context, the boolean values TRUE and FALSE are interpreted as 1 and 0 respectively, so September 2018 values will be included but others not.
```
```
Note: although SUMPRODUCT() handles arrays it returns a single value, so it is not necessary to enter it as an array function.
```
I trust this helps.

Brian Barker

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