I can't test it out currently, but maybe this will work:


Have calculated columns in your Table 1, but hide them. This ensures calculations are made when data are read in, but the Table will work like with 2 columns only;
Try to use FILTER() to get wanted report data displayed separately (I'm currently only aware there is such function in latest versions, but I have never used it. You have to test yourself, is it possible to get all 3 columns of data returned without applying any conditions).

Yes, with the likes of:
=INDEX(Table13,ROW(Table13)-ROW(Table13[#Headers]),{1,3,4})
See Sheet1 (2) of the attached.


If you want to include headers in the result:
=INDEX(Table13[#All],ROW(Table13[#All])-ROW(Table13[#Headers])+1,{1,3,4})

Attached Files

msofficeforums50314.xlsx (13.1 KB, 4 views)

I haven't Excel available currently to look at attached file.

Is this INDEX() formula entered into second Defined Table, or it behaves like those newest Excel features, where you enter the formula {like FILTER()} into single cell, and you get a range of values returned? When former, as I suspect, then the second Table will not expand automatically when new rows are added to first Table.

I think I found a probable solution with FILTER(). When the 1st Table has name e.g tData, with headers like Hdr1, Hdr2 for visible columns, and then for hidden columns headers like Hdr3 for column with formula
=tData[@Hdr1]
, Hdr4 and Gdr5 for columns for OP's calculations, and an additional column Hdr6 with formula
=1,
then formula
will probably work. And the returned range will expand automatically when Table tData expands.

No.

Yes.

Correct. All my suggestions require the results table NOT to be a proper Excel table.
Yes, it will probably work. The first argument of the FILTER function is the array you're going to filter, but in this case you're 'filtering in' all rows (not filtering at all), so really, you only need the first argument.
This should work:
=tData[Hdr3:Hdr5]
but in the OP's case the columns are not adjacent; he wants columns 1, 3 & 4

I would have expected
=INDEX(Table13,,{1,3,4})
or
=INDEX(Table13,0,{1,3,4})
to be enough but experimentation showed it wasn't (only one row returned).
It wants to see the likes of:
=INDEX(Table13,{1;2;3;4;5;6…},{1,3,4})
(not a real formula) hence the extra bits there in my last message (msg#10).