Quote:
Originally Posted by ArviLaanemets
Is this INDEX() formula entered into second Defined Table,
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No.
Quote:
Originally Posted by ArviLaanemets
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?
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Yes.
Quote:
Originally Posted by ArviLaanemets
When former, as I suspect, then the second Table will not expand automatically when new rows are added to first Table.
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Correct. All my suggestions require the results table NOT to be a proper Excel table.
Quote:
Originally Posted by p45cal
The result is NOT a proper Excel Table, and it won't work if you try to make it a table.
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Quote:
Originally Posted by ArviLaanemets
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
Code:
=FILTER(tData[Hdr3:Hdr5],tData[Hdr6]=1)
will probably work. And the returned range will expand automatically when Table tData expands. |
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).