Rude Awakening
New Member
- Joined
- Jul 10, 2014
- Messages
- 36
- Office Version
- 365
Hey All,
I have a conundrum,
I'm matching two values in a table and returning another value from that table to another
I have the below index match formula which works like a charm to a point. After the 37th value is returned, it wont return any more after that.
It would be easier to use the first formula as more data will be added to the main table and it saves changing formula numbering. Any ideas?
Formula1
Index/Match - Wont return values after the 37th
=IFERROR
(INDEX(Table251922242531346364[[#All],[CSI fields]] , <The column to Index>
MATCH(Table25192224253134636668697231676869120[@['#]]&$B$1, <matches a country in one table to a list of countries in another table>
Table251922242531346364[[#All],[Count]]&Table251922242531346364[[#All],[Country]]<matches a Column number in one table to a list of column numbers in other table>
,0)),"")
Formula2
Index/Match - Will return all required values
=IFERROR
(INDEX('All Country Fields'!E$3:E$943, <The column to Index>
MATCH(1,($B$1='All Country Fields'!D$3:D$943)<matches a country in one table to a list of countries in another table>
*(Belgium!B3='All Country Fields'!C$3:C$943)<matches a country in one table to a list of countries in another table>
,0)),"")
EDIT:
On retest, formula 2 will only return as far as value 47
I have a conundrum,
I'm matching two values in a table and returning another value from that table to another
I have the below index match formula which works like a charm to a point. After the 37th value is returned, it wont return any more after that.
It would be easier to use the first formula as more data will be added to the main table and it saves changing formula numbering. Any ideas?
Formula1
Index/Match - Wont return values after the 37th
=IFERROR
(INDEX(Table251922242531346364[[#All],[CSI fields]] , <The column to Index>
MATCH(Table25192224253134636668697231676869120[@['#]]&$B$1, <matches a country in one table to a list of countries in another table>
Table251922242531346364[[#All],[Count]]&Table251922242531346364[[#All],[Country]]<matches a Column number in one table to a list of column numbers in other table>
,0)),"")
Formula2
Index/Match - Will return all required values
=IFERROR
(INDEX('All Country Fields'!E$3:E$943, <The column to Index>
MATCH(1,($B$1='All Country Fields'!D$3:D$943)<matches a country in one table to a list of countries in another table>
*(Belgium!B3='All Country Fields'!C$3:C$943)<matches a country in one table to a list of countries in another table>
,0)),"")
EDIT:
On retest, formula 2 will only return as far as value 47
Last edited by a moderator: