Find/return the next value in the column list

New Contributor



I have a list of numbers in a column.


A have a number in another cell that has been selected from that list.


Is it possible to find this number in the list (a la vlookup) and then access/return the value that is in the next cell in the list, the value in the cell below the current one?



13 Replies
best response confirmed by wazza2040 (New Contributor)


You may find with MATCH() position of the number in the list, and with INDEX() return value from the cell in next position. If, for exmple, your list is in column A and number to search is in cell B1, it could be

=IFERROR(INDEX(A:A,MATCH(B1,A:A,0)+1),"no such number")



VLOOKUP always has been a flawed function for a number of reasons,  INDEX/MATCH would be better here.  Calling you list and number by those names, the simplest formula is

= INDEX( list, 1 + MATCH( number, list, 0 ) )

Instead of adding 1 to the index it is also possible to define 'offsetList' to be a range one cell down from the initial list:

= INDEX( offsetList, MATCH( number, list, 0 ) )

Switching to the latest versions of Office 365, one has the new XLOOKUP function which replaces other lookup function in almost all circumstances.  Using the offset list once more

= XLOOKUP( number, list, offsetList )

or if volatile functions are not a problem

= OFFSET( XLOOKUP( number, list, list ), 1, 0 )

Many thank.

@wazza2040 , you are welcome

@Sergei Baklan   What if you don't know the next value in the column? I have a column of random dates and next to it is a column of random values that don't populate every cell. I want to create a table that picks up just the date where there is a value, and then the corresponding value. Thanks!


7/28/2017     6,900,439.07
9/21/2017          52,618.28
10/20/2017        302,730.65
11/22/2017        293,637.39


Depends on your Excel version, for example



As @Sergei Baklan says, it depends on the version of Excel that you are using.

    IF(value>0, date),

I think XLOOKUP is Excel 2019 and on.  Th final '1' selects the next higher value rather than the next lower that would be returned by the old LOOKUP function.

@Sergei Baklan it is showing the function as invalid.




On which Excel version you are?

Try to open attached file if it works in your environment. It replicates your sample


Thanks @Sergei Baklan.

even your excel shows the formula as 

Mine is 2016. maybe the code is not supported.




Okay, for 2016 it could be


In E1
=IFERROR( INDEX( A:A,  AGGREGATE(15,6, 1/(B:B<>"")*(ROW(A:A)-ROW($A$1)+1), ROW()-ROW($E$1)+1 ) ), "" )

In F1
=IFERROR( INDEX( B:B,  AGGREGATE(15,6, 1/(B:B<>"")*(ROW(A:A)-ROW($A$1)+1), ROW()-ROW($E$1)+1 ) ), "" )

and drag them down till empty cells appear.  If you ranges start in another cells above shall be adjusted accordingly.

Yes this works! thank you.

@RADS_JAS1360 , you are welcome