IF [Employee SID] = [Row+1:Employee SID] OR [Employee SID] = [Row-1:Employee SID] AND [Error Type] = [Row+1:Error Type] OR [Error Type] = [Row-1:Error Type]
THEN 1
ELSE 0
ENDIF
This is capturing the following 2 entries as a match and returning a "1" when it should be a "0" because the Error Type does not match, only the Employee SID is a match:
Employee SID = 123456
Error Type = Complaint
Employee SID = 123456
Error Type = Privacy
Can someone please assist? Thank you!
Solved! Go to Solution.
@bh1789 Can you provide sample input data?
Seems to be ok as written. As binuacs mentioned, can you provide a sample data set?
Employee SID
Employee SID D444111
Error Type Complaint
Employee SID D444111
Error Type Privacy
@bh1789 The matches, 1, is due to the error type. The above line has the same error type, same with the bottom line, has the same error type. That is what is failing in your formula.
the end of your formula
OR [Error Type] = [Row-1:Error Type]
Replace the last OR with AND
that did not work and instead missed the 9 that should have returned a "1"
@bh1789 another way of finding the non-matching error type for the same sid