Using Mid and If function together

[Jeffy]

I'm trying to use a MID & IF function together, but it's not quite working.

In column J, I have a set of project numbers where the 2nd character of that project number indicates the year. In the destination column, I want to return the four digit year of that project number.

PROJECT #----------------YEAR

W63BT015141B-----------2016
F53BT015141C------------2015

The possible years would go back to 2010.

The function I'm trying to use is as follows, but I'm getting a return of "FALSE."

=IF(MID(J16,2,1)=6,2016,IF(MID(J16,2,1)=5,2015,IF(MID(J16,2,1)=4,2014,IF(MID(J16,2,1)=3,2013,IF(MID(J16,2,1)=2,2012,IF(MID(J16,2,1)=1,2011,IF(MID(J16,2,1)=0,2010)))))))

Furthermore, we have some project numbers which are unique in that they don't indicate a year but rather a alpha character, so I would have to manually check those instances. How can I indicate in my formula that if the 2nd digit doesn't contain a number value, return the word "Check" to indicate the need to manually check that instance?

Thanks!!

[ChemistB]

=IFERROR((201&MID(J16,2,1))+0, "Check")

[Gatti]

The problem is you are trying to use a Text Function and find a Value...

If you use MID(J16,2,1), the result will be "number" e.g.: J16 = 16, so the return of MID(J16,2,1) will be "6"... "6" is different then 6

you have to put a VALUE() function before each MID function...

By the way, use the @ChemistB's function... then you don't need to do all those "IF's"

[Jeffy]

This worked beautifully, thank you!