How to fix the #SPILL! Error by displaying only the second value?

I have a column with some info displayed like that:

I have a formula to extract the year in the above column that is:

=IFERROR(FILTERXML("<k><m>"&SUBSTITUTE([@[Product Name]]," ","</m><m>")&"</m></k>","//m[.=number() and string-length()=4]"),"")

The problem with this formula is that it works fine with the first case, but it gives me a #SPILL! Error on the other two cases. My ideal output would be:

• Basically, for the first case, just return the 4 digits. EVERY time that I only have one sequence of 4 digits, I want to return that sequence.
• For the second case, I want to return ONLY the second year. EVERY time I have 2 sequences of 4 digits, I want to return ONLY the second year.
• For the third case, I want to return nothing. EVERY time I have more than 2 sequences of 4 digits, I want to return blank.

The last thing I tried to add was position()>5 and that would cut off the 1995 in the second example, but I would continue having the Error on the third example. Also, my list is quite huge, and I am not sure if the position()>5 thing would work for ALL products that fall in the same second example.

I am not very good with XPATH, so any help would be greatly appreciated. Thank you!

_{Disclaimer: Below solution is written on the assumption that when 'count of years < 3', return the last given year. If 'count >= 3' then only return the last year if years come in pairs of two. Hence the use of 'modulus 2 == 0'.}^‡

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You can expand the xpath for sure if you so desire. However, I'd rewrite it a little bit. Each predicate, the structure between the opening and closing square brackets, is a filter of a given nodelist. To write multiple of these structures is in fact anding such predicates. To get a better understanding of what most common xpath 1.0 functions can do within FILTERXML(), I'd like to redirect you to this post.

So to write a consecutive pattern of predicates I'd opt for:

• [.*0=0] - First return a filtered nodelist of all numbers where a node multiplied by zero equals zero;
• [string-length()=4] - Then return only those that are 4 characters long^‡‡;
• [position() = last() and (position() = 1 or position() mod 2 = 0)] - The 3rd and last predicate is the trickiest for your query. This is done with a first check that position() = last() meaning the node needs to be the last node in the filtered nodelist of step 2 and (position() = 1 or position() mod 2 = 0) means we want to check that this node is also at the 1st index or the modulus 2 of the indexed position equals 0^‡‡‡.

Formula in B2:

=IFERROR(FILTERXML("<t><s>"&SUBSTITUTE(A2," ","</s><s>")&"</s></t>","//s[.*0=0][string-length()=4][position() = last() and (position() = 1 or position() mod 2 = 0)]"),"")

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Whilst the above would work for Excel 2013 and higher^‡‡‡‡, you do talk about spilled behaviour. If you happen to work with the current channel in ms365 you could also try:

=LET(x,TEXTSPLIT(A2," "),y,--FILTER(x,ISNUMBER(-(x&"**0"))*(LEN(x)=4),{1,2,3}),z,COUNT(y),IF(OR(z=1,MOD(z,2)=0),TAKE(y,,-1),""))

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^‡ _{If you need to simply return the last year if 'count < 3' then you can use xpath "//s[.*0=0][string-length()=4][position()<3 and position() = last()]" or ms365 formula =LET(x,TEXTSPLIT(A2," "),y,FILTER(x,ISNUMBER(-(x&"**0"))*(LEN(x)=4),""),IF(COUNTA(y)>2,"",TAKE(y,,-1))).}

^‡‡ _{Note that you can be more strict about this if you'd wish to validate that a year is between say 1900-2050 or so. One could replace the 1st and 2nd predicate with [.*1>1899][.*1<2051].}

^‡‡‡ _{Note that the order or writing your and/or statements in xpath do matter. We need to use explicit parentheses to control the precedence. See this}

^‡‡‡‡ _{This is not true for Excel Online or Excel for Mac}