last year occurrence from string

Question

I have strings like this:

ACB 01900 X1911D 1910 1955-2011 3424 2135 1934 foobar

I'm trying to get the last occurrence of a single year (from 1900 to 2050), so I need to extract only 1934 from that string.

I'm trying with:

 grep -P -o '\s(19|20)[0-9]{2}\s(?!\s(19|20)[0-9]{2}\s)'

or

grep -P -o '((19|20)[0-9]{2})(?!\s\1\s)'

But it matches: 1910 and 1934

Here's the Regex101 example:

https://regex101.com/r/UetMl0/3

https://regex101.com/r/UetMl0/4

Plus: how can I extract the year without the surrounding spaces without doing an extra grep to filter them?

Answer 1

Have you ever heard this saying:

Some people, when confronted with a problem, think
“I know, I'll use regular expressions.”   Now they have two problems. 

Keep it simple - you're interested in finding a number between 2 numbers so just use a numeric comparison, not a regexp:

$ awk -v min=1900 -v max=2050 '{yr=""; for (i=1;i<=NF;i++) if ( ($i ~ /^[0-9]{4}$/) && ($i >= min) && ($i <= max) ) yr=$i; print yr}' file
1934

You didn't say what to do if no date within your range is present so the above outputs a blank line if that happens but is easily tweaked to do anything else.

To change the above script to find the first instead of the last date is trivial (move the print inside the if), to use different start or end dates in your range is trivial (change the min and/or max values), etc., etc. which is a strong indication that this is the right approach. Try changing any of those requirements with a regexp-based solution.