find search item plus 4 lines before and after - regex

Question

I am using notepad++ and would like to find the context in which a particular string occurs.
So the search string is 0wh.*0subj and I would like to find this search item plus 4 lines immediately before and after it.
eg: xxx means whatever is on a new line. the search result should be:
xxx
xxx
xxx
xxx
0wh.*0subj
xxx
xxx
xxx
xxx
I have tried using \n\r but its not working. Any assistance afforded would be greatly appreciated.
Regards

Answer 1

This will work in Notepad++ (tested):
(?m)(^[^\r\n]*\R+){4}0wh\.\*0subj[^\r\n]*\R+(^[^\r\n]*\R+){4}
On the screenshot, note that the 555 line is not selected. It is just the current line.
Explain Regex
(?m) # set flags for this block (with ^ and $
# matching start and end of line) (case-
# sensitive) (with . not matching \n)
# (matching whitespace and # normally)
( # group and capture to \1 (4 times):
^ # the beginning of a "line"
[^\r\n]* # any character except: '\r' (carriage
# return), '\n' (newline) (0 or more times
# (matching the most amount possible))
\R+ # 'R' (1 or more times (matching the most
# amount possible))
){4} # end of \1 (NOTE: because you are using a
# quantifier on this capture, only the LAST
# repetition of the captured pattern will be
# stored in \1)
0wh # '0wh'
\. # '.'
\* # '*'
0subj # '0subj'
[^\r\n]* # any character except: '\r' (carriage
# return), '\n' (newline) (0 or more times
# (matching the most amount possible))
\R+ # 'R' (1 or more times (matching the most
# amount possible))
( # group and capture to \2 (4 times):
^ # the beginning of a "line"
[^\r\n]* # any character except: '\r' (carriage
# return), '\n' (newline) (0 or more times
# (matching the most amount possible))
\R+ # 'R' (1 or more times (matching the most
# amount possible))
){4} # end of \2 (NOTE: because you are using a
# quantifier on this capture, only the LAST
# repetition of the captured pattern will be
# stored in \2)