Print lines above and below regex 1

[cryptoadm]

abc = 1234
def = nnnn
ghi = XXX
jkl = yes

abc = 2345
def = nnoo
ghi = ZZZ
jkl = yes

abc = 6789
def = uuuu
ghi = XXX
jkl = no

I have output similar to the above and I can use sed to print lines above a regex or lines below. But how do I print above and below if I match a regex which in this case is XXX?

Example:

1234
nnnn
XXX
yes

6789
uuuu
XXX
no

Thanks!

 

[Annihilannic]

Try this:

awk '
        # strip off the beginning of each line and add to holding variable
        { sub(".*= *","");h=h $0 "\n" }
        # match found, print holding variable and subsequent line
        /XXX/ { printf h; getline; sub(".*= *",""); printf $0 "\n\n" }
        # blank line found, empty the holding variable for next record
        /^$/ { h="" }
' inputfile > outputfile

Annihilannic.

 

[cryptoadm]

Thank you.