How do i grep one line before and after a matching pattern in a file?

hawksbill

Man page for grep: https://www.lightnetics.com/post/3317

The contents of the file (named grepfile) is:

$ cat grepfile
apples
pears
This is the first block
oranges
plums

apples
pears
This is the second block
oranges
plums

apples
pears
This is the third block
oranges
plums

The -A and -B options as per the man page.

Context Line Control
       -A NUM, --after-context=NUM
	      Print  NUM  lines  of  trailing  context	after  matching lines.
	      Places  a  line  containing  a  group  separator	(--)   between
	      contiguous  groups  of  matches.	With the -o or --only-matching
	      option, this has no effect and a warning is given.

       -B NUM, --before-context=NUM
	      Print NUM  lines	of  leading  context  before  matching	lines.
	      Places   a  line	containing  a  group  separator  (--)  between
	      contiguous groups of matches.  With the  -o  or  --only-matching
	      option, this has no effect and a warning is given.

My search string is block.

$ grep -A 1 -B 1 block grepfile
pears
This is the first block
oranges
--
pears
This is the second block
oranges
--
pears
This is the third block
oranges

The before and after -A & -B values can be different. If you just a consistent number before and after you can also use -C

-C NUM, -NUM, --context=NUM
	      Print  NUM  lines of output context.  Places a line containing a
	      group separator (--) between contiguous groups of matches.  With
	      the  -o  or  --only-matching  option,  this  has no effect and a
	      warning is given.

$ grep -C 1 block grepfile
pears
This is the first block
oranges
--
pears
This is the second block
oranges
--
pears
This is the third block
oranges