How to grep for lines above and below a certain pattern

Question

I would like to search for a certain pattern (say Bar line) but also print lines above and below (i.e 1 line) the pattern or 2 lines above and below the pattern.

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Answer 1

Use grep with the parameters -A and -B to indicate the number a of lines After and Before you want to print around your pattern:

grep -A1 -B1 yourpattern file

• An stands for n lines "after" the match.
• Bm stands for m lines "before" the match.

If both numbers are the same, just use -C:

grep -C1 yourpattern file

Test

$ cat file
Foo  line
Bar line
Baz line
hello
bye
hello
Foo1 line
Bar line
Baz1 line

Let's grep:

$ grep -A1 -B1 Bar file
Foo  line
Bar line
Baz line
--
Foo1 line
Bar line
Baz1 line

To get rid of the group separator, you can use --no-group-separator:

$ grep --no-group-separator -A1 -B1 Bar file
Foo  line
Bar line
Baz line
Foo1 line
Bar line
Baz1 line

From man grep:

   -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.

   -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.