Add blank line after every result in grep

Question

my grep command looks like this zgrep -B bb -A aa \"pattern\" *

I would lke to have output as:

file1:line1
file1:line2
file1:line3
file1:pattern
file         


        

Answer 1

Pipe | any output to:

sed G

Example:

ls | sed G

If you man sed you will see

G Append's a newline character followed by the contents of the hold space to the pattern space.

Answer 2

You can also use the --group-separator parameter, with an empty value, so it'd just add a new-line.

some-stream | grep --group-separator=

Answer 3

pipe grep output through

awk -F: '{if(f!=$1)print ""; f=$1; print $0;}'

Answer 4

There is no option for this in grep and I don't think there is a way to do it with xargs or tr (I tried), but here is a for loop that will do it (for f in *; do grep -H "1" $f && echo; done):

[ 11:58 jon@hozbox.com ~/test ]$ for f in *; do grep -H "1" $f && echo; done

a:1

b:1

c:1

d:1

[ 11:58 jon@hozbox.com ~/test ]$ ll
-rw-r--r--  1 jon  people     2B Nov 25 11:58 a
-rw-r--r--  1 jon  people     2B Nov 25 11:58 b
-rw-r--r--  1 jon  people     2B Nov 25 11:58 c
-rw-r--r--  1 jon  people     2B Nov 25 11:58 d

The -H is to display file names for grep matches. Change the * to your own file glob/path expansion string if necessary.

Answer 5

I can't test it with the -A and -B parameters so I can't say for sure but you could try using sed G as mentioned here on Unix StackEx. You'll loose coloring though if that's important.

Answer 6

Try with -c 2; with printing a context I see grep is separating its found o/p