Come si grep un file e ottenere le 5 linee successive

Question

Come faccio a grep un file per 19:55 e ottenere la riga 1,2,3,4,5?

2013/10/08 19:55:27.471 
Line 1 
Line 2 
Line 3 
Line 4 
Line 5 

2013/10/08 19:55:29.566 
Line 1 
Line 2 
Line 3 
Line 4 
Line 5 

Answer 1

desiderato:

grep -A 5 '19:55' file 

Da man grep:

Context Line Control 

-A NUM, --after-context=NUM 

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

--group-separator=SEP 

Use SEP as a group separator. By default SEP is double hyphen (--). 

--no-group-separator 

Use empty string as a group separator. 

Answer 2

Alcuni awk versione.

awk '/19:55/{c=5} c-->0' 
awk '/19:55/{c=5} c && c--' 

Answer 3

Ecco una soluzione sed:

sed '/19:55/{ 
N 
N 
N 
N 
N 
s/\n/ /g 
}' file.txt