How do you grep a file and get the next 5 lines

Question

How do I grep a file for 19 55 and get the Line 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

User · Answer

Some awk version   awk   19 55  c 5  c-- gt 0  awk   19 55  c 5  c  amp  amp  c--    When pattern found  set c 5 If c is true  print and decrease number of c

User · Answer

Here is a sed solution   sed   19 55   N N N N N s  n   g    file txt

User · Answer

You want    grep -A 5  19 55  file   From 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

[shell] How do you grep a file and get the next 5 lines

Examples related to shell

Examples related to grep