11-01-2010
Print the above and below lines for the grep pattern.
Hi,
i would like to get the above and below lines of the grep pattern .
For ex :
file as below:
chk1- aaaa
1-Nov
chk2 -aaaa
##########
chk1-bbbbbb
1-Nov
chk2-bbbbbb
#########
my search pattern is date : 1-Nov
i need the o/p as below
chk1- aaaa
1-Nov
chk2 -aaaa
chk1-bbbbbb
1-Nov
chk2-bbbbbb
Thanks in Advance
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GREP(1) General Commands Manual GREP(1)
NAME
grep - search a file for a pattern
SYNOPSIS
grep [ option ... ] pattern [ file ... ]
DESCRIPTION
Grep searches the input files (standard input default) for lines (with newlines excluded) that match the pattern, a regular expression as
defined in regexp(6). Normally, each line matching the pattern is `selected', and each selected line is copied to the standard output.
The options are
-c Print only a count of matching lines.
-h Do not print file name tags (headers) with output lines.
-i Ignore alphabetic case distinctions. The implementation folds into lower case all letters in the pattern and input before interpre-
tation. Matched lines are printed in their original form.
-l (ell) Print the names of files with selected lines; don't print the lines.
-L Print the names of files with no selected lines; the converse of -l.
-n Mark each printed line with its line number counted in its file.
-s Produce no output, but return status.
-v Reverse: print lines that do not match the pattern.
Output lines are tagged by file name when there is more than one input file. (To force this tagging, include /dev/null as a file name
argument.)
Care should be taken when using the shell metacharacters $*[^|()= and newline in pattern; it is safest to enclose the entire expression in
single quotes '...'.
SOURCE
/sys/src/cmd/grep.c
SEE ALSO
ed(1), awk(1), sed(1), sam(1), regexp(6)
DIAGNOSTICS
Exit status is null if any lines are selected, or non-null when no lines are selected or an error occurs.
GREP(1)