10-09-2012
Get 20 lines above string found, and 35 below string
i want to search a log for a string. when that string is found, i want to grab the a set number of lines that came before the string, and a set number of lines that come after the string.
so if i search for the word "Error" in the /var/log/messages file, how can I output the 20 lines that came directly before it and the 35 lines that come after it..and also output the line that contained the specified string of "Error"?
im thinking a mixture of sed and awk can work here?
OS: Linux / SunOS
shell: bash
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GREP(1) General Commands Manual GREP(1)
NAME
grep - search a file for lines containing a given pattern
SYNOPSIS
grep [-elnsv] pattern [file] ...
OPTIONS
-e -e pattern is the same as pattern
-c Print a count of lines matched
-i Ignore case
-l Print file names, no lines
-n Print line numbers
-s Status only, no printed output
-v Select lines that do not match
EXAMPLES
grep mouse file # Find lines in file containing mouse
grep [0-9] file # Print lines containing a digit
DESCRIPTION
Grep searches one or more files (by default, stdin) and selects out all the lines that match the pattern. All the regular expressions
accepted by ed and mined are allowed. In addition, + can be used instead of * to mean 1 or more occurrences, ? can be used to mean 0 or 1
occurrences, and | can be used between two regular expressions to mean either one of them. Parentheses can be used for grouping. If a
match is found, exit status 0 is returned. If no match is found, exit status 1 is returned. If an error is detected, exit status 2 is
returned.
SEE ALSO
cgrep(1), fgrep(1), sed(1), awk(9).
GREP(1)