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Top Forums Shell Programming and Scripting Grep pattern and display all lines below Post 302906085 by venkidhadha on Tuesday 17th of June 2014 04:55:39 AM
Old 06-17-2014
Grep pattern and display all lines below

Hi I need to grep for a patter and display all lines below the pattern.

For ex: say my file has the below lines

file1
file2
file3
file4
file5

I NEED to grep for patter file3 and display all lines below the pattern. do we have an option to get this data. Let me know if you require any additional information on the same.
 

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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)
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