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Top Forums Shell Programming and Scripting shell script to remove all lines from a file before a line starting with pattern Post 302301903 by raksha.s on Sunday 29th of March 2009 06:43:52 AM
Old 03-29-2009
shell script to remove all lines from a file before a line starting with pattern

hi,,
i hav a file with many lines.i need to remove all lines before a line begginning with a specific pattern from the file because these lines are not required.
Can u help me out with either a perl script or shell script
example:-
if file initially contains lines:
a
b
c
d
.1.2
d
e
f
now i need the file after executing script to contain lines:
.1.2
d
e
f
 

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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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