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)