08-10-2011
print lines between line number
Hi,
Anyone help me to print the lines from the flat file between 879th line number and 1424th line number.
The 879 and 1424 should be passed as input to the shell script(It should be dynamic).
Can any one give me using sed or awk?
I tried using read, and print the lines..Its taking too much time for bigger files.
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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)