02-03-2009
print specific lines
I have a text file made of different blocks separated by blank lines. I need to print the blocks with odd indexes. How can I get it with awk?
For example i need to print the first and the third block of a file like this:
asgdg sadsd ssgsdgd
ass uff fedd sddddso
ieeduydd dddee deeo
ssancnc chhdh cchdd
aasi askf fkf kff
sskkkk dddu dduu
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TAIL(1) General Commands Manual TAIL(1)
NAME
tail - deliver the last part of a file
SYNOPSIS
tail [ +-number[lbc][rf] ] [ file ]
tail [ -fr ] [ -n nlines ] [ -c ncharacters ] [ file ]
DESCRIPTION
Tail copies the named file to the standard output beginning at a designated place. If no file is named, the standard input is copied.
Copying begins at position +number measured from the beginning, or -number from the end of the input. Number is counted in lines, 1K
blocks or characters, according to the appended flag or Default is -10l (ten ell).
The further flag causes tail to print lines from the end of the file in reverse order; (follow) causes tail, after printing to the end, to
keep watch and print further data as it appears.
The second syntax is that promulgated by POSIX, where the numbers rather than the options are signed.
EXAMPLES
tail file
Print the last 10 lines of a file.
tail +0f file
Print a file, and continue to watch data accumulate as it grows.
sed 10q file
Print the first 10 lines of a file.
SOURCE
/sys/src/cmd/tail.c
BUGS
Tails relative to the end of the file are treasured up in a buffer, and thus are limited in length.
According to custom, option +number counts lines from 1, and counts blocks and characters from 0.
TAIL(1)