Show only odd lines with cat

Question

HI,

probably is this something very easy, but I do not get the point now. In Linux, shell, given a text file, how can I do something like, cat filename so I only see the first line, the third line, 5, 7, 9 and so on?

if not with cat, with another command

Thanks

Answer 1

Using awk: awk 'NR % 2 == 0'

Edit: removed {print}, as shown in ghostdog74's answer and commented on by Roberto Bonvallet.

Answer 2

or something like

$ sed -n 'p;n' filename

For explanation look-up the man page for sed.

Answer 3

Use sed:

sed -n 1~2p filename

Answer 4

awk '{if(NR%2!=0)print}' myfile

Answer 5

A simple search in Google shows this link: http://bash.cyberciti.biz/academic/write-odd-even-line-file/

You only need to tweak it a little bit.

My 2 cents.

Answer 6

2 ways you can do it with awk

$ awk '{getline;print}' file # the equivalent of the selected sed answer
$ awk 'NR%2==0' file

OR with just the shell

i=1; while read -r line; do [ $((i++ % 2)) -eq 0 ] && echo $line; done <"file"

Answer 7

Unable to test, found a link: cat -n $1 | grep "$ODDRE" | cut -b 8- > oddfile.$$

Answer 8

Haven't tested this (doesn't seem to work on Mac OS X), but GNU sed supports first~step as an address selection, so for example:

sed -n 1~2p file

would start at the first line and print every other line after that.