Convert file to csv from bash with sed

Question

I have a file with a few lines like this:

-host hostname.domain.com -os "gpl x86_64 linux-5.3.18-24.52-default" -core A.10.20 -corepatch A.10.20 -ts_core A.10.20 -ts_corepatch A.10.20 -da A.10.20 -dapatch A.10.20

I would like to convert it in the shell to a file with tabs with just the first 2 fields and skip everything else:

hostname.domain.com     gpl x86_64 linux-5.3.18-24.52-default

Maybe with sed? I suck with these commands.

Answer 1

You can make it work with cut/awk. Here is example with cut:

[user@host~]$ echo '-host hostname.domain.com -os "gpl x86_64 linux-5.3.18-24.52-default" -core A.10.20 -corepatch A.10.20 -ts_core A.10.20 -ts_corepatch A.10.20 -da A.10.20 -dapatch A.10.20' | cut -d " "  -f2,4,5,6
hostname.domain.com "gpl x86_64 linux-5.3.18-24.52-default"

cut -d " " will use " " as delimeter and print fields in -f flag.

And to remove ", you can use tr.

tr -d "\""

Answer 2

$ cat your_file | awk '{ printf "%s\t%s\n", $1, $2 }'

Awk might do it: https://www.gnu.org/software/gawk/manual/html_node/Printf-Examples.html

Awk is super versatile with this kind of thing, particularly because it can leverage the amazing printf .