Grep content out of txt file with regex - grep

Question

I have a list in a txt file which looks like this.
10.9.0.18,tom,34.0.1.2:44395,Thu Apr 18 07:14:20 2019
10.9.0.10,jonas,84.32.45.2:44016,Thu Apr 18 07:16:06 2019
10.9.0.6,philip,23.56.222.3:55202,Thu Apr 18 07:16:06 2019
10.9.0.26,coolguy,12.34.56.7:53316,Thu Apr 18 07:16:06 2019
I would like to have a script which provides me with the following output:
tom jonas philip coolguy
I've been looking into something like this:
grep -oP "^10.9.0.*,wq$1-\K.*" | cut -d, -f1 | sort
But I am not quite getting there, getting no output at all.

Answer 1

Extract second field
Replace newlines with spaces
cat <<EOF |
10.9.0.18,tom,34.0.1.2:44395,Thu Apr 18 07:14:20 2019
10.9.0.10,jonas,84.32.45.2:44016,Thu Apr 18 07:16:06 2019
10.9.0.6,philip,23.56.222.3:55202,Thu Apr 18 07:16:06 2019
10.9.0.26,coolguy,12.34.56.7:53316,Thu Apr 18 07:16:06 2019
EOF
cut -d, -f2 | tr '\n' ' '

Answer 2

You're not getting output because grep doesn't return anything (you don't need perl regex for this).
You'll need to select the second field too:
grep '^10\.9\.0\.' data.txt | cut -d, -f

Answer 3

If awk is an option you could try:
awk -F, '{printf "%s ", $2} END {print ""}' file.txt
The {printf "%s ", $2 prevents using the default new line and instead uses a space.
The END {print ""} is to add a new line after finishing

Answer 4

This is the right answer if you want multi-line output:
$ awk -F, '/^10\.9\.0/{print $2}' file
tom
jonas
philip
coolguy
or this for single:
$ awk -F, '/^10\.9\.0/{o=o s $2; s=OFS} END{print o}' file
tom jonas philip coolguy
You need to escape the .s as they represent any character in a regexp and you don't need to add a .* at the end of the regexp as that'll literally match "something or nothing".