capturing words between double quotes in awk

Question

I'm trying to print the name of my Linux distribution. The output of cat /etc/os-release for my distro is:

NAME="Arch Linux"
PRETTY_NAME="Arch Linux"
ID=arch
BUILD_ID=rolling
ANSI_COLOR="0;36"
HOME_URL="https://www.archlinux.org/"
DOCUMENTATION_URL="https://wiki.archlinux.org/"
SUPPORT_URL="https://bbs.archlinux.org/"
BUG_REPORT_URL="https://bugs.archlinux.org/"
LOGO=archlinux

Now I want to grab the Arch Linux from the second line. I used this command:

cat /etc/os-release | awk '/PRETTY_NAME=\".+\"/ {print $1}'

But this prints PRETTY_NAME="Arch

How can I grab the words (including space) from this line?

suspectus · Accepted Answer · 2019-08-10T14:48:47.507

1

Setting the field delimiter to double quotes -F'"' to remove quotes from output.

 awk -F'"' '/^PRETTY_NAME=/{print $2}' /etc/os-release

edited Aug 10 '19 at 14:48

answered Aug 10 '19 at 10:51

suspectus

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This includes the double quotes, but I guess I could pipe it to `tr -d '"'`. Thanks. – Amir Shabani Aug 10 '19 at 10:52
Yes, or change the awk field delimiter to " – suspectus Aug 10 '19 at 10:56
2

You'll want to lose the [useless `cat`](/questions/11710552/useless-use-of-cat) too. – tripleee Aug 10 '19 at 11:00
You don't need `tr`, Awk can do that with `gsub("\"","")` – tripleee Aug 10 '19 at 11:01
1

`/^PRETTY_NAME=/{print $2}` or similar would be more robust. – Ed Morton Aug 10 '19 at 14:30

capturing words between double quotes in awk

1 Answers1