Using awk to print without double quotes

user_859227 0
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I would like to get the right value of the following command as a string without double quotes.

$ grep '^VERSION=' /etc/os-release
VERSION="20.04.3 LTS (Focal Fossa)"

When I pipe it with the following awk, I don't get the desired output.

$ grep '^VERSION=' /etc/os-release | awk '{print $0}'
VERSION="20.04.3 LTS (Focal Fossa)"
$ grep '^VERSION=' /etc/os-release | awk '{print $1}'
VERSION="20.04.3
$ grep '^VERSION=' /etc/os-release | awk '{print $2}'
LTS

How can I fix that?

awk
5 Answers

You may use this single awk command:

awk -F= '$1=="VERSION" {gsub(/"/, "", $2); print $2}' /etc/os-release

20.04.3 LTS (Focal Fossa)
user_5482250

1st solution: With your shown samples, please try following awk code.

awk 'match($0,/^VERSION="[^"]*/){print substr($0,RSTART+9,RLENGTH-9)' Input_file

Explanation: Simple explanation would be, using match function of awk to match starting VERSION=" till next occurrence of " and then printing the matched part(to get only desired output as per OP's shown samples).

2nd solution: Using GNU grep with PCRE regex enabled option try following.

grep -oP '^VERSION="\K[^"]*' Input_file

3rd solution: Using awk's capability to set different field separators and then check conditions accordingly and print values.

awk -F'"' '$1=="VERSION="{print $2}'  Input_file
user_58665800

Assuming that "the right value" you want output is 20.04.3:

$ awk -F'[" ]' '/^VERSION=/{print $2}' file
20.04.3

or if it's the whole quoted string:

$ awk -F'"' '/^VERSION=/{print $2}' file
20.04.3 LTS (Focal Fossa)
user_17450010

You can use an awk command like

awk 'match($0, /^VERSION="([^"]*)"/, m) {print m[1]}' /etc/os-release

Here, ^VERSION="([^"]*)" matches VERSION=" at the start of the string (^), then captures into Group 1 any zero or more chars other than " (with ([^"]*)) and then matches ". The match is saved in m where m[1] holds the Group 1 value.

Or, sed like

sed -n '/^VERSION="\([^"]*\)".*/s//\1/p' /etc/os-release

See an online test:

s='VERSION="20.04.3 LTS (Focal Fossa)"'
awk 'match($0, /^VERSION="([^"]*)"/, m) {print m[1]}' <<< "$s"
sed -n '/^VERSION="\([^"]*\)".*/s//\1/p' <<< "$s"

Here, -n option suppresses the default line output, /^VERSION="\([^"]*\)".*/ matches a string starting with VERSION=", then capturing into Group 1 any zero or more chars other than ", and then matching " and the rest of the string, and replacing the whole match with the Group 1 value. // means the previous regex pattern must be used. p only prints the result of the substition.

Both output 20.04.3 LTS (Focal Fossa).

user_38329700

Since the file /etc/os-release conforms to a variable assignment in bash or the shell in general (POSIX), sourcing it should do the job.

source /etc/os-release; echo "$VERSION"

Using a subshell just in case one does not want the pollute the current env variables.

( source /etc/os-release; echo "$VERSION" )

Assigning it to a variable.

version=$( source /etc/os-release; echo "$VERSION" )

If the shell you're using does not conform to POSIX.

sh -c '. /etc/os-release; echo "$VERSION"'

See your local man page if available.

man 5 os-release
user_44522650
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