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How to use awk to match only the given string and not anything else – CentOS

I have this file containing these contents

NAME="CentOS Linux"
VERSION="7 (Core)"
ID="centos"
ID_LIKE="rhel fedora"
VERSION_ID="7"
PRETTY_NAME="CentOS Linux 7 (Core)"
ANSI_COLOR="0;31"
CPE_NAME="cpe:/o:centos:centos:7"
HOME_URL="https://www.centos.org/"
BUG_REPORT_URL="https://bugs.centos.org/"

CENTOS_MANTISBT_PROJECT="CentOS-7"
CENTOS_MANTISBT_PROJECT_VERSION="7"
REDHAT_SUPPORT_PRODUCT="centos"
REDHAT_SUPPORT_PRODUCT_VERSION="7"

I need to awk only the ID="centos" part. I tried:

awk /ID=/{print} file.txt

But this gives the output both ID="centos" and VERSION_ID="7"

How to only awk the string given in the pattern?

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4

Answers


  1. 0

    Could you please try following(grep could be also used since question is about awk so going with it).

    awk '/^ID="/' Input_file
    • Use an anchor ^ to check if line starts from ID=.
    • Then add ID=" to make sure it only Matches ID= string NOT anything else.
    • Why string VERSION_ID= will not be matched because it is NOT starting from ID it contains ID, which is why we used ^ in our condition check.
    • awk works on method of pattern/condition then action so no action is mentioned here, that’s why default action printing of current line will happen.
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  2. 0

    You can use awk if you like, but the normal tool for the job is simply grep with the REGEX anchored at the beginning of the line, e.g.

    $ grep '^ID=' /etc/os-release
ID="centos"

    (that doesn’t mean there is anything wrong with using awk, but generally here you would think grep. I presume the file you are parsing is /etc/os-release)

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  3. 0

    Your awk code with ^ goes perfectly:

    awk /^ID=/{print} file
ID="centos"
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  4. 0

    Use this for exact matching:

    $ awk -F= '$1=="ID"' file

    Output:

    ID="centos"

    If you need the field separator to be something else, use:

    $ awk 'split($0,a,/=/)&&a[1]=="ID"' file
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