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ortiga
Asked: 2020-01-24 11:27:05 +0800 CST 2020-01-24 11:27:05 +0800 CST

How do I select the first number of a string?

  • 772

The command cat /etc/redhat-releasereturns:

 CentOS Linux release 5.6.1804 (Core)

How do I select only the 5?

i tried

 cat /etc/redhat-release | awk '{print $4}'

and it returns this:

5.6.1804

but I only want the first number.

bash

3 Answers

  1. Best Answer

    You can with a combination of awkandcut

    $ awk '{print $4}' /etc/redhat-release | cut -f 1 -d .

    Or withgrep

    $ grep -oP "(?<= )[[:digit:]]+(?=.)" /etc/redhat-release

    Which gets the digits between a space and a period.

    Another variant of grepis:

    $ grep -oE "[[:digit:]]+" /etc/redhat-release | head -n 1
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  2. If you want to use awkonly you would have to get the first of the digits, which you can easily solve using split():

    awk '{split($4, release, ".");print release[1]}' /etc/redhat-release

    This function will build an array separating the release value for each point. The first value of this array will be the version number you are looking for.

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  3. $ awk '{print $4}' /etc/redhat-release |awk -F'.' '{print $1}'

    The first awk allows the fourth field of the line to be selected, then another awk is added to process the output of the first and takes the period ( . ) as the separator. In this way, the first field of the first output is selected and thus give a solution to the problem posed.

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