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stackbiz
Asked: 2024-05-13 17:37:58 +0800 CST

How to get package version in one command line in bash [duplicate]

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I'm trying to get the installed version of systemd in the following command, but it returns all lines contain the keyword "systemd".

# dpkg -l | grep " systemd "

ii  gnome-logs                                       42.0-1                                                           amd64        viewer for the systemd journal
ii  libsystemd0:amd64                                249.11-0ubuntu3.12                                               amd64        systemd utility library
ii  systemd                                          249.11-0ubuntu3.12                                               amd64        system and service manager
ii  systemd-container                                249.11-0ubuntu3.12                                               amd64        systemd container/nspawn tools

How to make it return only the following line:

# dpkg -l | grep "__the_rule_for_systemd_"

ii  systemd                                          249.11-0ubuntu3.12                                               amd64        system and service manager

And then, I can use awk to get the version:

# dpkg -l | grep "__the_rule_for_systemd_" | awk '{print $3}'

Here is the expected output:

249.11-0ubuntu3.12

How to write the "__the_rule_for_systemd_" for grep or is there any other command can get the installed version of systemd?

Note:

# package_name="systemd"
# dpkg -l | grep " $package_name "

The "__the_rule_for_systemd_" should also work on other packages, it should be able to get any package by that rule, not just only for "systemd".

command-line

1 Answers

  1. Best Answer

    This should do it:

    apt-cache policy systemd | grep Installed

    If you only want to return the version number (for a script), search and cut the string with awk:

    apt-cache policy systemd | awk '/Installed/ {print $2}'
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