How to compare Debian package versions?

Perhaps because the title doesn't mention Python (though the tags do), Google brought me here when asking the same question but hoping for a bash answer. That seems to be:

$ dpkg --compare-versions 11a lt 100a && echo truetrue$ dpkg --compare-versions 11a gt 100a && echo true$ 

To install a version of rubygems that's at least as new as the version from lenny-backports in a way that gives no errors on lenny and squeeze installations:

sudo apt-get install rubygems &&VERSION=`dpkg-query --show --showformat '${Version}' rubygems` &&dpkg --compare-versions $VERSION lt 1.3.4-1~bpo50+1 &&sudo apt-get install -t lenny-backports rubygems

Perhaps I should have asked how to do that in a separate question, in the hope of getting a less clunky answer.

You could use apt_pkg.version_compare:

import apt_pkgapt_pkg.init_system()a = '1:1.3.10-0.3'b = '1.3.4-1'vc = apt_pkg.version_compare(a,b)if vc > 0:    print('version a > version b')elif vc == 0:    print('version a == version b')elif vc < 0:    print('version a < version b')        

yields

version a > version b

Thanks to Tshepang for noting in the comments thatfor newer versions: apt.VersionCompare is now apt_pkg.version_compare.

python-debian can do this too. It's used in an almost identical way to python-apt:

from debian import debian_support a = '1:1.3.10-0.3'b = '1.3.4-1'vc = debian_support.version_compare(a,b)if vc > 0:    print('version a > version b')elif vc == 0:    print('version a == version b')elif vc < 0:    print('version a < version b')

ouput:

version a > version b