# Christoph's last Weblog entries

Installing a python systemd service?
26th October 2016

As web search engines and IRC seems to be of no help, maybe someone here has a helpful idea. I have some service written in python that comes with a .service file for systemd. I now want to build&install a working service file from the software's setup.py. I can override the build/build_py commands of setuptools, however that way I still lack knowledge wrt. the bindir/prefix where my service script will be installed.

### Solution

Turns out, if you override the install command (not the install_data!), you will have self.root and self.install_scripts (and lots of other self.install_*). As a result, you can read the template and write the desired output file after calling super's run method. The fix was inspired by GateOne (which, however doesn't get the --root parameter right, you need to strip self.root from the beginning of the path to actually make that work as intended).

As suggested on IRC, the snippet (and my software) no use pkg-config to get at the systemd path as well. This is a nice improvement orthogonal to the original problem. The implementation here follows bley.


def systemd_unit_path():
try:
command = ["pkg-config", "--variable=systemdsystemunitdir", "systemd"]
path = subprocess.check_output(command, stderr=subprocess.STDOUT)
return path.decode().replace('\n', '')
except (subprocess.CalledProcessError, OSError):
return "/lib/systemd/system"

class my_install(install):
_servicefiles = [
'foo/bar.service',
]

def run(self):
install.run(self)

if not self.dry_run:
bindir = self.install_scripts
if bindir.startswith(self.root):
bindir = bindir[len(self.root):]

systemddir = "%s%s" % (self.root, systemd_unit_path())

for servicefile in self._servicefiles:
service = os.path.split(servicefile)[1]
self.announce("Creating %s" % os.path.join(systemddir, service),
level=2)
with open(servicefile) as servicefd: