service-bootstrap

Helper for starting up a service in python

Function start_service()

Usage:

from bootstrap.bootstrap import start_service

config, logger, timezone = start_service()

Configuration

start_service() will look for a config YAML file and load it into a dict, using the following algorithm:

• if the env variable CONFIG is defined, it will be used to determine the location of the config file
• otherwise a file ./config.yaml is expected to be present

The file will be parsed and returned as a first returned variable

Logging

start_service() will configure the logger will in the following way:

1. if ./logging.yaml exists, it will be loaded into a dict and used to configure logging (using dictConfig)
2. otherwise the default logging format will be applied, which is defined as writing logs on INFO level to stdout

Additionally, start_service() will look for ./loglevels.yaml file with the following structure:

'apscheduler.scheduler': 'ERROR',
'apscheduler.executors.default': 'WARN'

and use this file to configure per-logger log levels. Note that the settings two mentioned above ('apscheduler.scheduler': 'ERROR' and 'apscheduler.executors.default': 'WARN') will be used as a default.

Finally, start_service() will create a logger called main, log two messages

logger.info("Starting application!")
logger.info("Your timezone is %s" % timezone)

and return this logger as a second returned variable

Timezone

The third returned variable will contain the local timezone as a string (f.ex Europe/Warsaw), detected using the following algorithm:

• if the env variable TIMEZONE is defined, it will be used
• otherwise timezone will be autodetected using tzlocal

Note on YAML loading

All yaml files are loaded using load_yaml() and may contain env variables!

Function load_yaml()

Usage:

from bootstrap.utils import load_yaml

content = load_yaml("file.yaml")

It will load the content of file.yaml into a dict resolving ENV variables