bumblebee-status/bumblebee-status
Tobias Witek 9b0c9a49e8 [core] Cleanly abort on ctrl+c
During debugging, being able to cleanly (i.e. without backtrace) exit
using ctrl+c is a very welcome functionality :)
2017-05-13 20:28:11 +02:00

66 lines
1.7 KiB
Python
Executable file

#!/usr/bin/env python
import os
import sys
import logging
import bumblebee.theme
import bumblebee.engine
import bumblebee.config
import bumblebee.output
import bumblebee.input
import bumblebee.modules.error
def main():
config = bumblebee.config.Config(sys.argv[1:])
if config.debug():
logging.basicConfig(
level=logging.DEBUG,
format="[%(asctime)s] %(levelname)-8s %(message)s",
filename=os.path.expanduser(config.logfile())
)
else:
logging.basicConfig(
level=logging.DEBUG,
format="[%(asctime)s] %(levelname)-8s %(message)s",
stream=sys.stderr
)
theme = bumblebee.theme.Theme(config.theme())
output = bumblebee.output.I3BarOutput(theme=theme)
inp = bumblebee.input.I3BarInput()
engine = None
try:
engine = bumblebee.engine.Engine(
config=config,
output=output,
inp=inp,
)
engine.run()
except KeyboardInterrupt as error:
inp.stop()
sys.exit(0)
except BaseException as e:
logging.exception(e)
if output.started():
output.flush()
output.end()
else:
output.start()
import time
while True:
output.begin()
error = bumblebee.modules.error.Module(engine, config)
error.set("exception occurred: {}".format(e))
widget = error.widgets()[0]
widget.link_module(error)
output.draw(widget, error)
output.flush()
output.end()
time.sleep(1)
if __name__ == "__main__":
main()
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