Fix repeating task after timeout (#184)

This commit is contained in:
Stan Triepels
2024-06-23 22:36:29 +02:00
committed by GitHub
parent efa0d6d9e4
commit cb253357fb
5 changed files with 64 additions and 3 deletions
+16 -2
View File
@@ -17,7 +17,9 @@ except core.exceptions.AppRegistryNotReady:
django.setup()
from django_q.conf import Conf, error_reporter, logger, resource, setproctitle
from django_q.exceptions import TimeoutException
from django_q.signals import post_spawn, pre_execute
from django_q.timeout import TimeoutHandler
from django_q.utils import close_old_django_connections, get_func_repr
try:
@@ -89,25 +91,37 @@ def worker(
pre_execute.send(sender="django_q", func=f, task=task)
# execute the payload
timer.value = timer_value # Busy
if timer.value != -1:
timer.value += 3 # Add buffer so that guard doesn't kill the process on timeout before it gets processed
timeout_error = False
try:
if f is None:
# raise a meaningfull error if task["func"] is not a valid function
raise ValueError(f"Function {task['func']} is not defined")
res = f(*task["args"], **task["kwargs"])
with TimeoutHandler(timer_value):
res = f(*task["args"], **task["kwargs"])
result = (res, True)
except Exception as e:
except (Exception, TimeoutException) as e:
if isinstance(e, TimeoutException):
timeout_error = True
result = (f"{e} : {traceback.format_exc()}", False)
if error_reporter:
error_reporter.report()
if task.get("sync", False):
raise
with timer.get_lock():
# Process result
task["result"] = result[0]
task["success"] = result[1]
task["stopped"] = timezone.now()
result_queue.put(task)
if timeout_error:
# force destroy process due to timeout
timer.value = 0
break
timer.value = -1 # Idle
if setproctitle:
setproctitle.setproctitle(f"qcluster {proc_name} idle")