complete the request in a second if I insert in queue. and not a minute*
On Thursday 17 October 2013 12:45 AM, Niphlod wrote:
the main point is not that is not inefficient by itself. It's just
that you need to make sure you don't use that in a model because in
web2py models are executed at **every** request, and you don't seem to
want to queue a task for every request that comes in :-P
On Wednesday, October 16, 2013 8:55:34 PM UTC+2, Tushar Tuteja wrote:
thanks a ton,
I ll try this and what would be appropriate place to insert the
file , so that it is not efficient.
thanks,
regards,
Tushar Tuteja
On 17 October 2013 00:22, Niphlod <nip...@gmail.com <javascript:>>
wrote:
the scheduler is running fine.
In the data you pasted here there's the reason why the
scheduler isn't picking up any new task.
times_run is 3 and repeats is set to 3, so the task got
executed 3 times already.
Sidenote: append a "return 1" to your function, so you'll get
scheduler_run records holding more details about the
executions, until you figure out the issues you're having on
what to choose when you queue the task.
Also, please do
sched = Scheduler(db, dict(fun=func))
sched.queue_task('func', uuid='test_insert', repeats=3, period=10)
without any worker running, so you can see what is inserted
into the scheduler_task table correctly.
Execute the sched.queue_task just one time only (if you put
that in models, it will be executed at every request and that
would be inefficient)
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Civil Engineering
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