I have created a task queue that handles a service request because the
operation takes about 4 to 6 minutes to complete, and returns messages and
a file.
The task queue is working well when I start it manually. I am now trying
to create a cron @reboot entry so that the task queue will start when
web2py starts, but I cannot get it to work. The manual start for the queue
is this:
python web2py.py -S amlpoc -M -R
applications/amlpoc/private/aml_service.py
For the cron entry I have tried:
@reboot root web2py.py -S amlpoc2 -M -R
applications/amlpoc/private/aml_service.py
@reboot root -S amlpoc2 -M -R
applications/amlpoc/private/aml_service.py
@reboot root amlpoc2 -M -R
applications/amlpoc/private/aml_service.py
but none of these seem to work. Can I call the task queue directly like
this, with parameters? If so, what am I missing? If not, how should I make
the call?
tnx,
blutoh
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