How to create a django ViewFlow process programmatically

Question

Synopsis

I\'m developing a web application to learn Django (python 3.4 & Django 1.6.10). The web app has complex and often updated workflows. I decided

Answer 1

There are two additional Start build-in Tasks available for Flows

StartFunction - starts flow when the function called somewhere:

@flow_start_func
def create_flow(activation, **kwargs):
    activation.prepare()
    activation.done()
    return activation

class FunctionFlow(Flow):
    start = flow.StartFunction(create_flow) \
        .Next(this.end)

# somewhere in the code
FunctionFlow.start.run(**some_kwargs)

StartSignal - starts flow on django signal receive:

class SignalFlow(Flow):
    start = flow.StartSignal(some_signal, create_flow) \      
        .Next(this.end)

You can check the usage for them, and rest of build-in task in this viewflow test suite.

For manually process the task state, first you should get the task from the database, activate it, and call any activation method.

task  = MyFlow.task_cls.objects.get(...)
activation = task.activate()
if  activation.undo.can_proceed():
    activation.undo()

Any activation transition have .can_proceed() method, helps you to check, is the task in the state that allows the transition.