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

I needed to be able to manually or programmatically start a flow instance. The model I ended up with, based on the above reference to StartFunction looks like this:

class MyRunFlow(flow.Flow):
    process_class = Run

    start = flow.Start(ProcessCreate, fields=['schedule']). \
        Permission(auto_create=True). \
        Next(this.wait_data_collect_start)
    start2 = flow.StartFunction(process_create). \
        Next(this.wait_data_collect_start)

Note the important point is that process_create has the Process object and this code must programmatically set up the same fields that the manual form submission does via the fields specification to ProcessCreate:

@flow_start_func
def process_create(activation: FuncActivation, **kwargs):
    #
    # Update the database record.
    #
    db_sch = Schedule.objects.get(id=kwargs['schedule'])
    activation.process.schedule = db_sch # <<<< Same fields as ProcessCreate
    activation.process.save()
    #
    # Go!
    #
    activation.prepare()
    activation.done()
    return activation

Note that the activation subclass inside the flow_start_func is FuncActivation, which has the prepare() and save() methods. The kwargs come from the call to run, which goes something like:

start_node = 
activation = start_node.run(schedule=self.id)