Django views with Celery task: Render details page with two parameters

Question

I am processing a form in my Celery task and I can already render them after HttpResponseRedirect.

Here in my views.py:

from django_celery_results.models import TaskResult
from .forms import SampleForm
from .models import Sample
from analysis.tasks import process_sample_input

def sample_submit(request):
    if request.method == "POST":
        form = SampleForm(request.POST or None, request.FILES or None)
        if form.is_valid():
            #form = form.cleaned_data
            instance = form.save()
            task = process_sample_input.delay(instance.id)
            try:
                task_result = TaskResult(task_id = task.id)
                task_result.save()
                return HttpResponseRedirect(reverse("sample:results_detail", kwargs={"id":instance.id, "task_id":task.id}))
            except Exception as e: 
                raise e
            #return results_detail(request, instance.id, task.id) #show the results details

        else:
            print form.errors
            return not_processed(request)
    else:
        form = SampleForm()
    context = {
        "form": form,
    }
    return render(request, "analysis/sample_form.html", context)

My problem is how can I access the html rendered by results_detail from a results list page. I don't know how I can access my Celery task ID elsewhere from the views' sample_submit()

In my views.py:

def results_list(request):
    all_samples = Sample.objects.all()
    context = {'all_samples': all_samples}  
    return render(request, "analysis/results_list.html", context)

In my results_list.html

{% block content %}
<table>
    <thead>
        <tr>
            <th>Name [Local File or URL]</th>
        </tr>
    </thead>
    {% for sample in all_samples %}
        <td><a href='/sample/results/{{sample.id}}/'>{{ sample.get_sample_name }}</a></td>
    {% endfor %}
</table>
{% endblock %}

in the myapp/analysis/urls.py:

   url(r'^results/(?P<id>[0-9]+)/$', views.results_detail, name="results_detail"),

Here in urls.py and templates, I can pass only one parameter. How can I pass the Celery taskresult id (auto-generated) to the views if there is a way? Any help or ideas?

sparklights · Accepted Answer · 2017-03-26 05:47:27Z

Since it was difficult to pass two parameters, I just settled on having my app model's id same with taskresult.id. So that it is easier to access in the views:

def results_detail(request, id=1): 
    sample_instance = get_object_or_404(Sample, id = id)
    result_obj = None
    try:
        result_obj = TaskResult.objects.get(id=id)
    except Exception as e:
        raise e
    try:
        #AsyncResult waits for the Celery task results to finish and return results
        taskresult = AsyncResult(result_obj.task_id)
        # --- Accessing AsyncResult's results for Thug logs ---
        #ast.literal_eval for reading the list in a string
        if taskresult.get():
             #did this to make sure results are already available. Manually asking django-celery-taskresult's object.status returns PENDING. taskresult.get() will wait for the task.
             # include next code here

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