How to redirect to a dynamic url in django

Question

I am working on a django project. in the project I have a dynamic url as follows

app_name = 'test'

urlpatterns = [
    path('root', views.root, name='root'),
    path('output/<str:instance>', views.output_page, name='output_page'),
]

There exists two pages in the application. In the root page there exists a form which when submitted should redirect to the output_page. But because the output_page is a dynamic url, I am not able to redirect.

Here is my views file

def root(request):
    if request.method == 'POST':

        name = request.POST.get('name')
        job = request.POST.get('job')

        return redirect("test:output_page")

    return render(request, 'test/root.html')

def output_page(request, instance):

    record = Object.objects.all(Instance_id=instance)

    return render(request, 'test/output_page.html', {'record': record})

Here is the Model

class Object(models.Model):
    Name = models.CharField(max_length=200, null=True, blank=True)
    Job = models.CharField(max_length=200, default="")
    Instance_id = models.CharField(max_length=200)

When the redirect happens I want the url to be as follows

http://127.0.0.1:8000/output/test-001

where test-001 is the instance_id in the model.

The output_page should filter all the data in the model with instance_id test-001

Does this answer your question? [Django and urls.py: How do I HttpResponseRedirect via a named url?](https://stackoverflow.com/questions/1208802/django-and-urls-py-how-do-i-httpresponseredirect-via-a-named-url) — pygeek, Sep 08 '20 at 12:42

pygeek · Accepted Answer · 2020-09-08T12:33:12.317

3

Solution

The direct solution to your question would be the following:

from django.urls import reverse
from django.shortcuts import get_object_or_404
...
instance = get_object_or_404(Object, name=name, job=job)
redirect(reverse('test:output_page', args=instance))

However, it would be worth investigating class based views. I recommend using django’s built in RedirectView for this purpose.

References

Django Reverse: https://docs.djangoproject.com/en/3.1/ref/urlresolvers/

Django RedirectView: https://docs.djangoproject.com/en/3.1/ref/class-based-views/base/#redirectview

edited Sep 08 '20 at 12:33

answered Sep 08 '20 at 12:20

pygeek

7,356
1
20
41

Thanks for the reply. I am now getting this error `output_page() missing 1 required positional argument: 'instance'` – Sashaank Sep 08 '20 at 12:22
You can pass args and kwargs to reverse function. I’ve updated my answer to reflect this, but it’s also noted in referenced documentation. – pygeek Sep 08 '20 at 12:26
Hi. I am now getting this error `Reverse for 'output_page' with no arguments not found. 1 pattern(s) tried: ['output/(?P[^/]+)$']` – Sashaank Sep 08 '20 at 12:38
You need to ensure instance exists and is passed in before redirecting. – pygeek Sep 08 '20 at 12:40
Hi. I solved the error. I had made a reference to the page in the template. Once I removed that, the error was solved. Thanks a lot – Sashaank Sep 08 '20 at 12:43

Adithya · Answer 2 · 2020-09-08T12:41:30.840

3

You can do like this

return redirect(reverse("test:ouput_page",kwargs={'instance':str(instance_id)}))

edited Sep 08 '20 at 12:41

answered Sep 08 '20 at 12:25

Adithya

1,688
1
10
18

Hi. I am now getting this error `Reverse for 'output_page' with no arguments not found. 1 pattern(s) tried: ['output/(?P[^/]+)$']` – Sashaank Sep 08 '20 at 12:38
`instance` is given as a string in the URL so you need to send `str(instance_id)` in `kwargs` – Adithya Sep 08 '20 at 12:41
Hi. I solved the error. I had made a reference to the page in the template. Once I removed that, the error was solved. Thanks a lot – Sashaank Sep 08 '20 at 12:43

score -1 · Answer 3 · answered Sep 08 '20 at 12:32

-1

instance_model = "xyz"
return redirect('output_page', instance=str(instance_model))
(or)
return redirect('test:output_page', instance=str(instance_model))

answered Sep 08 '20 at 12:32

Surya Pratap Rana

423
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4
9

How to redirect to a dynamic url in django

3 Answers3

Solution

References