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django can't find url even though it is there

I have an issue where I try to go to my redirect page and get a NoReverseMatch when though the URL is there? Any idea how to fix this?
I have checked that the "schema" url works and it correctly supplies the openapi schema, but the other page simply can't understand the url.
URLS:
urlpatterns = [
path("schema/", SpectacularAPIView.as_view(), name="schema"),
# Optional UI:
path("docs/", SpectacularSwaggerView.as_view(url_name="schema"), name="swagger-ui"),
]
Errors:

For reverse url pathing, you have to use {% url api:schema %}. It's specified as namespace next to include('api.urls') or inside app urls, just above urlpatterns - like app_name = "api".

URL patterns in Django 2

I just have started my first project by using Django 2.0 in which I need to define a URL in a way as:
http://localhost:8000/navigator?search_term=arrow
But I couldn't know how to define a string parameter for a URL in Django 2.0
Here's what I have tried:
From ulrs.py:
from Django.URLs import path from. import views
urlpatterns = [
path('navigator/<str:search_term>', views.GhNavigator, name='navigator'),
]
Any help?

There is no need to define query params in URL. Below url is enough to work.
path('navigator/', views.GhNavigator, name='navigator')
Let you called URL http://localhost:8000/navigator/?search_term=arrow then you can get search_term by request.GET.get('search_term').

Request: GET
http://localhost:8000/navigator?search_term=arrow
urls.py
urlpatterns = [
path('navigator/', views.GhNavigator, name='navigator'),
]
views.py
search_term = request.GET.get('search_term', None)

Django redirect URL with parameter, inside urls.py file?

I'm using Django 1.9. Is there any way to redirect a URL with a parameter in my urls.py file?
I want to permanently redirect a URL like /org/123/ to the corresponding URL /neworg/123.
I know how to redirect within a view, but I'm wondering if there's any way to do it solely inside urls.py.

You can use RedirectView. As long as the old and new url patterns have the same args and kwargs, you can use pattern_name to specify the url pattern to redirect to.
from django.views.generic.base import RedirectView
urlpatterns = [
url(r'^neworg/(?P<pk>\d+)/$', new_view, name='new_view'),
url(r'^org/(?P<pk>\d+)/$', RedirectView.as_view(pattern_name='new_view'), name='old_view')
]

Django redirect

I have a loading page in Django while some server side processes are ongoing the view is;
def loading_page( request ):
testname = request.session['testname']
done_file = filepath_to_design_dir( testname + ".done" )
if os.path.exists( done_file ):
request.session["job_stat"] = "job_done"
return redirect( "single_output/")
else:
return render( request, 'single_design/loading.html' )
My problem is that the redirect goes to;
http://127.0.0.1:8000/single_design/loading_page/single_output/
Rather than
http://127.0.0.1:8000/single_design/single_output
What is the correct way to do this???
EDIT : issue resolved, thanks guys.
Urls as requested
from django.conf.urls import url , include
from . import views
from django.conf import settings
from django.conf.urls.static import static
urlpatterns = [
url(r'^$', views.get_single_input, name='single_design_input'),
url(r'^single_output/$', views.single_output, name='single_output'),
url(r'^loading_page/$', views.loading_page, name='loading_page'),
] + static(settings.STATIC_URL, document_root=settings.STATIC_ROOT)

The correct way is not to use hardcoded link. Use urlresolvers.
return redirect("some_view_name")

Without the leading slash, the current value single_output/ is treated as a relative url, which is appended to the current url /single_design/loading_page/ to give /single_design/loading_page/single_output/.
You could use the relative url ../single_output, but I wouldn't recommend it. It would be better to return the url you want to redirect to, including a leading slash.
return redirect('/single_design/single_output/' )
Ideally, you should use the name of the url pattern, then Django will reverse it for you. Since you have,
url(r'^single_output/$', views.single_output, name='single_output'),
you can use the name instead of the url,
return redirect('single_output')
The advantage of using the name single_output, is that you can now change the URL in your url patterns, without having to update the view.

How to get the url path of a view function in django

As an example:
view.py
def view1( request ):
return HttpResponse( "just a test..." )
urls.py
urlpatterns = patterns('',
url( r'^view1$', 'app1.view.view1'),
)
I want to get the URL path of view1. How can I do this.
I want to avoid hard coding any URL paths, such as "xxx/view1".

You need reverse.
from django.urls import reverse
reverse('app1.view.view1')
If you want to find out URL and redirect to it, use redirect
from django.urls import redirect
redirect('app1.view.view1')
If want to go further and not to hardcode your view names either, you can name your URL patterns and use these names instead.

This depends whether you want to get it, if you want to get the url in a view(python code) you can use the reverse function(documentation):
reverse('admin:app_list', kwargs={'app_label': 'auth'})
And if want to use it in a template then you can use the url tag (documentation):
{% url 'path.to.some_view' v1 v2 %}

If you want the url of the view1 into the view1 the best is request.get_path()

As said by others, reverse function and url templatetags can (should) be used for this.
I would recommend to add a name to your url pattern
urlpatterns = patterns('',
url( r'^view1$', 'app1.view.view1', name='view1'),
)
and to reverse it thanks to this name
reverse('view1')
That would make your code easier to refactor

Yes, of course you can get the url path of view named 'view1' without hard-coding the url.
All you need to do is - just import the 'reverse' function from Django urlresolvers.
Just look at the below example code:
from django.core.urlresolvers import reverse
from django.http import HttpResponseRedirect
def some_redirect_fun(request):
return HttpResponseRedirect(reverse('view-name'))

You can use the reverse function for this. You could specify namespaces and names for url-includes and urls respectively, to make refactoring easier.

Universal approach
install Django extensions and add it to INSTALLED_APPS
Generate a text file with all URLs with corresponding view functions
./manage.py show_urls --format pretty-json --settings=<path-to-settings> > urls.txt
like
./manage.py show_urls --format pretty-json --settings=settings2.testing > urls.txt
Look for your URL in the output file urls.txt
{
"url": "/v3/blockdocuments/<pk>/",
"module": "api.views.ganeditor.BlockDocumentViewSet",
"name": "block-documents-detail",
},