PAGE NOT FOUND- ERROR(404);GET request; django-python

Question:

Lately I have been trying to use the GET request in django-python. However I run into a 404 error when I do so. I want the program to print the parameter given to it.

URL PATTERN :

path('add<int:hello>/',views.add,name = 'Add')

VIEWS.PY:

def add(request,num1):
    val1 = request.GET["num1"]
    return HttpResponse(request,"Hello")

WHAT I AM SEARCHING ON BROWSER:

http://127.0.0.1:8000/add?1

Asked By: Tommy Sony

||

Answers:

http://127.0.0.1:8000/add?hello=1

This should work. you need to add the query param name as well.

To get the value you need to use this code:

request.GET["hello"]

Answered By: Ikram Khan Niazi

A couple of things are strange or broken here.

Given path('add<int:hello>/', views.add, ...),
Django expects that function views.add has the following signature:

def add(request, hello):

The pattern 'add<int:hello>/' expects urls with "add" followed by an integer, such as:

http://127.0.0.1:8000/add1
http://127.0.0.1:8000/add2
http://127.0.0.1:8000/add3
...

and so on.

request.GET["num1"] expects urls that have num1=... in their query string, such as:

http://127.0.0.1:8000/add1?num1=...

HttpResponse expects a string type as its first parameter,
not an HttpRequest type as in the posted code.

Fixing the problems

This might be close to what you want, and it will work:

path('add<int:num1>/', views.add, name='Add')

def add(request, num1):
    return HttpResponse(f"Hello {num1}")

# visit http://127.0.0.1:8000/add1

Another variation:

path('add', views.add, name='Add')

def add(request):
    num1 = request.GET['num1']
    return HttpResponse(f"Hello {num1}")

# visit http://127.0.0.1:8000/add?num1=123

Answered By: janos