can not render ' / ' in url in Django templates - django

Question

Update :: Problem solved.just follow the guy below.
in my urls.py
path('', store_view, name='store'),
path('category/<str:category_name>/', category_view, name='category'),
in views.py
def store_view(request):
categories = list(Category.objects.all())
context = {
'categories': categories,
}
return render(request, 'store/store.html', context)
def category_view(request, category_name):
category = Category.objects.get(name=category_name)
context = {
'category': category,
}
return render(request, 'store/single-category-view.html', context)
in my template : store.html , that is rendered by store_view >>
{% for item in categories %}
<a href="{% url 'category' item.name %}">
{{item.name}}
</a>
{% endfor %}
Now,the problem is, in the category column in my DB, i have got one category called 'Laptop/MacBook'.when ever this name is passed to the url, it says >>
"Reverse for 'category' with arguments '('Laptop/MacBook',)' not
found. 1 pattern(s) tried: ['category/(?P<category_name>[^/]+)/$']
But when i changed the category name from Laptop/MacBook to Laptop and MacBook , it worked fine and showed no error.
But i want to keep it as it was,'Laptop/MacBook'.How can i do that??and how do you guys deal with that?

Answer 1

Try encoding and decoding your DB values. Assuming its Python 3:
from urllib.parse import quote, unquote
encoded = quote("Laptop/Macbook", safe="")
decoded = unquote(encoded)
print(encoded, decoded)
Output:
Laptop%2FMacbook Laptop/Macbook
With this your route should take in the right param.
from django.http import HttpResponse, request
from django.shortcuts import render
def store_view(request):
name = "Laptop/Macbook"
return render(request, './store.html', context={"name": name})
def category_view(request, category_name):
print(category_name)
return HttpResponse(b"Here we go!")
templatetags/tags.py
from urllib.parse import quote, unquote
from django import template
register = template.Library()
#register.filter(name='encode')
def encode(name):
return quote(name, safe="")
#register.filter(name='decode')
def decode(name):
return unquote(name)
Template:
{% load tags %}
<a href="{% url 'category' name|encode %}">
{{name}}
</a>
Don't forget to add in settings:
'OPTIONS': {
'libraries':{
'tags': 'templatetags.tags',
}
},

Answer 2

When using a "/", django thinks that you are passing more than one parameter. To fix this, replace str by path in your urls like so:
path('', store_view, name='store'),
path('category/<path:category_name>/', category_view, name='category'),
This will make django understand that the / does not mean there are two separate parameters in your url.