Is there a way to pass URL template tag from database to template in Django?

Question

I apologize if my question is poorly worded. Basically, I am making a database of fictional characters and in the database, I have a biography of the character. I would like to link to other characters using the {% url 'viewname' otherchar.slug %} method. However, all I get back from the database is literally that line of code.

I understand it may not be possible, but is there a way to get Django to see that line and turn it into an absolute URL like it would if I manually added the URL tag into the page?

Something to add - I am using TinyMCE so the content in the database is being saved with HTML - I want to be able to save external links, subheadings, and whatnot, which is why I chose to use TinyMCE and its HTML field.

models.py

class Character(models.Model):
    name = models.CharField(max_length = 255)
    faction = models.IntegerField(default = 0)
    rankIMG = models.ImageField(upload_to = 'rankIMG/', blank = True)
    department = models.IntegerField(default = 0)
    content = HTMLField()
    slug = models.SlugField()

views.py

class CharacterFull(DetailView):
    model = Character
    context_object_name = 'character'

    def get_context_data(self, **kwargs):
        context = super(CharacterFull, self).get_context_data(**kwargs)
        context['factionDict'] = charFaction
        context['deptDict'] = charDepartment

        return context

urls.py

app_name = 'LCARS'

urlpatterns = [
    path('', LCARSView.LCARSHome.as_view(), name = 'lcarsHome'),
    path('Characters/', LCARSView.Characters.as_view(), name = 'characterHome'),
    path('Characters/p/<slug:slug>', LCARSView.CharacterPartialView, name = 'charPartialView'),
    path('Characters/<slug:slug>/', LCARSView.CharacterFull.as_view(), name = 'characterView'),
]

template (relevant code)

<div class="col-md-8 p-4 text-justify border border-secondary rounded shadow">
    {{ character.content|safe }}
</div>

page source example (if it's helpful)

<p dir="ltr">He was also assigned <a href="{% url 'LCARS:characterView' character.slug %}">Commander Shampoo</a> as his...

I appreciate any assistance you guys can be of. I can't seem to find anything on here or on Google. Thanks!

Answer 1

You can render the content in the get_context_data as a template:

from django.template import Template, RequestContext

class CharacterFull(DetailView):
    model = Character
    context_object_name = 'character'

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['factionDict'] = charFaction
        context['deptDict'] = charDepartment
        self.object = Template(
            self.object.content
        ).RequestContext(self.request, context)
        return context

where .other_attribute is an identifier you do not use as a field, property, method, etc.

and then render this in the "outer" template with:

<div class="col-md-8 p-4 text-justify border border-secondary rounded shadow">
    {{ character|safe }}
</div>