Python – Django templates stripping spaces

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I'm having trouble with Django templates and CharField models.

So I have a model with a CharField that creates a slug that replaces spaces with underscores. If I create an object, Somename Somesurname, this creates slug Somename_Somesurname and gets displayed as expected on the template.

However, if I create an object, Somename Somesurname (notice the second space), slug Somename__Somesurname is created, and although on the Django console I see this as <Object: Somename Somesurname>, on the template it is displayed as Somename Somesurname.

So do Django templates somehow strip spaces? Is there a filter I can use to get the name with its spaces?

Best Answer

Let me preface this by saying @DNS's answer is correct as to why the spaces are not showing.

With that in mind, this template filter will replace any spaces in the string with  

Usage:

{{ "hey there  world"|spacify }}

Output would be hey there  world

Here is the code:

from django.template import Library
from django.template.defaultfilters import stringfilter
from django.utils.html import conditional_escape
from django.utils.safestring import mark_safe
import re

register = Library()

@stringfilter
def spacify(value, autoescape=None):
    if autoescape:
    esc = conditional_escape
    else:
    esc = lambda x: x
    return mark_safe(re.sub('\s', '&'+'nbsp;', esc(value)))
spacify.needs_autoescape = True
register.filter(spacify)

For notes on how template filters work and how to install them, check out the docs.

Related Solutions

Access constants in settings.py from templates in Django

If it's a value you'd like to have for every request & template, using a context processor is more appropriate.

Here's how:

Make a context_processors.py file in your app directory. Let's say I want to have the ADMIN_PREFIX_VALUE value in every context:

from django.conf import settings # import the settings file

def admin_media(request):
    # return the value you want as a dictionnary. you may add multiple values in there.
    return {'ADMIN_MEDIA_URL': settings.ADMIN_MEDIA_PREFIX}

add your context processor to your settings.py file:

TEMPLATES = [{
    # whatever comes before
    'OPTIONS': {
        'context_processors': [
            # whatever comes before
            "your_app.context_processors.admin_media",
        ],
    }
}]

Use RequestContext in your view to add your context processors in your template. The render shortcut does this automatically:
```
from django.shortcuts import render

def my_view(request):
    return render(request, "index.html")
```

and finally, in your template:

...
<a href="{{ ADMIN_MEDIA_URL }}">path to admin media</a>
...

Python – How to do a not equal in Django queryset filtering

You can use Q objects for this. They can be negated with the ~ operator and combined much like normal Python expressions:

from myapp.models import Entry
from django.db.models import Q

Entry.objects.filter(~Q(id=3))

will return all entries except the one(s) with 3 as their ID:

[<Entry: Entry object>, <Entry: Entry object>, <Entry: Entry object>, ...]

Best Answer

Related Solutions

Access constants in settings.py from templates in Django

Python – How to do a not equal in Django queryset filtering

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