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I am trying to create a SlugField in Django.
I created this simple model:
from django.db import models
class Test(models.Model):
q = models.CharField(max_length=30)
s = models.SlugField()
I then do this:
>>> from mysite.books.models import Test
>>> t=Test(q="aa a a a", s="b b b b")
>>> t.s
'b b b b'
>>> t.save()
>>> t.s
'b b b b'
I was expecting b-b-b-b.
217
Adding Slugify to our project: Now we need to find a way to convert the title into a slug automatically. We want this script to be triggered every time a new instance of Post model is created. For this purpose, we will use signals. Note: Add new file util.py in the same directory where settings.py file is saved.
A slug is a short label for something, containing only letters, numbers, underscores or hyphens. They're generally used in URLs. ( as in Django docs) A slug field in Django is used to store and generate valid URLs for your dynamically created web pages.
A slug is a unique string (typically a normalized version of title or other representative string), often used as part of a URL.
You will need to use the slugify function.
>>> from django.template.defaultfilters import slugify
>>> slugify("b b b b")
u'b-b-b-b'
>>>
You can call slugify automatically by overriding the save method:
class Test(models.Model):
q = models.CharField(max_length=30)
s = models.SlugField()
def save(self, *args, **kwargs):
self.s = slugify(self.q)
super(Test, self).save(*args, **kwargs)
Be aware that the above will cause your URL to change when the q field is edited, which can cause broken links. It may be preferable to generate the slug only once when you create a new object:
class Test(models.Model):
q = models.CharField(max_length=30)
s = models.SlugField()
def save(self, *args, **kwargs):
if not self.id:
# Newly created object, so set slug
self.s = slugify(self.q)
super(Test, self).save(*args, **kwargs)
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