如何在 Django 中创建一个 slug？

Question

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 bbbb .

Answer 1

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)

Answer 2

There is corner case with some utf-8 characters

Example:

>>> from django.template.defaultfilters import slugify
>>> slugify(u"test ąęśćółń")
u'test-aescon' # there is no "l"

This can be solved with Unidecode

>>> from unidecode import unidecode
>>> from django.template.defaultfilters import slugify
>>> slugify(unidecode(u"test ąęśćółń"))
u'test-aescoln'

Answer 3

A small correction to Thepeer's answer: To override save() function in model classes, better add arguments to it:

from django.utils.text import slugify

def save(self, *args, **kwargs):
    if not self.id:
        self.s = slugify(self.q)

    super(test, self).save(*args, **kwargs)

Otherwise, test.objects.create(q="blah blah blah") will result in a force_insert error (unexpected argument).

Answer 4

If you're using the admin interface to add new items of your model, you can set up a ModelAdmin in your admin.py and utilize prepopulated_fields to automate entering of a slug:

class ClientAdmin(admin.ModelAdmin):
    prepopulated_fields = {'slug': ('name',)}

admin.site.register(Client, ClientAdmin)

Here, when the user enters a value in the admin form for the name field, the slug will be automatically populated with the correct slugified name .

Answer 5

In most cases the slug should not change, so you really only want to calculate it on first save:

class Test(models.Model):
    q = models.CharField(max_length=30)
    s = models.SlugField(editable=False) # hide from admin

    def save(self):
        if not self.id:
            self.s = slugify(self.q)

        super(Test, self).save()

Answer 6

Use prepopulated_fields in your admin class:

class ArticleAdmin(admin.ModelAdmin):
    prepopulated_fields = {"slug": ("title",)}

admin.site.register(Article, ArticleAdmin)

Answer 7

If you don't want to set the slugfield to Not be editable, then I believe you'll want to set the Null and Blank properties to False. Otherwise you'll get an error when trying to save in Admin.

So a modification to the above example would be::

class test(models.Model):
    q = models.CharField(max_length=30)
    s = models.SlugField(null=True, blank=True) # Allow blank submission in admin.

    def save(self):
        if not self.id:
            self.s = slugify(self.q)

        super(test, self).save()

Answer 8

I'm using Django 1.7

Create a SlugField in your model like this:

slug = models.SlugField()

Then in admin.py define prepopulated_fields ;

class ArticleAdmin(admin.ModelAdmin):
    prepopulated_fields = {"slug": ("title",)}

Answer 9

您可以查看SlugField的文档，以更具描述性的方式了解有关它的更多信息。