Django

Question

我在 Django 中有一个名为 ExampleModel 的模型，并希望每个模型对象都被唯一标识。虽然，我不希望 URL 中的用户可以看到对象的 ID；因此，出于这个原因，我希望对象 slug 是一个唯一的、随机生成的 8 位整数，它将出现在视图 URL 中。这与我见过的其他问题不同，因为这意味着不会生成基于模型对象名称//内容本身的 slug 字符串。

Models.py:

class ExampleModel(models.Model):
    user = models.ForeignKey(UserModel, related_name='examplemodel', on_delete=models.CASCADE, null=True)
    title = models.CharField(max_length=50, verbose_name='Title')
    slug = models.SlugField(unique=True, blank=True, null=True)

当前 slug 的值为空，因此我不必为所有当前 ExampleModel 对象设置默认 slug。

这是可以理解的相当模糊的内容，但是我还没有找到任何 guides/tutorials 可能适合我的具体情况。

感谢 help/guidance 提供

编辑这是我的 views.py:

def model_create(request):
    user=request.user.id
    if request.user.is_authenticated:
        try:
            example = request.user.examplemodel
        except ExampleProfile.DoesNotExist:
            example = ExampleProfile(user)
        if request.method == 'POST':
            form = NewForm(request.POST, request.FILES)
            if form.is_valid():
                form.save()
                return redirect('/dashboard/')
            else:
                return render(request, 'create.html', {'form': form})
        else:
            form = NewForm()
            return render(request, 'create.html', {'form': form})
    else:
        return redirect('/users/login/?next=')

编辑2 Models.py（保存方式）：

def save(self, *args, **kwargs):
        if self.user is None:  # Set default reference
            self.user = UserModel.objects.get(id=1)
        super(ExampleModel, self).save(*args, **kwargs)

Answer 1

覆盖保存：

def save(self, *args, **kwargs):
    try:
        self.slug = ''.join(str(random.randint(0, 9)) for _ in range(8))
        super().save(*args, **kwargs)
    except IntegrityError:
        self.save(*args, **kwargs)

虽然这可能需要一些针对 IntegrityError 的更多保护措施。如果你能接受两次保存：

def save(self, *args, **kwargs):
    super().save(*args, **kwargs)
    try:
        self.slug = ''.join(str(random.randint(0, 9)) for _ in range(8))
        super().save(*args, **kwargs)
    except IntegrityError:
        self.save(*args, **kwargs)

Answer 2

Django 有一个内置的 get_random_string 函数，可以生成 slug 所需的随机字符串。

正如 Sebastian Wozny 提到的，您想在覆盖保存方法时调用它。基础知识是：

from django.utils.crypto import get_random_string
# ...
the_slug = get_random_string(8,'0123456789') # 8 characters, only digits.

这不是实际的工作代码。更详细地说，真实的 models.py 如下所示。请注意，我并没有将自己限制在数字上，我正在检查是否正确并确保它不会拼写任何错误：

from django.db import models
from django.utils.crypto import get_random_string
# ...
class SomeModelWithSlug(models.Model):
  slug = models.SlugField(max_length=5,blank=True,) # blank if it needs to be migrated to a model that didn't already have this 
  # ...
  def save(self, *args, **kwargs):
    """ Add Slug creating/checking to save method. """
    slug_save(self) # call slug_save, listed below
    Super(SomeModelWithSlug, self).save(*args, **kwargs)
# ...
def slug_save(obj):
""" A function to generate a 5 character slug and see if it has been used and contains naughty words."""
  if not obj.slug: # if there isn't a slug
    obj.slug = get_random_string(5) # create one
    slug_is_wrong = True  
    while slug_is_wrong: # keep checking until we have a valid slug
        slug_is_wrong = False
        other_objs_with_slug = type(obj).objects.filter(slug=obj.slug)
        if len(other_objs_with_slug) > 0:
            # if any other objects have current slug
            slug_is_wrong = True
        naughty_words = list_of_swear_words_brand_names_etc
        if obj.slug in naughty_words:
            slug_is_wrong = True
        if slug_is_wrong:
            # create another slug and check it again
            obj.slug = get_random_string(5)

Answer 3

如果您重写保存方法，每次对象更新时 slug 都会发生变化，如果您不想那样做，那么这样做只会在第一次设置 slug：

def slug_generator():
    return ''.join(random.choices(string.ascii_lowercase + string.digits + string.ascii_uppercase, k=20))

def save(self, *args, **kwargs):
    if not self.slug:
        self.slug = slug_generator()
        super(Item, self).save()
    super(Item, self).save()

Django - 为每个模型对象生成随机、唯一的 slug 字段

Django - Generating random, unique slug field for each model object

python

slug