我在Django中有一个名为ExampleModel
的模型,并且希望每个模型对象都是唯一标识的。但是,我不希望URL中的用户可以看到对象的ID;因此我希望对象slug
是一个唯一的,随机生成的8位数整数,它将放在视图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)
目前slug的值为null,因此我不必为所有当前ExampleModel
个对象设置默认的slug。
这可以理解,但是我还没有能够找到任何可能适用于我的确切情况的指南/教程。
感谢您提供的任何帮助/指导
修改 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(保存方法):
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)
答案 0 :(得分:5)
Django内置了get_random_string
函数,可以生成slug所需的随机字符串。
正如Sebastian Wozny所提到的,你想要在覆盖save方法时调用它。基础是:
from django.utils.crypto import get_random_string
# ...
the_slug = get_random_string(8,'0123456789') # 8 characters, only digits.
那不是实际的工作代码。更详细地说,真正的 models.py 将如下所示。请注意,我并没有限制自己使用数字,而且我正在检查是否存在不正常情况,并确保它不会造成任何不良影响:
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)
答案 1 :(得分: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)
虽然可能需要对IntegrityError
提供更多保护措施。
如果你可以忍受两次保存:
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)
答案 2 :(得分:0)
如果你重写 save 方法,每次对象更新 slug 都会发生变化,如果你不想那样做,那么只在第一次设置 slug:
定义 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()