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Django : Generate unique slug
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# import signals and slugify | |
from django.db.models import signals | |
from django.template.defaultfilters import slugify | |
class Post(models.Model): | |
text = models.TextField("Post text") | |
# function for use in pre_save | |
def post_pre_save(signal, instance, sender, **kwargs): | |
if not instance.slug: | |
slug = slugify(instance.attribute) # change the attibute to the field that would be used as a slug | |
new_slug = slug | |
count = 0 | |
while Post.objects.filter(slug=new_slug).exclude(id=instance.id).count() > 0: | |
count += 1 | |
new_slug = '{slug}-{count}'.format(slug=slug, count=count) | |
instance.slug = new_slug | |
# Execute signals pre_save | |
signals.pre_save.connect(post_pre_save, sender=Post) |
Thanks @Alir3z4
thanks but how can i use it ??
this method can be useful when adding only one more similar slug because the count variable will be always 1 since it will be reset to 0 whenever the method is invoked.
You will face race condition when you have 3 similar attribute (instance.attribute).
@Radi85 Today I really think the best approach is use a external_id
field with last parameter of URL instead the count on post save.
e.g.
urlpatterns = [
path("<slug:str>-<external_id:str>/", PostDetailView.as_view(), name="detail"),
]
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Nice and simple <3