Django에서 슬러그를 어떻게 만듭니 까?

Django에서 슬러그를 어떻게 만듭니 까?

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SlugField장고에서 를 만들려고합니다 .

이 간단한 모델을 만들었습니다.

from django.db import models

class Test(models.Model):
    q = models.CharField(max_length=30)
    s = models.SlugField()

그런 다음이 작업을 수행합니다.

>>> 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'

나는 기대했다 b-b-b-b.

---

slugify 기능을 사용해야합니다.

>>> from django.template.defaultfilters import slugify
>>> slugify("b b b b")
u'b-b-b-b'
>>>

메소드 slugify를 재정 의하여 자동으로 호출 할 수 있습니다 save.

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)

위의 경우 q필드를 편집 할 때 URL이 변경되어 링크가 끊어 질 수 있습니다 . 새 객체를 만들 때 슬러그를 한 번만 생성하는 것이 좋습니다.

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)

---

utf-8 문자가있는 코너 케이스가 있습니다

예:

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

이것은 Unidecode 로 해결할 수 있습니다

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

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Thepeer의 답변에 대한 작은 수정 : save()모델 클래스의 함수 를 재정의하려면 인수를 더 잘 추가하십시오.

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).

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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.

---

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()

---

Use prepopulated_fields in your admin class:

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

admin.site.register(Article, ArticleAdmin)

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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()

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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",)}

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You can look at the docs for the SlugField to get to know more about it in more descriptive way.

참고URL : https://stackoverflow.com/questions/837828/how-do-i-create-a-slug-in-django