如何在 django 2.0.4 中从我的相册模型自动生成 slug

Question

我有一个包含歌曲列表的专辑字段

 class Album(models.Model):
    artist = models.CharField(max_length=250)
    album_title = models.CharField(max_length=250)
    genre = models.CharField(max_length=100)
    album_logo = models.CharField(max_length=1000,blank=True)
    slug = models.SlugField(unique=True)

    def __str__(self):
        return self.album_title               

class Song(models.Model):
    album = models.ForeignKey(Album, on_delete=models.CASCADE)
    artist = models.CharField(max_length=250, blank=True)
    file_type = models.CharField(max_length=10)
    song_title = models.CharField(max_length=100)    

    def __str__(self):
        return self.artist

我想知道如何从专辑名称生成 slug。我正在关注使用 django 1.8 的教程，该教程使用正则表达式来实现此任务。但是通过查看文档，他们引入了一种更简单的方法（''）。那么，您能否帮助解释一下我如何实施它，让初学者不仅在这种情况下，而且在可能的情况下都能全面理解。提前致谢。

Answer 1

Django 会根据您传递给 slug 字段的字符串对象自动生成 slug。

# Import slugify to generate slugs from strings
from django.utils.text import slugify 

class Album(models.Model):
    artist = models.CharField(max_length=250)
    album_title = models.CharField(max_length=250)
    genre = models.CharField(max_length=100)
    album_logo = models.CharField(max_length=1000,blank=True)
    slug = models.SlugField(unique=True)

    def save(self, *args, **kwargs):
        self.slug = slugify(self.album_title)
        super(Album, self).save(*args, **kwargs)

    def __str__(self):
        return self.album_title    


class Song(models.Model):
    album = models.ForeignKey(Album, on_delete=models.CASCADE)
    artist = models.CharField(max_length=250, blank=True)
    file_type = models.CharField(max_length=10)
    song_title = models.CharField(max_length=100)
    slug = models.SlugField(max_length=100, unique=True)

    def save(self, *args, **kwargs):
        self.slug = slugify(self.song_title)
        super(Song, self).save(*args, **kwargs)    

    def __str__(self):
        return self.artist

Answer 2

要使用信号，这是 tutorial 我找到解决方案的地方

models.py

导入这些东西

from django.utils.text import slugify
from django.dispatch import receiver
from django.db.models.signals import pre_save

在你的模型下添加这个。

class Album(models.Model):
    artist = models.CharField(max_length=250)
    album_title = models.CharField(max_length=250)
    genre = models.CharField(max_length=100)
    album_logo = models.CharField(max_length=1000,blank=True)
    slug = models.SlugField(unique=True)

    def __str__(self):
        return self.album_title   

@receiver(pre_save, sender=Store)
def store_pre_save(sender, instance, *args, **kwargs):
    if not instance.slug:
        instance.slug = slugify(instance.song_title)

Make sure it's under the class and the def, otherwise you will get an error

如何在 django 2.0.4 中从我的相册模型自动生成 slug

How to auto generate slug from my Album model in django 2.0.4

python

django

slug