动态文件字段路径
我正在尝试使用 get_file_path 函数来生成动态路径。我可以在 get_file_path 中使用 Album slug 字段代替此 str(instance.id) 吗？谢谢
这是模型
def get_file_path(instance, filename):
return os.path.join('files', str(instance.id), filename)
class Album(models.Model):
name = models.CharField(max_length=100)
slug = models.SlugField(unique = True,max_length=100,help_text="Suggested value automatically generated from name. Must be unique.")
path = models.CharField(max_length=100,null=True, blank=True)
language = models.ForeignKey(Category)
albumid = models.CharField(max_length=100)
class Song(models.Model):
title = models.CharField(max_length=100)
artist = models.ManyToManyField(Artist)
music = models.ForeignKey(Music)
album = models.ForeignKey(Album)
file = models.FileField(upload_to=get_file_path)
更新：我尝试了 instance.slug 。它不起作用。歌曲模型中不存在 instance.slug。它仅存在于专辑模型中（想要使用 Album Slug 字段） Update2：这是模型快照
I'm trying to use get_file_path function to generate dynamic path. Can I use Album slug field instead of this str(instance.id) in get_file_path? thanks
Here is models
def get_file_path(instance, filename):
return os.path.join('files', str(instance.id), filename)
class Album(models.Model):
name = models.CharField(max_length=100)
slug = models.SlugField(unique = True,max_length=100,help_text="Suggested value automatically generated from name. Must be unique.")
path = models.CharField(max_length=100,null=True, blank=True)
language = models.ForeignKey(Category)
albumid = models.CharField(max_length=100)
class Song(models.Model):
title = models.CharField(max_length=100)
artist = models.ManyToManyField(Artist)
music = models.ForeignKey(Music)
album = models.ForeignKey(Album)
file = models.FileField(upload_to=get_file_path)
Update: I tried instance.slug . Its not working. instance.slug does not exist in Song Model. Its only exists in Album model ( Want to use Album Slug field)
Update2: Here is model snapshot
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评论（2）
非常简单：
str(instance.album.slug)Quite straightforward:
str(instance.album.slug)是的，只需使用
instance.slug而不是instance.id您可以在帖子的答案中找到另一个示例 在 Django 中保存文件之前更改文件名
更新：如果不是所有实例都有 slug 字段，那么您可能会有兴趣像这样的解决方案：
Yes, just use
instance.sluginstead ofinstance.idAnother example you can find in the answer of post Change filename before save file in Django
Update: If not all instances have a slug field than you could be interested in a solution like this: