Django FileField with upload_to 在运行时确定

您可能已经阅读过 the documentation，所以这里有一个简单的例子来说明它的含义：

def content_file_name(instance, filename):
    return '/'.join(['content', instance.user.username, filename])

class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to=content_file_name)

如您所见，您甚至不需要使用给定的文件名 - 如果您愿意，您也可以在您的 upload_to 可调用文件中覆盖它。

这真的很有帮助。为了更简洁起见，决定在我的情况下使用 lambda：

file = models.FileField(
    upload_to=lambda instance, filename: '/'.join(['mymodel', str(instance.pk), filename]),
)

关于使用“实例”对象的 pk 值的说明。根据文档：

在大多数情况下，这个对象还没有保存到数据库中，所以如果它使用默认的 AutoField，它可能还没有主键字段的值。

因此，使用 pk 的有效性取决于您的特定模型是如何定义的。

如果您在迁移时遇到问题，您可能应该使用 @deconstructible 装饰器。

import datetime
import os
import unicodedata

from django.core.files.storage import default_storage
from django.utils.deconstruct import deconstructible
from django.utils.encoding import force_text, force_str


@deconstructible
class UploadToPath(object):
    def __init__(self, upload_to):
        self.upload_to = upload_to

    def __call__(self, instance, filename):
        return self.generate_filename(filename)

    def get_directory_name(self):
        return os.path.normpath(force_text(datetime.datetime.now().strftime(force_str(self.upload_to))))

    def get_filename(self, filename):
        filename = default_storage.get_valid_name(os.path.basename(filename))
        filename = force_text(filename)
        filename = unicodedata.normalize('NFKD', filename).encode('ascii', 'ignore').decode('ascii')
        return os.path.normpath(filename)

    def generate_filename(self, filename):
        return os.path.join(self.get_directory_name(), self.get_filename(filename))

用法：

class MyModel(models.Model):
    file = models.FileField(upload_to=UploadToPath('files/%Y/%m/%d'), max_length=255)

如果您有用户实例，请快速设置以生成

<model-slug>/<username>-<first_name>-<last_name>/filename-random.png

例如：/medias/content/ft0004-john-doe/filename-lkl9237.png

def upload_directory_name(instance, filename):

    user = getattr(instance, 'user', None)
    if user:
        name = f"{user.username}-{user.get_full_name().replace(' ', '-')}"
    else:
        name=str(instance)
    model_name = instance._meta.verbose_name.replace(' ', '-')
    return str(os.path.pathsep).join([model_name, name, filename])


class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to=upload_directory_name)

[@SmileyChris 的修改版]

我想在运行时更改上传路径，没有一个解决方案适合这种需求。

这就是我所做的：

class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to=DynamicUploadPath.get_file_path)


class ContentSerializer(serializers.ModelSerializer):
    class Meta:
        model = Content
        fields = '__all__'


class UploadDir(models.TextChoices):
    PRODUCT = 'PRD', _('Product')
    USER_PROFILE = 'UP', _('User Profile')


class DynamicUploadPath:
    dir: UploadDir = None

    @classmethod
    def get_file_path(cls, instance, filename):
        return str(cls.dir.name.lower() + '/' + filename)


def set_DynamicUploadPath(dir: UploadDir):
    DynamicUploadPath.dir = dir


class UploadFile(APIView):
    parser_classes = (MultiPartParser, FormParser)

    def post(self, request):
        # file save path: MEDIA_ROOT/product/filename
        set_DynamicUploadPath(UploadDir.PRODUCT)

        # file save path: MEDIA_ROOT/user_profile/filename
        # set_DynamicUploadPath(UploadDir.USER_PROFILE)

        serializer = ContentSerializer(data=request.data)
        serializer.is_valid(raise_exception=True)
        serializer.save()

        return Response(serializer.data, status=status.HTTP_200_OK)