Question

我有一个带有多个ImageField的Django模型，并使用callable来确定上传路径。我想在上传路径中包含原始上传字段的名称，在本例中为tiny，small，medium或press。

我能想到的唯一方法是创建一个pre_save接收器，用uuid替换file.name。然后upload_to callable通过将其与filename进行比较来查找匹配项。这样做不是一种不那么粗俗的方式吗？

class SomeDjangoModel(models.Model):

    IMAGE_SIZES = ('tiny', 'small', 'medium', 'press')

    def image_path(self, filename):
        """ Example return: [some-django-model]/[medium]/[product1].[jpg] """
        size = None
        for field_name in self.IMAGE_SIZES:
            field_fn = getattr(getattr(self, field_name), 'name', '')
            if field_fn == filename.rpartition('/')[2]:
                size = field_name
                break

        return u'{}/{}/{}.{}'.format(
            slugify(self._meta.verbose_name),
            size or 'undetermined',
            self.slug,
            filename.rpartition('.')[2].lower(),
        )

    tiny = models.ImageField(upload_to=image_path, blank=True, null=True)
    small = models.ImageField(upload_to=image_path, blank=True, null=True)
    medium = models.ImageField(upload_to=image_path, blank=True, null=True)
    press = models.ImageField(upload_to=image_path, blank=True, null=True)

pre_save接收者：

@receiver(pre_save, sender=SomeDjangoModel)
def set_unique_fn(sender, instance, **kwargs):
    """ Set a unique (but temporary) filename on all newly uploaded files. """

    for size in instance.IMAGE_SIZES:
        field = getattr(instance, '{}_img'.format(size), None)
        if not field:
            continue
        fieldfile = getattr(field, 'file', None)
        if isinstance(fieldfile, UploadedFile):
            fieldfile.name = u'{}.{}'.format(
                uuid.uuid4().hex,
                fieldfile.name.rpartition('.')[2],
            )

Answer 1

您可以更改image_path()，以便它返回已知道大小的可调用项：

def image_path(size):
    def callback(self, filename)
        """ Example return: [some-django-model]/[medium]/[product1].[jpg] """
        return u'{}/{}/{}.{}'.format(
            slugify(self._meta.verbose_name),
            size,
            self.slug,
            filename.rpartition('.')[2].lower(),
        )
    return callback

class SomeDjangoModel(models.Model):
    tiny = models.ImageField(upload_to=image_path('tiny'), blank=True, null=True)
    small = models.ImageField(upload_to=image_path('small'), blank=True, null=True)
    medium = models.ImageField(upload_to=image_path('medium'), blank=True, null=True)
    press = models.ImageField(upload_to=image_path('press'), blank=True, null=True)

动态上传路径 - 包括原始字段

1 个答案: