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

我能想到的唯一方法是创建一个file.name用 uuid 替换的 pre_save 接收器。然后 upload_to 可调用通过将其与filename. 难道没有一种不那么老套的方法吗？

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)

接收方pre_save：

@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],
            )
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1 回答 1

6

您可以进行更改image_path()，使其返回一个已经知道大小的可调用对象：

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)
于 2011-07-31T12:07:09.097 回答