Question

我有一个模型，我的目标是生成主键（字符串），然后重命名上传的文件以匹配该字符串。这是我缩短的模型和我使用的upload_to函数。

class Thing(models.Model):
    id = models.CharField(primary_key=True, max_length=16)
    photo = ImageField('Photo', upload_to=upload_path, null=False, blank=False)
    ...



def upload_path(instance, filename):
    if not instance.id:
        randid = random_id(16)    # This returns a 16 character string of ASCII characters
        while Thing.objects.filter(id=randid).exists():
            logger.error("[Thing] ThingID of %s already exists" % randid)
            randid = random_id(16)
        instance.id = randid
    return "%s%s" % ("fullpath/",randid)

这会导致图像在适当的路径中正确地重命名为随机字符串。但是，主键设置为空字符串。

如何使用生成的主键重命名ImageField文件和正确保存生成的主键？

Answer 1

您可以在Model上定义save方法。设置True null 和空白 ImageField，如果必须控制空字段，则可以在模型表单上控制它们。因此，您可以轻松地在DB上为模型生成唯一ID。然后，您可以使用唯一ID或自定义生成ID保存图像。最好使用我建议的主键的默认唯一整数id。

class Thing(models.Model):
    fake_id = models.CharField(max_length=16)
    photo = ImageField('Photo', upload_to=upload_path, null=True, blank=True)

    def save(self, *args, **kwargs):

        imagefile = self.photo

        self.photo = ''
        super(Thing, self).save(*args, **kwargs)

    """ after superclass save, you can use self.id also its unique integer """

        randid = random_id(16)    # This returns a 16 character string of ASCII characters
        while Thing.objects.filter(id=randid).exists():
            logger.error("[Thing] ThingID of %s already exists" % randid)
            randid = random_id(16)

       self.fake_id = randid

   """manipulate rename and upload your file here you object is 'imagefile' """

       save_dir = "fullpath/" + new_file_name_with_randid

       self.photo = save_dir

       super(Thing, self).save(*args, **kwargs)

Answer 2

我最终删除了upload_to回拨，并在Thing的{{1}}方法中执行了此操作。

save()

在ImageField upload_to调用上访问生成的主键

2 个答案: