Question

我使用此代码创建重命名功能以在模型中重命名文件。

PGSSLMODE=verify-full PGSSLROOTCERT=server-ca.pem psql -h your_host -U your_user -W -d your_db

我使用这样的功能。

def path_and_rename(path):
    def path_and_rename_func(instance, filename):
        upload_to = path
        ext = filename.split('.')[-1]
        # get filename
        if instance.pk:
            filename = '{}.{}'.format(instance.pk, ext)
        else:
            # set filename as random string
            filename = '{}.{}'.format(uuid4().hex, ext)
        # return the whole path to the file
        return os.path.join(upload_to, filename)
    return path_and_rename_func

当我使用命令进行迁移image=models.ImageField(upload_to=path_and_rename("test_image")) video=models.FileField(upload_to=path_and_rename("test_file"),blank=True, null=True)

时

它显示这样的错误。

python manage.py makemigrations

如何解决？

Answer 1

您不能使用这种包装函数，因为这意味着结果函数没有限定名称。

但是，您可以定义两个辅助函数并使用它们，例如：

def path_and_rename_func(instance, filename, path):
    upload_to = path
    ext = filename.split('.')[-1]
    # get filename
    if instance.pk:
        filename = '{}.{}'.format(instance.pk, ext)
    else:
        # set filename as random string
        filename = '{}.{}'.format(uuid4().hex, ext)
    # return the whole path to the file
    return os.path.join(upload_to, filename)

def path_and_rename_test_image(instance, filename):
    return path_and_rename(instance, filename, 'test_image')

def path_and_rename_test_file(instance, filename):
    return path_and_rename(instance, filename, 'test_file')

class MyModel(models.Model):
    image = models.ImageField(upload_to=path_and_rename_test_image)
    video = models.FileField(
        upload_to=path_and_rename_test_file,
        blank=True,
        null=True
    )

创建重命名文件的功能后，Django无法迁移模型

1 个答案: