36

我刚刚将我的应用程序升级到 1.7（实际上仍在尝试）。

这就是我在models.py中的内容：

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

class UserProfile(AbstractUser):
    #...
    avatar = models.ImageField(upload_to=path_and_rename("avatars/"),
                               null=True, blank=True,
                               default="avatars/none/default.png",
                               height_field="image_height",
                               width_field="image_width")

当我尝试时makemigrations，它会抛出：

ValueError: Could not find function wrapper in webapp.models.
Please note that due to Python 2 limitations, you cannot serialize unbound method functions (e.g. a method declared
and used in the same class body). Please move the function into the main module body to use migrations.
4

4 回答 4

82

我不确定是否可以回答我自己的问题，但我只是想通了（我认为）。

根据这个错误报告，我编辑了我的代码：

from django.utils.deconstruct import deconstructible

@deconstructible
class PathAndRename(object):

    def __init__(self, sub_path):
        self.path = sub_path

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

path_and_rename = PathAndRename("/avatars")

然后，在字段定义中：

avatar = models.ImageField(upload_to=path_and_rename,
                               null=True, blank=True,
                               default="avatars/none/default.png",
                               height_field="image_height",
                               width_field="image_width")

这对我有用。

于 2014-09-10T14:27:27.557 回答
3

我有同样的问题，但我的模型中有很多 ImageFile

head = ImageField(upload_to=upload_to("head")
icon = ImageField(upload_to=upload_to("icon")
...etc

我不想为我拥有的每个 ImageField 列创建 upload_to 包装函数。

所以我只是创建了一个名为 wrapper 的函数，它就可以了

def wrapper():
    return

我认为它工作得很好，因为我打开了迁移文件，我发现了这个：

('head', models.ImageField(upload_to=wrapper)),

我想这不会影响迁移过程

于 2014-11-06T16:08:07.500 回答
0

您可以像这样使用 kwargs 创建函数：

def upload_image_location(instance, filename, thumbnail=False):
    name, ext = os.path.splitext(filename)
    path = f'news/{instance.slug}{f"_thumbnail" if thumbnail else ""}{ext}'
    n = 1
    while os.path.exists(path):
        path = f'news/{instance.slug}-{n}{ext}'
        n += 1
    return path

并在您的模型中将此方法与functools.partial一起使用：

image = models.ImageField(
    upload_to=upload_image_location,
    width_field='image_width',
    height_field='image_height'
)
thumbnail_image = models.ImageField(upload_to=partial(upload_image_location, thumbnail=True), blank=True)

你会得到这样的迁移：

class Migration(migrations.Migration):

    dependencies = [
        ('news', '0001_initial'),
    ]

    operations = [
        migrations.AddField(
            model_name='news',
            name='thumbnail_image',
            field=models.ImageField(blank=True, upload_to=functools.partial(news.models.upload_image_location, *(), **{'thumbnail': True})),
        ),
        migrations.AlterField(
            model_name='news',
            name='image',
            field=models.ImageField(height_field='image_height', upload_to=news.models.upload_image_location, width_field='image_width'),
        ),
    ]
于 2018-10-17T11:35:37.197 回答
0 
field=models.ImageField(blank=True, upload_to="/smth", *(), **{'thumbnail': True})),

代替

field=models.ImageField(blank=True, upload_to=functools.partial(news.models.upload_image_location, *(), **{'thumbnail': True}))
于 2022-02-08T07:37:41.147 回答