Django 模型在同一类的属性中引用另一个属性答案

【问题标题】：Django models reference another attribute within an attribute of the same classDjango 模型在同一类的属性中引用另一个属性
【发布时间】：2020-07-16 11:22:48
【问题描述】：

我正在尝试为我的 django 网络应用程序的用户的个人资料详细信息构建一个模型，例如：

class UserDetails(models.Model):
    user = models.ForeignKey(User, on_delete=models.SET_NULL, null=True)
    profilePicture = models.ImageField(blank = True, upload_to='profile_pics/'+self.user.id+'/')
    country = models.CharField(max_length = 50, default='India')
    gender = models.CharField(max_length=10, default='NA')
    birthday = models.DateField(default=datetime.now())

    phone = models.CharField(max_length=15)

我在上述模型中有一个图像字段，我想将传入的图像上传到我的媒体存储路径中的路径profile_pics/<id of the user whose profile is being set up>/。我试图通过将图像字段的upload_to 属性指定为upload_to = 'profile_pics/'+self.user.id+'/' 来做到这一点。我将 AWS S3 用于我的媒体存储，并且在我的设置中设置了必要的设置：

AWS_ACCESS_KEY_ID = 'myaccesskeyid'
AWS_SECRET_ACCESS_KEY = 'mysecretaccesskey'
AWS_STORAGE_BUCKET_NAME = 'mybucketname'

AWS_S3_FILE_OVERWRITE = False
AWS_DEFAULT_ACL = None

DEFAULT_FILE_STORAGE = 'storages.backends.s3boto3.S3Boto3Storage'

当我尝试进行迁移时，我收到以下错误：

Traceback (most recent call last):
  File "manage.py", line 21, in <module>
    main()
  File "manage.py", line 17, in main
    execute_from_command_line(sys.argv)
  File "/home/suraj/Work/treeapp/treeapp-backend/treeEnv/lib/python3.6/site-packages/django/core/management/__init__.py", line 401, in execute_from_command_line
    utility.execute()
  File "/home/suraj/Work/treeapp/treeapp-backend/treeEnv/lib/python3.6/site-packages/django/core/management/__init__.py", line 377, in execute
    django.setup()
  File "/home/suraj/Work/treeapp/treeapp-backend/treeEnv/lib/python3.6/site-packages/django/__init__.py", line 24, in setup
    apps.populate(settings.INSTALLED_APPS)
  File "/home/suraj/Work/treeapp/treeapp-backend/treeEnv/lib/python3.6/site-packages/django/apps/registry.py", line 114, in populate
    app_config.import_models()
  File "/home/suraj/Work/treeapp/treeapp-backend/treeEnv/lib/python3.6/site-packages/django/apps/config.py", line 211, in import_models
    self.models_module = import_module(models_module_name)
  File "/home/suraj/Work/treeapp/treeapp-backend/treeEnv/lib/python3.6/importlib/__init__.py", line 126, in import_module
    return _bootstrap._gcd_import(name[level:], package, level)
  File "<frozen importlib._bootstrap>", line 994, in _gcd_import
  File "<frozen importlib._bootstrap>", line 971, in _find_and_load
  File "<frozen importlib._bootstrap>", line 955, in _find_and_load_unlocked
  File "<frozen importlib._bootstrap>", line 665, in _load_unlocked
  File "<frozen importlib._bootstrap_external>", line 678, in exec_module
  File "<frozen importlib._bootstrap>", line 219, in _call_with_frames_removed
  File "/home/suraj/Work/treeapp/treeapp-backend/userMgmt/models.py", line 7, in <module>
    class UserDetails(models.Model):
  File "/home/suraj/Work/treeapp/treeapp-backend/userMgmt/models.py", line 9, in UserDetails
    profilePicture = models.ImageField(blank = True, upload_to='profile_pics/'+self.user.id+'/')
NameError: name 'self' is not defined

请帮我将上传到此模型的图像的默认路径设置为profile_pics/<id of the user whose profile is being set up>/。

【问题讨论】：

标签： python django postgresql amazon-web-services amazon-s3

【解决方案1】：

您可以将可调用对象传递给upload_to=… parameter [Django-doc]：

class UserDetails(models.Model):
    def profile_picture_upload(self, filename):
        return 'profile_pics/{}/{}'.format(self.user_id, filename)

    # …
    profilePicture = models.ImageField(blank=True, upload_to=profile_picture_upload)

注意：通常Django模型中的字段名是用snake_case写的，而不是PerlCase，所以应该是：@ 987654324@ 而不是 ~~profilePicture~~。

【讨论】：