更改应用程序名称和数据库表后，Django 无法进行迁移

Question

我正在使用 Django 1.11.5 并使用 PyCharm 作为我的 IDE。我一直在尝试将我的应用程序名称从“clinicaltrials”重构为“cancer_trials”。 PyCharm 相应地更新了我的所有项目文件。然后我按照 this SO answer 中的步骤操作更新适当的数据库表。但是，当我尝试运行 makemigration 时，出现以下错误。我似乎无法弄清楚这意味着什么和/或我在这里缺少什么部分。

    > python manage.py makemigrations
Traceback (most recent call last):
  File "manage.py", line 22, in <module>
    execute_from_command_line(sys.argv)
  File "C:\Python_3.6.1\lib\site-packages\django\core\management\__init__.py", line 364, in execute_from_command_line
    utility.execute()
  File "C:\Python_3.6.1\lib\site-packages\django\core\management\__init__.py", line 356, in execute
    self.fetch_command(subcommand).run_from_argv(self.argv)
  File "C:\Python_3.6.1\lib\site-packages\django\core\management\base.py", line 283, in run_from_argv
    self.execute(*args, **cmd_options)
  File "C:\Python_3.6.1\lib\site-packages\django\core\management\base.py", line 330, in execute
    output = self.handle(*args, **options)
  File "C:\Python_3.6.1\lib\site-packages\django\core\management\commands\makemigrations.py", line 150, in handle
    loader.project_state(),
  File "C:\Python_3.6.1\lib\site-packages\django\db\migrations\loader.py", line 323, in project_state
    return self.graph.make_state(nodes=nodes, at_end=at_end, real_apps=list(self.unmigrated_apps))
  File "C:\Python_3.6.1\lib\site-packages\django\db\migrations\graph.py", line 409, in make_state
    project_state = self.nodes[node].mutate_state(project_state, preserve=False)
  File "C:\Python_3.6.1\lib\site-packages\django\db\migrations\migration.py", line 92, in mutate_state
    operation.state_forwards(self.app_label, new_state)
  File "C:\Python_3.6.1\lib\site-packages\django\db\migrations\operations\fields.py", line 142, in state_forwards
    for name, instance in state.models[app_label, self.model_name_lower].fields:
KeyError: ('cancer_trials', 'cancer_trials')

这是抛出错误的函数

def state_forwards(self, app_label, state):
    new_fields = []
    old_field = None
    for name, instance in state.models[app_label, self.model_name_lower].fields:
        if name != self.name:
            new_fields.append((name, instance))
        else:
            old_field = instance
    state.models[app_label, self.model_name_lower].fields = new_fields
    # Delay rendering of relationships if it's not a relational field
    delay = not old_field.is_relation
    state.reload_model(app_label, self.model_name_lower, delay=delay)

Answer 1

根据我的经验，最简单的解决方案是创建新应用程序并复制代码:

1. 使用所需名称创建新应用并将其添加到设置
2. 将代码从旧应用复制/粘贴到新应用，将引用从旧应用更改为新应用，运行 makemigrations 并迁移
3. 打开数据库并将数据从旧表复制到新表
4. 检查新应用中的一切是否正常

5. 搜索 stackoverflow.com 或 google 如何从项目中删除应用程序或直接保留在那里。不幸的是，我对这些步骤并不是 100% 肯定，如果我错了，请有人纠正我，但据我记忆:

   • 运行python manage.py migrate old_app zero (这不会应用应用程序的所有迁移)
   • 从设置中删除应用
   • 删除文件