我已经有一个使用PostgreSQL作为db的应用程序。我想用Django构建一个服务。尝试过“inspectdb”...但它给人一种期待:在Django中使用inspectdb时出错
Traceback (most recent call last):
File "manage.py", line 10, in <module>
execute_from_command_line(sys.argv)
File "/usr/local/lib/python3.4/dist-packages/django/core/management/__init__.py", line 350, in execute_from_command_line
utility.execute()
File "/usr/local/lib/python3.4/dist-packages/django/core/management/__init__.py", line 342, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/usr/local/lib/python3.4/dist-packages/django/core/management/base.py", line 348, in run_from_argv
self.execute(*args, **cmd_options)
File "/usr/local/lib/python3.4/dist-packages/django/core/management/base.py", line 399, in execute
output = self.handle(*args, **options)
File "/usr/local/lib/python3.4/dist-packages/django/core/management/commands/inspectdb.py", line 25, in handle
for line in self.handle_inspection(options):
File "/usr/local/lib/python3.4/dist-packages/django/core/management/commands/inspectdb.py", line 70, in handle_inspection
constraints = connection.introspection.get_constraints(cursor, table_name)
File "/usr/local/lib/python3.4/dist-packages/django/db/backends/postgresql/introspection.py", line 172, in get_constraints
"foreign_key": tuple(used_cols[0].split(".", 1)) if kind.lower() == "foreign key" else None,
IndexError: list index out of range
嗯,它是一个非常大的数据库?你能设法自己写模型吗? – Kris
嗨克里斯,很多关系都在那里。我认为这将是错误倾向 – gagangupt16
我听到你的声音,但我不会依赖这样的捷径。无论如何,它肯定会犯错 - 我相信你遇到过一个哈哈。我会自己编写模型,编写一个数据库迁移脚本,将数据库中的内容复制到django的体积适中的块中。它的乏味,但我不知道你还能做什么。 – Kris