Python 我的模特怎么了
我正在使用Django 1.4和PostgreSQL 9.1.4 我的模特怎么了Python 我的模特怎么了,python,django,django-models,django-1.4,Python,Django,Django Models,Django 1.4,我正在使用Django 1.4和PostgreSQL 9.1.4 我的模特怎么了 class UserProfile(models.Model): ROLES_TYPE = ( ('Employee', 'Employee'), ('Employor', 'Employor'), ) user = models.OneToOneField(User, related_name='Profile') role = models.Cha
class UserProfile(models.Model):
ROLES_TYPE = (
('Employee', 'Employee'),
('Employor', 'Employor'),
)
user = models.OneToOneField(User, related_name='Profile')
role = models.CharField(max_length=15, choices=ROLES_TYPE)
class Student(models.Model):
id = models.AutoField(db_column='studentid', primary_key=True)
username = models.OneToOneField(UserProfile, related_name='Employee')
first_name = models.CharField(db_column='fname', max_length=100,
null=True, blank=True)
last_name = models.CharField(db_column='lname', max_length=100,
null=True, blank=True)
运行./manage.py syncdb时,出现错误
Creating tables ...
Creating table Student
Traceback (most recent call last):
File "manage.py", line 10, in <module>
execute_from_command_line(sys.argv)
File "/home/jayapaldharmamani/Projects/jayapal/aplink/aplink_env/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 443, in execute_from_command_line
utility.execute()
File "/home/jayapaldharmamani/Projects/jayapal/aplink/aplink_env/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 382, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/home/jayapaldharmamani/Projects/jayapal/aplink/aplink_env/local/lib/python2.7/site-packages/django/core/management/base.py", line 196, in run_from_argv
self.execute(*args, **options.__dict__)
File "/home/jayapaldharmamani/Projects/jayapal/aplink/aplink_env/local/lib/python2.7/site-packages/django/core/management/base.py", line 232, in execute
output = self.handle(*args, **options)
File "/home/jayapaldharmamani/Projects/jayapal/aplink/aplink_env/local/lib/python2.7/site-packages/django/core/management/base.py", line 371, in handle
return self.handle_noargs(**options)
File "/home/jayapaldharmamani/Projects/jayapal/aplink/aplink_env/local/lib/python2.7/site-packages/django/core/management/commands/syncdb.py", line 102, in handle_noargs
cursor.execute(statement)
File "/home/jayapaldharmamani/Projects/jayapal/aplink/aplink_env/local/lib/python2.7/site-packages/django/db/backends/util.py", line 40, in execute
return self.cursor.execute(sql, params)
File "/home/jayapaldharmamani/Projects/jayapal/aplink/aplink_env/local/lib/python2.7/site-packages/django/db/backends/postgresql_psycopg2/base.py", line 52, in execute
return self.cursor.execute(query, args)
django.db.utils.DatabaseError: column "id" referenced in foreign key constraint does not exist
创建表“学生”(
请帮帮我
谢谢和问候,
Jayapal D尝试从
Student
模型中的id
列中删除db\u column='studentid'
为什么要更改列名
另外,为什么不使用自动id?这是由数据库而不是django本身产生的问题。请尝试重新创建数据库架构(您可以删除所有相关列并执行
/manage.py syncdb
我尝试了这两种方法,但错误相同。如果删除字段用户名,则效果良好。为什么需要id为的异常db名称?请尝试删除此行:id=models.AutoField(db\u column='studentid',primary\u key=True)
。您不需要它(django将自动pk设置为每个模型)我尝试了所有可能的方法,但都出现了相同的错误。我删除了ID字段和相关名称。您能否发布django尝试执行的sql查询?请回答您的问题。
"id" serial NOT NULL PRIMARY KEY,
"user_id" integer NOT NULL UNIQUE REFERENCES "auth_user" ("id") DEFERRABLE INITIALLY DEFERRED,
"role" varchar(15) NOT NULL
);
"id" serial NOT NULL PRIMARY KEY,
"username_id" integer NOT NULL UNIQUE REFERENCES "admin_userprofile" ("id") DEFERRABLE INITIALLY DEFERRED
);