[英]Django migrating foreign keys django.db.utils.OperationalError: (1005, "Can't create table 'asp052mysqlpy.#sql-6ac_56' (errno: 150)") error
[英]django.db.utils.OperationalError: 1005, 'Can't create table `xyz`.`#sql-600_237` (errno: 150 "Foreign key constraint is incorrectly formed")
我正在尝试将一对一密钥添加到我的 Django 应用程序中,但是当我尝试“迁移”过程时总是会出现该错误(makemigrations 效果很好)。
Applying xyzapp.0007_personne_extended_foreign...Traceback (most recent call last):
File "manage.py", line 29, in <module>
execute_from_command_line(sys.argv)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 354, in execute_from_command_line
utility.execute()
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 346, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/core/management/base.py", line 394, in run_from_argv
self.execute(*args, **cmd_options)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/core/management/base.py", line 445, in execute
output = self.handle(*args, **options)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/core/management/commands/migrate.py", line 222, in handle
executor.migrate(targets, plan, fake=fake, fake_initial=fake_initial)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/db/migrations/executor.py", line 110, in migrate
self.apply_migration(states[migration], migration, fake=fake, fake_initial=fake_initial)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/db/migrations/executor.py", line 148, in apply_migration
state = migration.apply(state, schema_editor)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/db/backends/base/schema.py", line 91, in __exit__
self.execute(sql)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/db/backends/base/schema.py", line 111, in execute
cursor.execute(sql, params)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/db/backends/utils.py", line 79, in execute
return super(CursorDebugWrapper, self).execute(sql, params)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/db/backends/utils.py", line 64, in execute
return self.cursor.execute(sql, params)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/db/utils.py", line 98, in __exit__
six.reraise(dj_exc_type, dj_exc_value, traceback)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/db/backends/utils.py", line 64, in execute
return self.cursor.execute(sql, params)
File "/opt/xyz/env/local/lib/python2.7/site-packages/django/db/backends/mysql/base.py", line 124, in execute
return self.cursor.execute(query, args)
File "/usr/lib/python2.7/dist-packages/MySQLdb/cursors.py", line 226, in execute
self.errorhandler(self, exc, value)
File "/usr/lib/python2.7/dist-packages/MySQLdb/connections.py", line 36, in defaulterrorhandler
raise errorvalue
django.db.utils.OperationalError: (1005, 'Can\'t create table `xyz`.`#sql-600_297` (errno: 150 "Foreign key constraint is incorrectly formed")')
这就是我的模型的样子:
class PersonVolunteer(models.Model):
person = models.OneToOneField(Person)
class Person(models.Model):
name = models.CharField(max_length=100)
以及导致崩溃的迁移过程:
class 迁移(migrations.Migration):
dependencies = [
('xyzapp', '0014_member_to_person'),
]
operations = [
migrations.CreateModel(
name='PersonVolunteer',
fields=[
('id', models.AutoField(verbose_name='ID', serialize=False, auto_created=True, primary_key=True)),
('personne', models.OneToOneField(to='xyzapp.Personne')),
],
),
]
但是,即使在迁移错误之后,当我对其进行测试时,一切正常。 但是,如果在“迁移”期间出现该错误消息,那应该不是好事。 如果没有问题,是否可以跳过使我的迁移崩溃的最后一步?
有人可以说我为什么会收到该错误消息以及如何解决它?
非常感谢您,祝您有美好的一天!
我找到了一个解决方案,可能不是更清洁,但该死的它有效,这对我来说是完美的。
# -*- coding: utf-8 -*-
from __future__ import unicode_literals
from django.db import migrations, models
class Migration(migrations.Migration):
dependencies = [
('cvmapp', '0006_person_extended'),
]
operations = [
migrations.AddField(
model_name='PersonVolunteer',
name='personne',
field=models.OneToOneField(related_name='info_volunteer', to='cvmapp.Person', db_constraint=False),
),
]
诀窍是添加“db_constraint=False”作为 OneToOne 字段的参数。
请确保您的所有 MySQL 表都使用相同的存储引擎(即 MyISM v. InnoDB)。 特别是它们之间有外键的表。
如果您需要更多关于 MySQL 存储引擎的信息以及如何知道您使用的是哪一个,您需要阅读 MySQL 文档,尽管我们数据库文档中的 MySQL 注释也有一些介绍性信息。
我怀疑您使用 MyISAM 存储引擎(MySQL < 5.5 的默认值)创建了表,并且由于 MySQL 5.5 默认为 InnoDB 并且您从那时起创建了新表,您以混合结束
您可以从 phpMyadmin 更改表存储引擎。 单击您的 teble 名称,转到“操作”选项卡,然后在“表操作”框中进行更改。 将所有或您的存储引擎转换为相同的。
Look at the output of the command SHOW ENGINE INNODB STATUS
in MySql or MariaDB console client ( mariadb link , mysql link ). 它比错误消息提供更多信息。 例如,它向我显示了这条消息:
...
------------------------
LATEST FOREIGN KEY ERROR
------------------------
2021-09-20 18:27:08 7faea3ad1700 Error in foreign key constraint of table `my_db`.`django_admin_log`:
Alter table `my_db`.`django_admin_log` with foreign key constraint failed. Referenced table `my_db`.`profiles_userprofile` not found in the data dictionary near ' FOREIGN KEY (`user_id`) REFERENCES `profiles_userprofile` (`id`)'.
...
它对我说我忘了创建迁移。
首先要做的应该是检查表的存储引擎。 但是,在我的情况下,存储引擎是相同的(都是 INNODB)。 我的问题是无与伦比的表格collation
。 设置数据库的默认字符集后,我的问题已经解决。
您可以通过以下方式查看表格详细信息;
SHOW TABLE STATUS WHERE Name = "yyyyyyy";
一旦你弄清楚你的表(将是 ForeignKey)排序规则,你可以设置你的数据库默认字符集,如下所示:
ALTER DATABASE my_db COLLATE = 'latin1_swedish_ci';
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