我试图添加一个新的数据库模型,让我能够"分组"费用,但在运行python manage.py makemigrations时遇到了这个问题
我的虚拟环境如下所示:
Django==1.7.3
argparse==1.2.1
django-braces==1.4.0
django-chartit==0.1
django-crispy-forms==1.4.0
django-debug-toolbar==1.2.2
psycopg2==2.6
six==1.9.0
sqlparse==0.1.14
wsgiref==0.1.2
这是追溯:
Traceback (most recent call last):
File "manage.py", line 10, in <module>
execute_from_command_line(sys.argv)
File "/home/jeff/.virtualenvs/ggc/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 385, in execute_from_command_line
utility.execute()
File "/home/jeff/.virtualenvs/ggc/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 377, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/home/jeff/.virtualenvs/ggc/local/lib/python2.7/site-packages/django/core/management/base.py", line 288, in run_from_argv
self.execute(*args, **options.__dict__)
File "/home/jeff/.virtualenvs/ggc/local/lib/python2.7/site-packages/django/core/management/base.py", line 338, in execute
output = self.handle(*args, **options)
File "/home/jeff/.virtualenvs/ggc/local/lib/python2.7/site-packages/django/core/management/commands/makemigrations.py", line 111, in handle
convert_apps=app_labels or None,
File "/home/jeff/.virtualenvs/ggc/local/lib/python2.7/site-packages/django/db/migrations/autodetector.py", line 42, in changes
changes = self._detect_changes(convert_apps, graph)
File "/home/jeff/.virtualenvs/ggc/local/lib/python2.7/site-packages/django/db/migrations/autodetector.py", line 109, in _detect_changes
self.old_apps = self.from_state.render(ignore_swappable=True)
File "/home/jeff/.virtualenvs/ggc/local/lib/python2.7/site-packages/django/db/migrations/state.py", line 67, in render
model.render(self.apps)
File "/home/jeff/.virtualenvs/ggc/local/lib/python2.7/site-packages/django/db/migrations/state.py", line 316, in render
body,
File "/home/jeff/.virtualenvs/ggc/local/lib/python2.7/site-packages/django/db/models/base.py", line 229, in __new__
'base class %r' % (field.name, name, base.__name__)
django.core.exceptions.FieldError: Local field u'id' in class 'ExpenseGroup' clashes with field of similar name from base class 'Asset
&#39;
这是模型代码 - 我尝试将ExpenseGroup模型切割为仅作为字段的groupName但我得到相同的错误。我错过了什么?
class Asset(TimeStampedModel):
assetName = models.CharField(max_length=255)
assetAddress = models.CharField(max_length=255)
slug = models.SlugField(max_length=255, blank=True)
class Meta:
unique_together = ('assetName', 'assetAddress')
def __str__(self):
return self.assetName
def save(self, *args, **kwargs):
self.slug = slugify(self.assetName)
super(Asset, self).save(*args, **kwargs)
class ExpenseGroup(TimeStampedModel):
groupName = models.CharField(max_length=255, blank=False)
expenseName = models.ForeignKey(Expense, related_name='expenseGroup')
class Meta:
unique_together = ('expenseName', 'groupName')
def __str__(self):
return self.groupName
def save(self, *args, **kwargs):
return super(ExpenseGroup, self).save(*args, **kwargs)
class Expense(TimeStampedModel):
assetName = models.ForeignKey(Asset, related_name='assetExpense')
category = models.ForeignKey(ExpenseCategory, related_name='expenseCategory')
expensePeriod = models.DateTimeField(blank=False)
expenseName = models.CharField(max_length=255, blank=False)
expenseAmount = models.DecimalField(max_digits=20, decimal_places=2, blank=True)
class Meta:
unique_together = ('expenseName', 'expensePeriod')
def __str__(self):
return self.expenseName
def save(self, *args, **kwargs):
super(Expense, self).save(*args, **kwargs)
答案 0 :(得分:1)
你可以发布TimeStampedModel定义吗?我怀疑你没有将基本模型声明为Abstract。这就是“id”字段与彼此冲突的原因。
class TimeStampedModel(models.Model):
created = models.DateTimeField(auto_now_add=True)
modified = models.DateTimeField(auto_now=True)
class Meta:
abstract = True
答案 1 :(得分:1)
这是Django 1.7.3的一个错误,升级到最新版本,你会没事的(当时是1.7.5)
答案 2 :(得分:0)
这比你发布的更多。你可以发布整个模型文件吗?或者一个要点的链接?我很惊讶你没有从'ExpenseGroup'模型中获得关于FK引用'Expense'的错误,因为它在文件的后面定义。
您是否有任何现有迁移?我建议删除任何现有的迁移和所有pyc文件,然后重试。