Django:表不存在 [英] Django : Table doesn't exist
本文介绍了Django:表不存在的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
python manage.py syncdb
它显示错误,如
django.db.utils.ProgrammingError: (1146,表someapp.feed不存在)
models.py
class feed(models.Model):
user = models.ForeignKey(User,null = True,blank = True)
feed_text = models.CharField(max_length = 2000)
date = models.CharField(max_length = 30)
upvote = models.IntegerField(default = 0)
downvote = models。 IntegerField(default = 0)
def __str __(self):
return feed.content
我可以做什么来获取该应用的表?
解决方案
- drop tables(你已经做过),
- 注释掉model.py中的模型,
- 和..
如果django版本> = 1.7:
python manage.py makemigrations
python manage.py migrate --fake
其他
python manage.py schemamigration someapp --auto
python manage.py migrate someapp --fake
- 注释你模型中的models.py
- 转到步骤3. 但是这次没有 strong> - fake
I dropped some table related to an app. and again tried the syncdb command
python manage.py syncdb
It shows error like
django.db.utils.ProgrammingError: (1146, "Table 'someapp.feed' doesn't exist")
models.py
class feed(models.Model):
user = models.ForeignKey(User,null=True,blank=True)
feed_text = models.CharField(max_length=2000)
date = models.CharField(max_length=30)
upvote = models.IntegerField(default=0)
downvote = models.IntegerField(default=0)
def __str__(self):
return feed.content
What I can do to get the tables for that app ?
解决方案
- drop tables (you already did),
- comment-out the model in model.py,
- and ..
if django version >= 1.7:
python manage.py makemigrations
python manage.py migrate --fake
else
python manage.py schemamigration someapp --auto
python manage.py migrate someapp --fake
- comment-in your model in models.py
- go to step 3. BUT this time without --fake
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