2015-10-19 64 views
0

我用现有的行在我的表user_table中添加了一个新的uuid字段。由于竞争条件,我在迁移时收到“重复输入”错误。然后我遵循的步骤在这里:Django:在uuid4中重复输入

http://django.readthedocs.org/en/latest/howto/writing-migrations.html

现在我的第一和第二迁移运行,但我最后的迁徙给了我同样的错误。是我的每个链接3迁移如下:

迁移 '0009_label'

from __future__ import unicode_literals 

from django.db import models, migrations 
import uuid 


class Migration(migrations.Migration): 

    dependencies = [ 
     ('app', '0008_label'), 
    ] 

    operations = [ 
     migrations.AddField(
      model_name='user_table', 
      name='UUID_loc', 
      field=models.UUIDField(default=uuid.uuid4, null=True), 
     ), 
     migrations.AlterField(
      model_name='another_table', 
      name='Time', 
      field=models.CharField(default=0, max_length=3), 
     ), 
    ] 

迁移 '0010_label'

from __future__ import unicode_literals 

from django.db import migrations, models 
import uuid 

def gen_uuid(apps, schema_editor): 
    MyModel = apps.get_model('app', 'user_table') 
    for row in MyModel.objects.all(): 
     row.uuid = uuid.uuid4() 
     row.save() 

class Migration(migrations.Migration): 

    dependencies = [ 
     ('app', '0009_label'), 
    ] 

    operations = [ 
     # omit reverse_code=... if you don't want the migration to be reversible. 
     migrations.RunPython(gen_uuid, reverse_code=migrations.RunPython.noop), 
    ] 

迁移 '0011_label'

from __future__ import unicode_literals 

from django.db import models, migrations 
import uuid 


class Migration(migrations.Migration): 

    dependencies = [ 
     ('app', '0010_label'), 
    ] 

    operations = [ 
     migrations.AlterField(
      model_name='user_table', 
      name='UUID_loc', 
      field=models.UUIDField(default=uuid.uuid4, unique=True), 
     ), 

    ] 

的联系是与我非常相关,但不幸的是我得到了同样的错误。现在我陷入了这里,我的表中有UUID_loc字段,但它还不是唯一的,即第三个迁移尚未运行。任何人都可以提供一些见解吗?谢谢。

回答

0

我认为你应该稍微编辑你的第二个迁移,以防止django用重复的uuid值更新你的模型。

def gen_uuid(apps, schema_editor): 
    MyModel = apps.get_model('app', 'user_table') 
    for row in MyModel.objects.all(): 
     while True: 
      row.uuid = uuid.uuid4() 
      if not MyModel.objects.filter(uuid=row.uuid).exists(): 
       break 

     row.save() 

的,你应该再重新运行第二个迁移,然后你应该能够运行没有任何问题第三次迁移。

+0

谢谢你的回答,但我仍然得到相同的错误:/ – plumSemPy

+0

更新:我意识到我有一个列名称的差异,它现在必须工作:) row.uuid = uuid.uuid4()应该是行。 UUID_loc = uuid.uuid4();那对我来说很愚蠢 – plumSemPy