Django OneToOne字段用户编程错误

时间:2017-01-05 11:49:36

标签: python django django-models

我尝试首先使用OneToOneField关系向用户添加ManyToManyField。这是我的代码:

ENTERPRISE_TYPE_CHOICES = (
    ('a', "A"),
    ('b', "B"),
    ('c', "C"),
)

class Enterprise(models.Model):
    et_type = models.CharField(
        max_length = 10,
        choices=ENTERPRISE_TYPE_CHOICES,
        default='a'
        )
    name = models.TextField(max_length=100)
    number = models.TextField(max_length=7)

    def __str__(self):
        return "%s %s %s" % (self.et_type, self.name, self.number)

class Secretaire(models.Model):
    user = models.OneToOneField(User, on_delete=models.CASCADE)
    workshop = models.ManyToManyField(Enterprise)

@receiver(post_save, sender=User)
def create_user_secretaire(sender, instance, created, **kwargs):
    if created:
        Secretaire.objects.create(user=instance)

@receiver(post_save, sender=User)
def save_user_secretaire(sender, instance, **kwargs):
    instance.profile.save()

然后在python manage.py shell中我尝试:

from django.contrib.auth.models import User
u = User.objects.create(username="a", email="a@a.fr", password="a")

它抛出错误:

ProgrammingError:relation" helpdesk_secretaire"不存在 第1行:插入" helpdesk_secretaire" (" user_id")VALUES(11)RE ...

(用户仍然被创建,但没有人能够访问user.secretaire,它会抛出同样的错误)

您对我收到此错误的原因有任何疑问吗?

1 个答案:

答案 0 :(得分:1)

根据对原始问题的评论:您需要先运行makemigrationsmigrate。您收到的错误消息表明该表在数据库中不存在。