在Django(2.x)中,我有一个输入表单,模型在这里:
from django.db import models
from django.contrib.auth.models import User
from django.conf import settings
class Sample(models.Model):
sample_id = models.AutoField(primary_key=True)
area_easting = models.IntegerField()
area_northing = models.IntegerField()
context_number = models.IntegerField()
sample_number = models.IntegerField()
# taken_by = models.IntegerField()
taken_by = models.ForeignKey(settings.AUTH_USER_MODEL, on_delete = models.PROTECT)
def __str__(self):
return str(self.sample_id)
class Meta:
db_table = 'samples\".\"sample'
#ordering = ["sample_id"]
managed = False
#verbose_name_plural = "samples"
这按预期工作,用户名列表下降(而我想格式化-名字姓氏)。但是,当我返回主查看页面时,看到一个错误。
django.db.utils.ProgrammingError: column sample.taken_by_id does not exist
LINE 1: ...text_number", "samples"."sample"."sample_number", "samples"....
^
HINT: Perhaps you meant to reference the column "sample.taken_by".
很明显,Django正在将_id
添加到导致错误的表名中,我希望是因为它是外键。
有什么想法可以纠正这种行为吗?
答案 0 :(得分:1)
您可以通过db_column
属性来显式设置基础数据库列:
taken_by = models.ForeignKey(settings.AUTH_USER_MODEL, db_column='taken_by', on_delete=models.PROTECT)
答案 1 :(得分:0)
https://docs.djangoproject.com/en/2.1/ref/models/fields/#database-representation
^指向文档的链接,在该文档中它指定创建_id
字段。
基于您已发布的错误消息。看来您的数据库架构没有更新。
您可能需要管理makemigrations
和migrate
才能将模型更改应用于数据库模式
例如
$ python manage.py makemigrations
# to apply the new migrations file
$ python manage.py migrate