我正在尝试创建具有多个地址的地址表格,用户可以在其中选择家庭住址或送货地址。我有当前模型:
from django.db import models
from django.contrib.auth.models import User
from PIL import Image
class Address(models.Model):
name = models.CharField(max_length=30)
address = models.CharField(max_length=50)
city = models.CharField(max_length=60, default="Miami")
state = models.CharField(max_length=30, default="Florida")
zipcode = models.CharField(max_length=5, default="33165")
country = models.CharField(max_length=50)
class Meta:
verbose_name = 'Address'
verbose_name_plural = 'Address'
def __str__(self):
return self.name
所以我想知道那是否正确。
无论如何,我想知道如何使用当前模型创建视图,以便获得地址表格。使用普通模型会很容易,但是如何使用模型中的贯穿选项呢?
有人可以帮我吗?
谢谢
答案 0 :(得分:0)
使用外键指向您的地址模型:
class Profile(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
nick_name = models.CharField('Nick name', max_length=30, blank=True, default='')
bio = models.TextField(max_length=500, blank=True)
image = models.ImageField(default='default.jpg', upload_to='profile_pics')
addresses = models.ForeignKey(Address) # <-- fix here
希望这会有所帮助!
答案 1 :(得分:0)
您应以>>>import newspaper
>>>cbs_paper = newspaper.build('http://cbs.com', memoize_articles=False)
>>>cbs_paper.size()1030
格式声明ForeignKey
:
'<app>.<model>'
或直接给班级:
class AddressType(models.Model):
address = models.ForeignKey('yourapp.Address', on_delete=models.CASCADE)
profile = models.ForeignKey('yourapp.Profile', on_delete=models.CASCADE)
答案 2 :(得分:0)
其他两个答案都不正确,我最终修改了所有内容并创建了一个新模型,这里是:
Summarised_QuercusRobur1NewData<-structure(list(Stand_density_index = structure(c(1,1,2,2), .Label = c("1",
"2"), class = "factor"), Species = structure(c(1L,
1L, 1L, 1L), .Label = "Quercus robur", class = "factor"), Obs_no = c(9L,
82L, 40L, 58L), Mean_Tree_Diameter = c(86.9222222222222, 121.717073170732,
82, 72.4275862068965), SD_Tree_Diameter = c(57.2766046867693,
134.510951231506, 60.202253131019, 61.1575440200358)), row.names = c(NA,
-4L), class = "data.frame")