Question

我有模型

class Supplier_items(models.Model):
    si_name = models.CharField(max_length=200, blank=True, null= True, verbose_name='Название поставщика')
    si_url = models.CharField(max_length=200, blank=True, null= True, verbose_name='URL поставщика')
    si_ean = models.CharField(max_length=200, blank=True, null= True)
    si_price = models.FloatField( blank=True, null= True, verbose_name='Price')

    def __str__(self):
      return self.si_name

class Order(models.Model):
  order_id = models.CharField(max_length=50, blank=True, null= True)
  shop = models.CharField(max_length=50, blank=True, null= True)
  created = models.DateTimeField('Дата добавления', default=timezone.now)
  price = models.CharField(max_length=50, blank=True, null= True)
  price_real = models.CharField(max_length=50, blank=True, null= True)
  order_ean = models.CharField(max_length=100, blank=True, null= True)
  variant = models.CharField(max_length=50, blank=True, null= True)
 
  def __str__(self):
      return self.item_name

def variant(self, *args, **kwargs):
      queryset = Supplier_items.objects.filter(si_ean__contains = self.order_ean)
      return queryset

无法按键链接

在 list_display 中它显示文本是如何正常的，list_editable - 给出错误 (admin.E121) list_editable [0]' refers to 'variant' 的值

我想将 admin 中的 variant 字段作为选择字段输出，并将选择写入模型中的相应字段。

感谢您的帮助

没有fk的list_editable中的Django Admin自定义字段

0 个答案: