Question

class Sample(models.model):
sampleID = models.AutoField(primary_key=True)
    something = models.IntegerField()
    .
    .
    .

def save(self, *args, **kwargs):
    is_new = self.pk is None
    super(Sample, self).save(*args, **kwargs)
    if is_new:
        alpha = AnotherSample()
        alpha.asampleID =  self.pk
        alpha.say = "Lolz"
        alpha.save()

我无法将主键分配给Sample的AnotherSample引用键。总是说

Cannot assign "1L": "AnotherSample.asampleID" must be a "Sample" instance.

Answer 1

来自the docs：

在幕后，Django将"_id"附加到字段名称以创建其数据库列名。

alpha.asampleID_id = self.pk

这就是为什么后缀为“ID”是不合适的。

Answer 2

根据给定的错误消息，您似乎正在尝试保存关系。您应该传递实际对象，而不是其主键。

def save(self, *args, **kwargs):
    is_new = self.pk is None
    super(Sample, self).save(*args, **kwargs)
    if is_new:
        alpha = AnotherSample()
        alpha.asampleID =  self
        alpha.say = "Lolz"
        alpha.save()

获得外键的价值

2 个答案: