Django 1.9 中的内部连接

Question

我有这两个模型：

class ModelInteractions(models.Model):
    id1 = models.IntegerField(primary_key=True)
    id2 = models.IntegerField()
    comm = models.TextField(blank=True, null=True)

    class Meta:
        managed = False
        unique_together = (('id1', 'id2'),)


class Models(models.Model):
    id = models.IntegerField(primary_key=True)
    name = models.TextField()

    class Meta:
        managed = False

我还想选择comm。在我看来，我使用以下代码从Models获取数据：

condition = Q(name__icontains=names[0])
for name in names[1:]:
    condition &= Q(name__icontains=name)
# ↓↓↓ this line is for what I need but it doesn't work
condition &= Q(ModelInteractions__id2=id)

models = Models.objects.filter(condition)

应请求接收id (def details(request, id):)。

我需要从ModelInteractions 中选择comm，其中id2 = id（根据要求收到ID）。

当前代码返回：

Cannot resolve keyword 'ModelInteractions' into field. Choices are: id, name

Answer 1

class ModelInteractions(models.Model):
    id1 = models.IntegerField(primary_key=True)
    id2 = models.IntegerField()
    comm = models.TextField(blank=True, null=True)

    class Meta:
        managed = False
        unique_together = (('id1', 'id2'),)


class Models(models.Model):
    id = models.IntegerField(primary_key=True)
    name = models.TextField()
    interaction= models.ForeignKey(ModelInteractions,on_delete=models.CASCADE)

像这样制作你的模型，然后进行迁移和迁移

然后运行相同的查询直到这里

condition = Q(name__icontains=names[0])
for name in names[1:]:
    condition &= Q(name__icontains=name)

然后改变这一行

condition &= Q(interaction__id2=id)

然后

models = Models.objects.filter(condition)