python - 内部加入 Django 1.9

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)

id 根据请求接收 (def details(request, id):)。

我需要从 ModelInteractions 中选择 comm，其中 id2 = id(根据请求收到 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)