Question

我们假设我们有以下模型：

from django.db import models

class Foo(models.Model):
    name = models.CharField(max_length=50)

class Bar(models.Model):
    name = models.CharField(max_length=50)

class Zomg(models.Model):
    foo = models.ForeignKey(Foo)
    bar = models.ForeignKey(Bar)

在Zomg模型中foo和bar字段用作复合键，也就是说，对于任何一对（foo，bar），只有一个Zomg对象。

在我的项目中，我需要不时更新数千条Zomg条记录，因此查找记录的效率非常重要。问题是我无法将foo_id和bar_id传递给Zomg.objects.get()方法：

>>> z = models.Zomg.objects.get(foo_id=1, bar_id=1)
Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "/usr/lib/python2.7/site-packages/django/db/models/manager.py", line 132, in get
    return self.get_query_set().get(*args, **kwargs)
  File "/usr/lib/python2.7/site-packages/django/db/models/query.py", line 341, in get
    clone = self.filter(*args, **kwargs)
  File "/usr/lib/python2.7/site-packages/django/db/models/query.py", line 550, in filter
    return self._filter_or_exclude(False, *args, **kwargs)
  File "/usr/lib/python2.7/site-packages/django/db/models/query.py", line 568, in _filter_or_exclude
    clone.query.add_q(Q(*args, **kwargs))
  File "/usr/lib/python2.7/site-packages/django/db/models/sql/query.py", line 1194, in add_q
    can_reuse=used_aliases, force_having=force_having)
  File "/usr/lib/python2.7/site-packages/django/db/models/sql/query.py", line 1069, in add_filter
    negate=negate, process_extras=process_extras)
  File "/usr/lib/python2.7/site-packages/django/db/models/sql/query.py", line 1260, in setup_joins
    "Choices are: %s" % (name, ", ".join(names)))
FieldError: Cannot resolve keyword 'foo_id' into field. Choices are: bar, foo, id

但是，当对象被实例化时，foo_id和bar_id几乎可以访问：

>>> z.foo_id
1
>>> z.bar_id
1

现在我不得不首先获取Foo和Bar个对象，然后使用它们获取所需的Zomg对象，从而进行两次不必要的数据库查询：

>>> f = models.Foo.objects.get(id=1)
>>> b = models.Bar.objects.get(id=1)
>>> z = models.Zomg.objects.get(foo=f, bar=b)

所以问题是，如何通过相关字段的ID获取Zomg个对象？

Answer 1

试试这个： z = models.Zomg.objects.get(foo__id=1, bar__id=1)

Django - 通过相关字段的ID获取对象

1 个答案: