Django错误.无法分配必须是实例

Question

当我尝试在其中一个表中运行插入时,出现以下错误.

无法分配"1":"Team.department_id"必须是"部门"实例

不可否认,我有点不确定我是否正确使用外键概念.我正在尝试运行的插件和我的models.py中的一个片段如下所示.

我想要做的是当有人想要创建一个新团队时.他们必须将它附加到一个部门.因此,部门ID应该在两组表中.

new_team = Team(
    nickname = team_name,
    employee_id = employee_id,
    department_id = int(Department.objects.get(password = password, department_name = department_name).department_id)
)

models.py

class Department(models.Model):  
    department_id = models.AutoField(auto_created=True, primary_key=True, default=1)  
    department_name = models.CharField(max_length=60)
    head_id = models.CharField(max_length=30)
    password = models.CharField(max_length=128)


class Team(models.Model):  
    team_id = models.AutoField(primary_key=True)
    department_id = models.ForeignKey('Department', related_name = 'Department_id')
    employee_id = models.CharField(max_length=30)
    nickname = models.CharField(max_length=60)
    team_image = models.ImageField(upload_to=get_image_path, blank=True, null=True)

在此处输入代码

Answer 1

您不需要传递部门ID,实例本身就足够了.以下应该工作得很好:

new_team = Team(
    nickname = team_name,
    employee_id = employee_id,
    department_id = Department.objects.get(password = password, department_name = department_name)
)

只是注意,不要老是命名洋田东西 _id.那东西就足够了.Django旨在从用户的角度简化操作,_id后缀意味着您正在考虑数据库层.实际上,如果您为列命名department,django将自动department_id为您创建数据库中的列.事情的方式,你正在制作django创造department_id_id,这是相当愚蠢的.

Answer 2

这首先出现在谷歌搜索中，因此为新手提供了一种选择。如果您可以方便地访问 id 并且不想进行其他查询，您也可以这样做：

new_team = Team(
    nickname = team_name,
    employee_id = employee_id,
    department_id_id = Department.objects.get(password = password, department_name = department_name).department_id
)

简而言之，{foreign_key_name}_id如果你想直接分配id。