模型中的 django 循环导入

Question

我知道我在这里有一个循环导入，但其中一个用于定义一些外键字段，另一个仅用于另一个模型中的属性。

型号：

收据/models.py

from django.db import models
from employees.models import Employee

class Receipt(models.Model):
    Supervisor = models.ForeignKey(Employee, on_delete=models.CASCADE, related_name='receipt_supervisor')
    Employee = models.ForeignKey(Employee, on_delete=models.CASCADE, related_name='receipt_employee')
    amount = models.DecimalField(max_digits=6, decimal_places=2)

员工/models.py

from django.db import models
from receipts.models import Receipt

class Employee(models.Model):
    number = models.CharField(max_length=50, unique=True)
    first_name = models.CharField(max_length=255)
    last_name = models.CharField(max_length=255)

    @property
    def receipt_sum(self):
        receipts = Receipt.objects.filter(Employee=self.pk)
        sum = 0
        for receipt in receipts:
            sum += receipt.amount
        return sum

我收到此错误：

无法从部分初始化的模块“employees.models”导入名称“Employee”（很可能是由于循环导入）

有没有办法解决这个问题，以便我可以在 Employee 属性中进行收据查询？还是我需要在别处进行此计算？

谢谢！

Answer 1

有没有办法解决这个问题，以便我可以在 Employee 属性中进行收据查询？

可以，可以在方法中导入模块：

from django.db import models

class Employee(models.Model):
    number = models.CharField(max_length=50, unique=True)
    first_name = models.CharField(max_length=255)
    last_name = models.CharField(max_length=255)

    @property
    def receipt_sum(self):
        from receipts.models import Receipt
        receipts = Receipt.objects.filter(Employee=self.pk)
        sum = 0
        for receipt in receipts:
            sum += receipt.amount
        return sum

但其实根本不需要导入Receipt，可以利用receipt_employee关系来访问相关的Receipt对象，也可以用.aggregate(…) [Django-doc]来总结数据库方面，这可能更有效：

from django.db import models
from django.db.models import Sum

class Employee(models.Model):
    number = models.CharField(max_length=50, unique=True)
    first_name = models.CharField(max_length=255)
    last_name = models.CharField(max_length=255)

    @property
    def receipt_sum(self):
        return self.receipt_employee.aggregate(
            total=Sum('amount')
        )['total'] or 0