django - 使用 ForeignKey 对链接模型进行排序

Question

我需要Credit通过CreditPayment.

模型.py

class Credit(models.Model):
    hot = models.BooleanField(default=False)


class CreditPayment(models.Model):
    credit = models.ForeignKey(Credit)
    period_to = models.PositiveIntegerField()
    rate = models.DecimalField(max_digits=7, decimal_places=2)

视图.py

credits = credits.filter(hot=False).distinct().order_by(...)

输入数据示例：

Credit #1:
CreditPayment #1:
period_to = 12
rate = 10

CreditPayment #2:      (minimal)
period_to = 10
rate = 8

CreditPayment #3:      
period_to = 9
rate = 10

---

Credit #2:
CreditPayment #1:        (minimal)
period_to = 6
rate = 20

CreditPayment #2:
period_to = 9
rate = 20

---

Credit #3:
CreditPayment #1:
period_to = 12
rate = 8

CreditPayment #2:
period_to = 9
rate = 11

CreditPayment #3:       (minimal)
period_to = 9
rate = 8

结果，样本减少到：

Credit #1:    
CreditPayment #2:
period_to = 10
rate = 8

---

Credit #2:
CreditPayment #1:
period_to = 6
rate = 20

---

Credit #3:
CreditPayment #3:
period_to = 9
rate = 8

结果：

信用 #3 -> 信用 #1 -> 信用 #2

如您所见，最初CreditPayment为每个Credit（credits在views.py中）选择了最小值。然后，根据这些最小值CreditPayment，全部Credit排序。如果两个条目的比率相等，则将这些条目比较period to。据我了解，在这里您需要以某种方式应用聚合。

标签： djangodjango-modelsdjango-views

Answer 1

使用annotate()&prefetch_related()

模型.py

class CreditPayment(models.Model):
    credit = models.ForeignKey(Credit, related_name='creditpayments')
    ....

视图.py

from django.db.models import Max, Min, Prefetch

prefetch = Prefetch('creditpayments', CreditPayment.objects.all())
sorted_credits = Credit.objects.all().prefetch_related(prefetch).annotate(min_rate=Min('creditpayments__rate')).annotate(max_period_to=Max('creditpayments__period_to')).order_by('min_rate', '-max_period_to')