Django 按点列表的最小距离排序

Question

我想获取在给定距离内的所有作业的查询集，这些作业至少到许多提供的位置之一，按最小距离排序，并且不显示重复的作业。

from django.db import models

from cities.models import City    

class Job(models.Model):

    title = models.CharField(max_length=255)

    cities = models.ManyToManyField(City)

如果只有一点，我可以这样做：

from django.contrib.gis.db.models.functions import Distance

from django.contrib.gis.geos import Point

point = Point(x, y, srid=4326)  

Job.objects.filter(cities__location__dwithin=(point, dist)) \

           .annotate(distance=Distance("cities__location", point) \

           .order_by('distance')

但是当我有很多点时，我为过滤器构建了一个 Q 表达式，但不确定是否有一种干净的方法来注释作业到所有点的最小距离

query = Q()

for point in points:

    query |= Q(cities__location__dwithin=(point, dist))

Job.objects.filter(query).annotate(distance=Min(...)).order_by('distance')

仅供参考，使用带有 PostGIS 扩展的 postgres 12.1

Answer 1

query = Q()distances = []for point in points:    query |= Q(cities__location__dwithin=(point, dist))    distances.append(Distance("cities__location", point))# LEAST requires 2 or more expressions, MIN works for single expressionif len(distances) == 1:    MIN_FUNC = Minelse:    MIN_FUNC = LeastJob.objects.filter(query).annotate(distance=MIN_FUNC(*distances)).order_by('distance')MIN是一个聚合函数，它采用单个表达式（例如列名）并将多个输入减少为单个输出值LEAST是一个条件表达式，它通过从任意数量的表达式列表中选择最小值来发挥作用https://docs.djangoproject.com/en/3.0/ref/models/querysets/#min https://docs.djangoproject.com/en/3.0/ref/models/database-functions/#least