有可能执行raw SQL queries in Django。
我的建议是，编写查询来拉取一个ID列表（看起来你现在正在做），然后使用ID来拉取关联的模型（在一个常规的、非原始的SQL Django查询中）。尽量保持SQL与方言无关，这样如果你不得不切换数据库，你就不必再担心一件事了。
为了澄清，这里有一个如何做到这一点的例子：

def get_models_within_25 (self):
    from django.db import connection, transaction
    cursor = connection.cursor()

    cursor.execute("""SELECT id, ( 
        3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) * 
        cos( radians( lng ) - radians(-122) ) + sin( radians(37) ) * 
        sin( radians( lat ) ) ) )
        AS distance FROM stores HAVING distance < 25
        ORDER BY distance LIMIT 0 , 20;""")
    ids = [row[0] for row in cursor.fetchall()]

    return MyModel.filter(id__in=ids)

字符串
作为免责声明，我不能保证这段代码，因为我已经有几个月没有写任何Django了，但它应该是沿着正确的路线。

为了跟进Tom的回答，它在SQLite中默认不工作，因为SQLite默认缺少数学函数。没问题，添加它非常简单：

class LocationManager(models.Manager):
    def nearby_locations(self, latitude, longitude, radius, max_results=100, use_miles=True):
        if use_miles:
            distance_unit = 3959
        else:
            distance_unit = 6371

        from django.db import connection, transaction
        from mysite import settings
        cursor = connection.cursor()
        if settings.DATABASE_ENGINE == 'sqlite3':
            connection.connection.create_function('acos', 1, math.acos)
            connection.connection.create_function('cos', 1, math.cos)
            connection.connection.create_function('radians', 1, math.radians)
            connection.connection.create_function('sin', 1, math.sin)

        sql = """SELECT id, (%f * acos( cos( radians(%f) ) * cos( radians( latitude ) ) *
        cos( radians( longitude ) - radians(%f) ) + sin( radians(%f) ) * sin( radians( latitude ) ) ) )
        AS distance FROM location_location WHERE distance < %d
        ORDER BY distance LIMIT 0 , %d;""" % (distance_unit, latitude, longitude, latitude, int(radius), max_results)
        cursor.execute(sql)
        ids = [row[0] for row in cursor.fetchall()]

        return self.filter(id__in=ids)

字符串

为了跟进Tom，如果你想有一个在postgresql中也能工作的查询，你不能使用AS，因为你会得到一个错误，说'distance'不存在。
你应该把整个球面定律表达式放在WHERE子句中，像这样（它在mysql中也有效）：

import math
from django.db import connection, transaction
from django.conf import settings

from django .db import models

class LocationManager(models.Manager):
    def nearby_locations(self, latitude, longitude, radius, use_miles=False):
        if use_miles:
            distance_unit = 3959
        else:
            distance_unit = 6371

        cursor = connection.cursor()

        sql = """SELECT id, latitude, longitude FROM locations_location WHERE (%f * acos( cos( radians(%f) ) * cos( radians( latitude ) ) *
            cos( radians( longitude ) - radians(%f) ) + sin( radians(%f) ) * sin( radians( latitude ) ) ) ) < %d
            """ % (distance_unit, latitude, longitude, latitude, int(radius))
        cursor.execute(sql)
        ids = [row[0] for row in cursor.fetchall()]

        return self.filter(id__in=ids)

字符串
请注意，您必须选择纬度和经度，否则您不能在WHERE子句中使用它。

为了跟进jboxer的回答，下面是作为自定义管理器的一部分的整个过程，其中一些硬编码的东西变成了变量：

class LocationManager(models.Manager):
    def nearby_locations(self, latitude, longitude, radius, max_results=100, use_miles=True):
        if use_miles:
            distance_unit = 3959
        else:
            distance_unit = 6371

        from django.db import connection, transaction
        cursor = connection.cursor()

        sql = """SELECT id, (%f * acos( cos( radians(%f) ) * cos( radians( latitude ) ) *
        cos( radians( longitude ) - radians(%f) ) + sin( radians(%f) ) * sin( radians( latitude ) ) ) )
        AS distance FROM locations_location HAVING distance < %d
        ORDER BY distance LIMIT 0 , %d;""" % (distance_unit, latitude, longitude, latitude, int(radius), max_results)
        cursor.execute(sql)
        ids = [row[0] for row in cursor.fetchall()]

        return self.filter(id__in=ids)

字符串

JBoxer的回应

def find_cars_within_miles_from_postcode(request, miles, postcode=0):

    # create cursor for RAW query
    cursor = connection.cursor()

    # Get lat and lon from google
    lat, lon = getLonLatFromPostcode(postcode)

    # Gen query
    query = "SELECT id, ((ACOS(SIN("+lat+" * PI() / 180) * SIN(lat * PI() / 180) + COS("+lat+" * PI() / 180) * COS(lat * PI() / 180) * COS(("+lon+" - lon) * PI() / 180)) * 180 / PI()) * 60 * 1.1515) AS distance FROM app_car HAVING distance<='"+miles+"' ORDER BY distance ASC"

    # execute the query
    cursor.execute(query)

    # grab all the IDS form the sql result
    ids = [row[0] for row in cursor.fetchall()]

    # find cars from ids
    cars = Car.objects.filter(id__in=ids)

    # return the Cars with these IDS
    return HttpResponse( cars )

字符串
这返回了我的汽车从x数量的英里，这工作得很好。然而，原始查询返回多远，他们从某个位置，我认为fieldname是'距离'。
我怎样才能返回这个字段'距离'与我的汽车对象？

使用上面提出的一些答案，我得到了不一致的结果，所以我决定再次使用[这个链接] http://www.movable-type.co.uk/scripts/latlong.html作为参考来检查方程，方程是d = acos(sin(lat1)*sin(lat2) + cos(lat1)*cos(lat2)*cos(lon2-lon1) ) * 6371，其中d是要计算的距离，
lat1,lon1是基点的坐标，lat2,lon2是其他点的坐标，在我们的例子中，这些点是数据库中的点。
从上面的答案来看，LocationManager类如下所示

class LocationManager(models.Manager):
def nearby_locations(self, latitude, longitude, radius, max_results=100, use_miles=True):
    if use_miles:
        distance_unit = 3959
    else:
        distance_unit = 6371

    from django.db import connection, transaction
    from mysite import settings
    cursor = connection.cursor()
    if settings.DATABASE_ENGINE == 'sqlite3':
        connection.connection.create_function('acos', 1, math.acos)
        connection.connection.create_function('cos', 1, math.cos)
        connection.connection.create_function('radians', 1, math.radians)
        connection.connection.create_function('sin', 1, math.sin)

    sql = """SELECT id, (acos(sin(radians(%f)) * sin(radians(latitude)) + cos(radians(%f))
          * cos(radians(latitude)) * cos(radians(%f-longitude))) * %d)
    AS distance FROM skills_coveragearea WHERE distance < %f
    ORDER BY distance LIMIT 0 , %d;""" % (latitude, latitude, longitude,distance_unit, radius, max_results)
    cursor.execute(sql)
    ids = [row[0] for row in cursor.fetchall()]

    return self.filter(id__in=ids)

字符串
使用网站[链接] http://www.movable-type.co.uk/scripts/latlong.html作为检查，我的结果一致。

使用Django的数据库函数也可以做到这一点，这意味着你可以使用.annotate()调用添加一个distance_miles列，然后按它排序。下面是一个例子：

from django.db.models import F
from django.db.models.functions import ACos, Cos, Radians, Sin

locations = Location.objects.annotate(
    distance_miles = ACos(
        Cos(
            Radians(input_latitude)
        ) * Cos(
            Radians(F('latitude'))
        ) * Cos(
            Radians(F('longitude')) - Radians(input_longitude)
        ) + Sin(
            Radians(input_latitude)
        ) * Sin(Radians(F('latitude')))
    ) * 3959
).order_by('distance_miles')[:10]

字符串

@classmethod def nearby_locations（cls，latitude，longitude，radius，max_results=1000，use_miles=False）：if use_miles：distance_unit = 3959 else：distance_unit = 6371000

from django.db import connection, transaction
from django.conf import settings
cursor = connection.cursor()

sql = """SELECT id, (%f * acos( cos( radians(%f) ) * cos( radians( latitude ) ) *
cos( radians( longitude ) - radians(%f) ) + sin( radians(%f) ) * sin( radians( latitude ) ) ) )
AS distance FROM yourapp_yourmodel 
GROUP BY id, latitude, longitude
HAVING (%f * acos( cos( radians(%f) ) * cos( radians( latitude ) ) *
cos( radians( longitude ) - radians(%f) ) + sin( radians(%f) ) * sin( radians( latitude ) ) ) ) < %d
ORDER BY distance OFFSET 0 LIMIT %d;""" % (distance_unit, latitude, longitude, latitude, distance_unit, latitude, longitude, latitude, int(radius), max_results)

cursor.execute(sql)
ids = [row[0] for row in cursor.fetchall()]

return cls.objects.filter(id__in=ids)

字符串