Question

我指的是下面列出的问题中的答案，因为我正在尝试运行一个查询，该查询获取与我的django应用程序中给定点最接近的对象。我的django版本是1.9.2。

How to return a record with the lowest distance from a point using geodjango?

在我看来，我传递一个Zipcode并使用GeoPy来获取位置和纬度/经度。

from django.contrib.gis.geos import *
from django.contrib.gis.measure import D
from geopy.geocoders import Nominatim

geolocator = Nominatim()

zipcode = self.request.query_params.get('zipcode', None)
distance_m = 20000

if zipcode is not None:
    location = geolocator.geocode(zipcode, timeout=None)
    origin = Point((location.latitude, location.longitude))

    queryset = Chapter.objects.filter(location__distance_lte=(origin, D(m=distance_m))).distance(origin).order_by('distance')[:1][0] 
    return queryset

章模型有一个名为 location 的字段，这是一个点字段。

from django.contrib.gis.db import models

class Chapter(models.Model):
    name = models.CharField(max_length=500)
    school = models.ForeignKey(School)
    verification = models.CharField(max_length=50)
    address1 = models.TextField()
    address2 = models.TextField(null=True, blank=True)
    city = models.TextField()
    state = models.TextField()
    zipcode = models.CharField(max_length=5)
    location = models.PointField(srid=4326, null=True, blank=True)
    is_school = models.BooleanField(default=False)
    is_association = models.BooleanField(default=False)
    authorized = models.NullBooleanField(default=False)
    stripe_account_active = models.NullBooleanField(default=False)

    def __unicode__(self):
        return self.name

    def get_nickname(self):
        return self.school.nickname

执行此操作时出现以下错误：

AttributeError: 'QuerySet' object has no attribute 'distance'

修改 -

当我尝试在上面的问题中使用替代方法获得距离时，我得到一个不同的错误，这让我觉得我的模型有问题。

point = Chapter.objects.get(name='Chapter of the Year').location
distance_m = 20000

for chapter in Chapter.objects.distance(point):
    print(chapter.name, chapter.distance)

错误：

'Manager' object has no attribute 'distance'

Answer 1

让这个工作

Chapter.objects.filter(location__distance_lte=(origin, D(m=distance_m))).annotate(distance=Distance('location', origin)).order_by('distance')[:1][0]

Answer 2

如下所述：

queryset = Chapter.objects.filter(location__distance_lte = (origin, D(m=distance_m))).distance(origin).order_by('distance')[:1][0]

您按距离排序结果，但这不是章节模型中的字段/属性。我认为你想要的是 location__distance 的顺序，所以实际的过滤查询应该是：

queryset = Chapter.objects.filter(location__distance_lte= (origin, D(m=distance_m))).distance(origin).order_by('location__distance')[:1][0]

此行也完全不正确：

for chapter in Chapter.objects.distance(point):

因为此处距离不是过滤器或获取等方法。它应该是这样的：

point = Chapter.objects.get(name='Chapter of the Year')
distance_m = 20000

for chapter in Chapter.objects.filter(id = point.id ):
    print(chapter.name, chapter.location.distance)

对象没有属性'distance' - GeoDjango

2 个答案: