9

I have an app that takes an address string, sends it to Google Maps API and gets lat/long co-ordinates, I then want to show the all users within X meteres of this point (there lat/long is stored in my database), I then want to filter the result to only show users with certain pets

So first off, I have my Models

class User(UserMixin, Base):
    first_name = Column(Unicode)

    address = Column(Unicode)
    location = Column(Geometry('POINT'))

    pets = relationship('Pet', secondary=user_pets, backref='pets') 

class Pet(Base):
    __tablename__ = 'pets'   
    id = Column(Integer, primary_key=True)
    name = Column(Unicode)

user_pets = Table('user_pets', Base.metadata,
    Column('user_id', Integer, ForeignKey('users.id')),
    Column('pet_id', Integer, ForeignKey('pets.id'))
)

I get my lat/long from Google API and store it in my database, so from the address string "London England" I get

POINT (-0.1198244000000000 51.5112138999999871)

this stores in my database like:

0101000000544843D7CFACBEBF5AE102756FC14940

Now that all works fine, now reading the Geoalchemy2 docs I cant seem to find an exmaple query to resolve my problem.

What I want to pass is another set of lat/long co-ordinates to Geoalchemy2 and then return the nearest say 10 users. Whilst querying this I will also filter only users that have certain pets (this isn't essential for my query to work, but I wanted to show what the query will actually do in its entirety).

I don't really like to answer a question without providing a sample query, but I really don't know what functions I should be using to achieve my required result.

I am guessing I will need to use "ST_DWithin" or "ST_DFullyWithin" but I cannot find a full example of either function. Thank's.

So I know have a working query

distance = 10
address_string = "London, England"
results = Geocoder.geocode(address_string)
# load long[1], lat[0] into shapely
center_point = Point(results.coordinates[1], results.coordinates[0])
print center_point
# 'POINT (-0.1198244000000000 51.5112138999999871)'
wkb_element = from_shape(center_point)
users = DBSession.query(User).\
    filter(func.ST_DWithin(User.location,  wkb_element, distance)).all()

Which generates the following SQL

2013-12-30 15:12:06,445 INFO  [sqlalchemy.engine.base.Engine][Dummy-2] SELECT users.first_name AS users_first_name, users.last_name AS users_last_name, users.phone AS users_phone, users.address AS users_address, users.about AS users_about, ST_AsBinary(users.location) AS users_location, users.profile_image_id AS users_profile_image_id, users.searchable AS users_searchable, users.user_password AS users_user_password, users.registered_date AS users_registered_date, users.id AS users_id, users.last_login_date AS users_last_login_date, users.status AS users_status, users.user_name AS users_user_name, users.email AS users_email, users.security_code AS users_security_code 
FROM users 
WHERE ST_DWithin(users.location, ST_GeomFromWKB(%(ST_GeomFromWKB_1)s, %(ST_GeomFromWKB_2)s), %(param_1)s)
2013-12-30 15:12:06,445 INFO  [sqlalchemy.engine.base.Engine][Dummy-2] {'ST_GeomFromWKB_1': <read-only buffer for 0x7f7d10258f70, size -1, offset 0 at 0x7f7d10258db0>, 'param_1': 10, 'ST_GeomFromWKB_2': -1}

Now this always returns all my users, regardless of the distance variable, so I am guessing something is not quite, right, but I cannot work out why.

4

2 回答 2

10

回答：

这些单位以度为半径，因此我必须将英里转换为度数以获得最佳（粗略）估计。不需要很精确：

d = 90
distance = d * 0.014472
#1 mile = 0.014472 degrees  

r1 = -0.1198244
r2 = 51.5112139

# load long[1], lat[0] into shapely
center_point = Point(r1, r2)
# 'POINT (-0.1198244000000000 51.5112138999999871)'

wkb_element = from_shape(center_point)

users = DBSession.query(User).\
    filter(func.ST_DFullyWithin(User.location,  wkb_element, distance)).all()
于 2013-12-30T16:11:58.180 回答
2

纬度/经度数据不是特别适合距离计算。

每个纬度相距约 69 英里（111 公里）。范围从赤道的 68.703 英里（110.567 公里）到两极的 69.407（111.699 公里）不等（由于地球略呈椭圆形）。这很方便，因为每分钟（1/60 度）大约是一英里。

一个经度在赤道处最宽，为 69.172 英里 (111.321)，并在两极逐渐缩小至零。在北纬 40° 或南纬 40° 经度之间的距离为 53 英里（85 公里）。

您可以使用ST_Transform将坐标转换为使用米或英里的不同投影。这些确实往往是局部的，因为它们将地球球体投影到光栅化平面上。英国国家电网 (SRID 27700) 可能适合您的需求。

于 2014-01-05T16:39:43.713 回答