Sql 使用Ruby Geocoder检索附近用户和返回距离

是否可以使用地理编码器检索（例如）某个半径内的附近用户，并使其返回距离计算结果而无需重复显示

例如，假设我们使用这个（在Rails应用程序中）：

它将返回一个

ActiveRecord:：Relation

与距离user

@me

10公里以内的所有用户的关系

如果我现在想显示一个包含所有返回用户及其到

@me

的距离的列表，我是否需要使用

distance\u to

逐个重新计算距离，或者是否有办法让

Geocoder

在添加的模型属性中返回计算出的距离（例如）

运行位置感知查询时，返回的对象有两个 添加到它们的属性：

• obj.distance—从搜索点到该对象的英里数
• obj.方位-从搜索点到该对象的方向

---

我使用的是这个嵌入式Ruby版本的逻辑，它在逻辑上类似于上面@maxcal给出的答案。我之所以包含它，是因为下面的跟踪与此代码相关

<% for location in @location.nearbys(20) %>
  <li> <span class="badge"><%= location.distance.round(2) %> miles</span> from here you'll find <strong>
  <%= link_to location.name, location %> </strong> at <%= location.full_address %>.</li>
<% end %> 

---

谢谢你@Valerie，现在一切都有意义了！

User.near(@me, 10, units: :km).each do |user|
  puts "#{user.name} is #{user.distance} clicks to the #{user.bearing}"
end

<% for location in @location.nearbys(20) %>
  <li> <span class="badge"><%= location.distance.round(2) %> miles</span> from here you'll find <strong>
  <%= link_to location.name, location %> </strong> at <%= location.full_address %>.</li>
<% end %> 

  [1m[36m (0.0ms)[0m  [1mSELECT COUNT(*) FROM "reviews" WHERE "reviews"."user_id" = $1[0m  [["user_id", 8]]
  Rendered shared/_user_posted_badge.html.erb (15.6ms)
// Comment Geocoding starts here
  Rendered locations/_map_show.html.erb (0.0ms)
  [1m[35mLocation Load (0.0ms)[0m  SELECT locations.*, 3958.755864232 * 2 * ASIN(SQRT(POWER(SIN((30.267153 - locations.latitude) * PI() / 180 / 2), 2) + COS(30.267153 * PI() / 180) * COS(locations.latitude * PI() / 180) * POWER(SIN((-97.7430608 - locations.longitude) * PI() / 180 / 2), 2))) AS distance, MOD(CAST((ATAN2( ((locations.longitude - -97.7430608) / 57.2957795), ((locations.latitude - 30.267153) / 57.2957795)) * 57.2957795) + 360 AS decimal), 360) AS bearing FROM "locations" WHERE (locations.latitude BETWEEN 29.977689433778306 AND 30.556616566221695 AND locations.longitude BETWEEN -98.07821040202988 AND -97.40791119797011 AND (3958.755864232 * 2 * ASIN(SQRT(POWER(SIN((30.267153 - locations.latitude) * PI() / 180 / 2), 2) + COS(30.267153 * PI() / 180) * COS(locations.latitude * PI() / 180) * POWER(SIN((-97.7430608 - locations.longitude) * PI() / 180 / 2), 2)))) BETWEEN 0.0 AND 20 AND locations.id != 65)  ORDER BY distance ASC
// Geocoding section complete
  [1m[36mPhotograph Exists (0.0ms)[0m  [1mSELECT  1 AS one FROM "photographs" WHERE "photographs"."location_id" = $1 LIMIT 1[0m  [["location_id", 65]]
  [1m[35mPhotograph Load (0.0ms)[0m  SELECT "photographs".* FROM "photographs" WHERE "photographs"."location_id" = $1  ORDER BY "photographs"."created_at" DESC  [["location_id", 65]]