在postgresql中按经纬度查找最近的位置

Question

您好，我正在尝试在 postgresql database.But 中按纬度和经度查找最近的位置，当我运行下面的查询显示列距离不存在时。

ERROR:  column "distance" does not exist
LINE 1: ... ) ) ) AS distance FROM station_location   HAVING distance <...
                                                             ^
********** Error **********

ERROR: column "distance" does not exist
SQL state: 42703
Character: 218

CREATE TABLE station_location
(
  id bigint NOT NULL DEFAULT nextval('location_id_seq'::regclass),
  state_name character varying NOT NULL,
  country_name character varying NOT NULL,
  locality character varying NOT NULL,
  created_date timestamp without time zone NOT NULL,
  is_delete boolean NOT NULL DEFAULT false,
  lat double precision,
  lng double precision,
  CONSTRAINT location_pkey PRIMARY KEY (id)
)

SELECT  *,( 3959 * acos( cos( radians(6.414478) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(12.466646) ) + sin( radians(6.414478) ) * sin( radians( lat ) ) ) ) AS distance 
FROM station_location
HAVING distance < 5
ORDER BY distance
LIMIT 20;

Answer 1

select * from (
SELECT  *,( 3959 * acos( cos( radians(6.414478) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(12.466646) ) + sin( radians(6.414478) ) * sin( radians( lat ) ) ) ) AS distance 
FROM station_location
) al
where distance < 5
ORDER BY distance
LIMIT 20;

Answer 2

请参阅 this gist，您将了解如何在 point 类型上声明域以及如何将 distance 运算符覆盖为 return 顺向距离。

声明从 point 继承的 latlong 类型：

CREATE DOMAIN latlong AS point CHECK (VALUE[0] BETWEEN -90.0 AND 90.0 AND VALUE[1] BETWEEN -180 AND 180);

以公里为单位的正向距离（地球半径球体上的距离）：

CREATE OR REPLACE FUNCTION orthodromic_distance(latlong, latlong) RETURNS float AS $_$
     SELECT acos(
              sin(radians([0])) 
            * 
              sin(radians([0]))
            + 
              cos(radians([0])) 
            * 
              cos(radians([0]))
            * 
              cos(radians([1]) 
            - 
              radians([1]))
            ) * 6370.0;
$_$ LANGUAGE sql IMMUTABLE;

在与 latlongs 一起使用时使用此函数覆盖距离运算符 <->：

CREATE OPERATOR <-> ( PROCEDURE = orthodromic_distance
, LEFTARG = latlong, RIGHTARG = latlong
);

现在在您的 SQL 查询中，查找最近的实体：

WITH
  station_distance AS (
    SELECT
      id AS station_id,
      point(lat, long)::latlong <-> point(6.414478, 12.466646)::latlong AS distance
    FROM station_location
    WHERE NOT is_deleted
  )
  SELECT
    sl.state_name,
    sl.country_name,
    sl.locality,
    point(sl.lat, sl.long)::latlong AS coordinates,
    sd.distance
  FROM
    station_location sl
    JOIN station_distance sd
      ON sd.station_id = sl.id
  ORDER BY
    distance ASC
  LIMIT 10

您可能希望使用 latlong 类型将位置 lat 和 long 存储在同一字段中。

Answer 3

PostGIS

不要像这样在 table 上存储纬度和经度。而是使用 PostGIS 几何或 geography type.

CREATE EXTENSION postgis;

CREATE TABLE foo (
  geog geography;
);

CREATE INDEX ON foo USING gist(geog);

INSERT INTO foo (geog)
  VALUES (ST_MakePoint(x,y));

现在，当您需要查询它时，您可以使用 KNN (<->)，它实际上会在索引上执行此操作。

SELECT *
FROM foo
ORDER BY foo.geog <-> ST_MakePoint(x,y)::geography;

在您的查询中，您明确 HAVING distance < 5。您也可以在索引上执行此操作。

SELECT *
FROM foo
WHERE ST_DWithin(foo.geog, ST_MakePoint(x,y)::geography, distance_in_meters)
ORDER BY foo.geog <-> ST_MakePoint(x,y)::geography;

这确保如果所有点都位于 distance_in_meters 之外，则不会返回任何内容。

另外x和y是十进制数ST_MakePoint(46.06, 14.505)

Answer 4

手册阐明：

An output column's name can be used to refer to the column's value in ORDER BY and GROUP BY clauses, but not in the WHERE or HAVING clauses; there you must write out the expression instead.

Answer 5

您可以使用 PostgreSQL 的 cube and earthdistance 扩展。

像这样启用它们：

CREATE EXTENSION cube;
CREATE EXTENSION earthdistance;

假设您当前的位置是 35.697933, 139.707318。那么您的查询将是这样的：

SELECT *, point(35.697933, 139.707318) <@>  (point(longitude, latitude)::point) as distance
FROM station_location
-- WHERE (point(35.697933, 139.707318) <@> point(longitude, latitude)) < 3
ORDER BY distance;

请注意 distance 以英里为单位（默认）。

在postgresql中按经纬度查找最近的位置

find the nearest location by latitude and longitude in postgresql

sql

postgresql

spatial

PostGIS