Question

我正在尝试找到获得半径属性计数的最快方法。我见过其他一些类似的问题。有没有人知道为什么第三个查询给出了不同的结果？它非常快。

慢

SELECT count(*)
FROM property
WHERE ST_Distance_Sphere(geom_pt, ST_MakePoint(-104.989879,39.736355))<=2000;

计数= 2665

非常慢

SELECT count(*)
FROM property
WHERE ST_Distance(geom_pt, ST_SetSRID(ST_MakePoint(-104.989879, 39.736355),4326)::geography)<=2000;

计数= 2665

非常快

SELECT count(*)
FROM property
WHERE ST_Within(geom_pt,ST_Transform(ST_Buffer(ST_Transform(ST_SetSRID(ST_MakePoint(-104.989879, 39.736355), 4326), 3857), 2000), 4326));

计数= 1794

Answer 1

不，您希望ST_DWithin使用GIST索引。

SELECT count(*)
FROM property
WHERE ST_DWithin(
  geom_pt, -- make sure this is also geography
  ST_MakePoint(-104.989879, 39.736355)::geography,
  2000     -- note distance in meters.
);

您也不必在地理位置上将srid设置为4326。这是默认设置。

如果您没有索引，

CREATE INDEX ON property USING gist(geom_pt);
VACUUM property;

您也可以在CLUSTER上geom_pt。

最佳postgis距离查询

1 个答案: