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我想问如何用radius=4km. 我已经尝试过该ST_Buffer功能，但它会创建一个更大的圆圈。（我通过将多边形插入新的 kml 文件来查看创建的圆。）

这就是我正在尝试的。

INSERT INTO camera(geom_circle) VALUES(geometry(ST_Buffer(georgaphy(ST_GeomFromText('POINT(21.304116745663165 38.68607570952619)')), 4000)))

圆的中心是一个 lon lat 点，但我不知道它SRID，因为我已经从 kml 文件中导入了它。SRID为了转换几何形状等，我需要吗？

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1 回答 1

25

KML 文件始终为 lat/long 并使用 SRID=4326。如果您使用geography. 地理是在纬度/经度数据上混合 4 公里公制度量的好方法……太好了，你试过了！

试试这个语句来修复强制转换，并使用参数化点构造函数：

SELECT ST_Buffer(ST_MakePoint(21.304116745663165, 38.68607570952619)::geography, 4000);

如果您需要将其转换回几何体，::geometry请在末尾添加一个转换。

更新准确性

上一个答案在内部将几何图形（通常）重新投影到该点适合的 UTM 区域（参见ST_Buffer）。如果该点位于两个 UTM 边界的边缘，这可能会导致轻微的失真。大多数人不会关心这些错误的大小，但通常会是几米。但是，如果您需要亚毫米精度，请考虑构建动态方位角等距投影。这需要 PostGIS 2.3 ST_Transform，并且改编自另一个答案：

CREATE OR REPLACE FUNCTION geodesic_buffer(geom geometry, dist double precision,
                                           num_seg_quarter_circle integer)
  RETURNS geometry AS $$
  SELECT ST_Transform(
    ST_Buffer(ST_Point(0, 0), $2, $3),
      ('+proj=aeqd +x_0=0 +y_0=0 +lat_0='
       || ST_Y(ST_Centroid($1))::text || ' +lon_0=' || ST_X(ST_Centroid($1))::text),
      ST_SRID($1))
  $$ LANGUAGE sql IMMUTABLE STRICT COST 100;
CREATE OR REPLACE FUNCTION geodesic_buffer(geom geometry, dist double precision)
  RETURNS geometry AS 'SELECT geodesic_buffer($1, $2, 8)'
  LANGUAGE sql IMMUTABLE STRICT COST 100;
-- Optional warppers for geography type
CREATE OR REPLACE FUNCTION geodesic_buffer(geog geography, dist double precision)
  RETURNS geography AS 'SELECT geodesic_buffer($1::geometry, $2)::geography'
LANGUAGE sql IMMUTABLE STRICT COST 100;
CREATE OR REPLACE FUNCTION geodesic_buffer(geog geography, dist double precision,
                                           num_seg_quarter_circle integer)
  RETURNS geography AS 'SELECT geodesic_buffer($1::geometry, $2, $3)::geography'
  LANGUAGE sql IMMUTABLE STRICT COST 100;

运行其中一个功能的简单示例是：

SELECT geodesic_buffer(ST_MakePoint(21.304116745663165, 38.68607570952619)::geography, 4000);

为了比较到每个缓冲点的距离，这里是每个测地线的长度（旋转椭圆体上的最短路径，即 WGS84）。首先这个函数：

SELECT count(*), min(buff_dist), avg(buff_dist), max(buff_dist)
FROM (
  SELECT ST_Distance((ST_DumpPoints(geodesic_buffer(poi, dist)::geometry)).geom, poi) AS buff_dist
  FROM (SELECT ST_MakePoint(21.304116745663165, 38.68607570952619)::geography AS poi, 4000 AS dist) AS f
) AS f;

 count |      min       |       avg       |      max
-------+----------------+-----------------+----------------
    33 | 3999.999999953 | 3999.9999999743 | 4000.000000001

将此与 ST_Buffer （答案的第一部分）进行比较，这表明它偏离了大约 1.56 m：

SELECT count(*), min(buff_dist), avg(buff_dist), max(buff_dist)
FROM (
  SELECT ST_Distance((ST_DumpPoints(ST_Buffer(poi, dist)::geometry)).geom, poi) AS buff_dist
  FROM (SELECT ST_MakePoint(21.304116745663165, 38.68607570952619)::geography AS poi, 4000 AS dist) AS f
) AS f;

 count |      min       |       avg        |      max
-------+----------------+------------------+----------------
    33 | 4001.560675049 | 4001.56585986067 | 4001.571105793
于 2012-12-14T04:56:12.720 回答