postgresql 计算距离的实例(单位直接生成米)

select st_distance(ST_GeomFromText('POINT(120.451737 36.520975)',900913),ST_GeomFromText('POINT(120.455636 36.520885)',900913))*60*1.852;

select ST_Length(Geography(ST_GeomFromText('LINESTRING(120.451737 36.520975,120.455636 36.520885)')));

select ST_Contains(St_Astext(ST_Buffer(geography(geomfromtext('MULTILINESTRING((线的坐标点))')),25.00{以米为单位的距离})),st_astext(geography(geomfromtext('POINT(121.37805 37.54142)')))) as result

CREATE TABLE mylocation ( 
 id SERIAL PRIMARY KEY,
 geom GEOMETRY(Point, 4326),
 name VARCHAR(128),
 x double precision,
 y double precision
); 
 
INSERT INTO mylocation (geom,name,x,y) VALUES (
 ST_GeomFromText('POINT(0.0001 0)', 4326),'zhangsan',0.0001,0
);
INSERT INTO mylocation (geom,name,x,y) VALUES (
 ST_GeomFromText('POINT(0.001 0)', 4326),'zhangsan',0.001,0
);
INSERT INTO mylocation (geom,name,x,y) VALUES (
 ST_GeomFromText('POINT(0.001 0)', 4326),'zhangsan',0.001,0
);
INSERT INTO mylocation (geom,name,x,y) VALUES (
 ST_GeomFromText('POINT(0.1 0)', 4326),'zhangsan',0.1,0
); 
 
SELECT id, name,geom,x,y,  ST_DistanceSphere(
           geom,
           ST_GeometryFromText('POINT(0 0)')) distance
FROM mylocation
WHERE ST_DWithin(
 geom, 
 ST_GeomFromText('POINT(0 0)', 4326),
 0.001
)ORDER BY distance asc;;

SELECT id, name,geom,x,y,  ST_DistanceSphere(
           geom,
           ST_GeometryFromText('POINT(0 0)')) distance
FROM mylocation
WHERE ST_DWithin(
 geom::geography, 
 ST_GeomFromText('POINT(0 0)', 4326)::geography,
 1000
) ORDER BY distance asc;