ST_Difference

定义

ST_Difference 获取两个 ST_Geometry 对象，然后返回表示两个源对象之差的 ST_Geometry 对象。

语法

sde.st_difference (g1 sde.st_geometry, g2 sde.st_geometry)

返回类型

ST_Geometry

示例

城市工程师需要知道没有被建筑物覆盖的城市地块的总面积。实际上，她想要知道减去建筑物面积之后的地块面积的总和。

CREATE TABLE footprints (building_id integer,
footprint sde.st_geometry);

CREATE TABLE lots (lot_id integer,
lot sde.st_geometry);

INSERT INTO footprints (building_id, footprint) VALUES (
1, 
sde.st_polygon ('polygon ((0 0, 0 10, 10 10, 10 0, 0 0))', 0)
);

INSERT INTO footprints (building_id, footprint) VALUES (
2,
sde.st_polygon ('polygon ((20 0, 20 10, 30 10, 30 0, 20 0))', 0)
);

INSERT INTO footprints (building_id, footprint) VALUES (
3,
sde.st_polygon ('polygon ((40 0, 40 10, 50 10, 50 0, 40 0))', 0)
);

INSERT INTO lots (lot_id, lot) VALUES (
1,
sde.st_polygon ('polygon ((-1 -1, -1 11, 11 11, 11 -1, -1 -1))', 0)
);

INSERT INTO lots (lot_id, lot) VALUES (
2,
sde.st_polygon ('polygon ((19 -1, 19 11, 29 9, 31 -1, 19 -1))', 0)
);

INSERT INTO lots (lot_id, lot) VALUES (
3,
sde.st_polygon ('polygon ((39 -1, 39 11, 51 11, 51 -1, 39 -1))', 0)
);

城市工程师通过 lot_id 相等连接 footprints 表和 lots 表，然后获取地块减去覆盖区之后的差值面积之和。

Oracle

SELECT SUM (sde.st_area (sde.st_difference (lot, footprint)))
FROM FOOTPRINTS bf, LOTS
WHERE bf.building_id = lots.lot_id;

SUM(ST_AREA(ST_DIFFERENCE(LOT,FOOTPRINT)))

114

PostgreSQL

SELECT SUM (sde.st_area (sde.st_difference (lot, footprint)))
FROM footprints bf, lots
WHERE bf.building_id = lots.lot_id;

sum

114