说，我有一个名为buildings的表，可以通过以下查询创建：
create table buildings(
building_id number primary key,
building_name varchar2(32),
shape sdo_geometry
);
我可以通过以下查询在其中插入一个矩形：
insert into buildings values(
4, -- index
'Reading Room', -- building_name
sdo_geometry(
2003, --SDO_GTYPE: dltt - 2(2D)0(linear referencing)03(polygon)
8307, --SDO_SRID: coordinate system
null, --SDO_POINT: it is for point inserting, if the next two field = null, then it could not be null.
sdo_elem_info_array( --SDO_ELEM_INFO:
1, --SDO_STARTING_OFFSET: indicates from which index of the next param of SDO_GEOMETRY would be considered, starts from 1.
1003, --SDO_ETYPE: 1(exterior, interior - 2)003(this digits usually comes from SDO_GTYPE)
3), --SDO_INTERPRETATION: 1 - simple polygon, 2 - polygon connecting arcs, 3 - rectangle, 4 - circle etc.
sdo_ordinate_array(
24.916312, 91.832393,
24.916392, 91.832678
) --SDO_ORDINATES: co-ordinates of the geometry
-- two corner points of the main diagonal
)
);
这里，两个大地测量点来自真实数据，作为sdo_ordinate_array的对象。以下两点直接插入上述查询中：
现在，我想从两个不同的子查询中插入这两个点。
子查询如下：
SELECT 180+SDO_GEOM.SDO_CENTROID(c.shape, m.diminfo).SDO_POINT.X,
180-SDO_GEOM.SDO_CENTROID(c.shape, m.diminfo).SDO_POINT.Y
FROM buildings c, user_sdo_geom_metadata m
WHERE m.table_name = 'BUILDINGS' AND m.column_name = 'SHAPE'
AND c.building_name = 'IICT';
因此，查询的结果如下：
X Y
---------- ----------
24.9181097 91.83097409
如何将此结果转换为逗号分隔值，例如：24.9181097, 91.83097409？
这样我就可以替换以下代码：
sdo_ordinate_array(
24.916312, 91.832393,
24.916392, 91.832678
) --SDO_ORDINATES: co-ordinates of the geometry
使用：
sdo_ordinate_array(
(/*sub-query*/),
(/*another-subquery*/)
) --SDO_ORDINATES: co-ordinates of the geometry
我有谷歌它并探索了几个博客，但没有运气。
标题似乎不合适，但子查询的简单版本返回SDO_GEOMETRY的对象。如果你研究了oracle空间查询，那么你很清楚我只是从返回的对象中检索X和Y的值。
答案 0 :(得分：0)
SDO_CENTROID（）函数返回一个SDO_GEOMETRY对象，然后您可以将其用于插入结果表。
假设这是你的结果表
List<MyMarker> markers = new ArrayList<>();
MyMarker myMarker = new MyMarker(
new LatLng(1.123456, -2.123456),
BitmapDescriptorFactory.fromResource(R.drawable.icon2)),
"title1",
"snippet1");
markers.add(myMarker);
然后以下内容将填充所有建筑物的质心
create table building_centroids (
building_id number primary key,
centroid sdo_geometry
);
我不明白为什么你需要改变坐标（180 + x，180-y）。这毫无意义。
无论如何，你说以上不是你想要的。我将继续猜测，并假设您要构建一个矩形，其中两个角被计算为两个建筑物中的质心。这需要一些PL / SQL，如下所示：
首先定义一个从两个输入点构建矩形的函数
insert into building_centroids (building_id, centroid)
select building_id, sdo_geom.sdo_centroid(shape, 0.05)
from buildings;
现在，像这样使用它：
create or replace function rectangle_from_points (
point_1 sdo_geometry,
point_2 sdo_geometry
)
return sdo_geometry
as
rectangle sdo_geometry;
begin
-- Initialize resulting rectangle
rectangle := sdo_geometry (2003, point_1.sdo_srid, null,
sdo_elem_info_array (1,1003,3),
sdo_ordinate_array()
);
-- Fill it with the two point points
rectangle.sdo_ordinates.extend(4);
rectangle.sdo_ordinates(1) := point_1.sdo_point.x;
rectangle.sdo_ordinates(2) := point_1.sdo_point.y;
rectangle.sdo_ordinates(3) := point_2.sdo_point.x;
rectangle.sdo_ordinates(4) := point_2.sdo_point.y;
-- Return it
return rectangle;
end;
/
如果这不符合您的预期，请重新提出问题（并澄清您要解决的实际业务问题）
答案 1 :(得分：0)
我发现了一种插入方式。可能存在有效的方法，但这个工作正常。
sdo_ordinate_array(
--this sub-query returns the Longitude of the first point
(SELECT SDO_GEOM.SDO_CENTROID(c.shape, m.diminfo).SDO_POINT.X X
FROM buildings c, user_sdo_geom_metadata m
WHERE m.table_name = 'BUILDINGS' AND m.column_name = 'SHAPE'
AND c.building_name = 'IICT'),
--this sub-query returns the Latitude of the first point
(SELECT SDO_GEOM.SDO_CENTROID(c.shape, m.diminfo).SDO_POINT.Y Y
FROM buildings c, user_sdo_geom_metadata m
WHERE m.table_name = 'BUILDINGS' AND m.column_name = 'SHAPE'
AND c.building_name = 'IICT'),
--this sub-query returns the Longitude of the second point
(SELECT SDO_GEOM.SDO_POINTONSURFACE(c.shape, m.diminfo).SDO_POINT.X X
FROM buildings c, user_sdo_geom_metadata m
WHERE m.table_name = 'BUILDINGS' AND m.column_name = 'SHAPE'
AND c.building_name = 'IICT'),
--this sub-query returns the Latitude of the second point
(SELECT SDO_GEOM.SDO_POINTONSURFACE(c.shape, m.diminfo).SDO_POINT.Y Y
FROM buildings c, user_sdo_geom_metadata m
WHERE m.table_name = 'BUILDINGS' AND m.column_name = 'SHAPE'
AND c.building_name = 'IICT')
) --SDO_ORDINATES: co-ordinates of the geomentry
sdo_ordinate_array只接受点数，而不是sdo_geometry的对象。所以，我必须分别获得经度和纬度。
因此，完整的插入查询将如下所示：
insert into buildings values(
4, -- index
'Reading Room', -- building_name
sdo_geometry(
2003, --SDO_GTYPE: dltt - 2(2D)0(linear referencing)03(polygon)
8307, --SDO_SRID: coordinate system
null, --SDO_POINT: it is for point inserting, if the next two field = null, then it could not be null.
sdo_elem_info_array( --SDO_ELEM_INFO:
1, --SDO_STARTING_OFFSET: indicates from which index of the next param of SDO_GEOMETRY would be considered, starts from 1.
1003, --SDO_ETYPE: 1(exterior, interior - 2)003(this digits usually comes from SDO_GTYPE)
3), --SDO_INTERPRETATION: 1 - simple polygon, 2 - polygon connecting arcs, 3 - rectangle, 4 - circle etc.
sdo_ordinate_array(
--this sub-query returns the Longitude of the first point
(SELECT SDO_GEOM.SDO_CENTROID(c.shape, m.diminfo).SDO_POINT.X X
FROM buildings c, user_sdo_geom_metadata m
WHERE m.table_name = 'BUILDINGS' AND m.column_name = 'SHAPE'
AND c.building_name = 'IICT'),
--this sub-query returns the Latitude of the first point
(SELECT SDO_GEOM.SDO_CENTROID(c.shape, m.diminfo).SDO_POINT.Y Y
FROM buildings c, user_sdo_geom_metadata m
WHERE m.table_name = 'BUILDINGS' AND m.column_name = 'SHAPE'
AND c.building_name = 'IICT'),
--this sub-query returns the Longitude of the second point
(SELECT SDO_GEOM.SDO_POINTONSURFACE(c.shape, m.diminfo).SDO_POINT.X X
FROM buildings c, user_sdo_geom_metadata m
WHERE m.table_name = 'BUILDINGS' AND m.column_name = 'SHAPE'
AND c.building_name = 'IICT'),
--this sub-query returns the Latitude of the second point
(SELECT SDO_GEOM.SDO_POINTONSURFACE(c.shape, m.diminfo).SDO_POINT.Y Y
FROM buildings c, user_sdo_geom_metadata m
WHERE m.table_name = 'BUILDINGS' AND m.column_name = 'SHAPE'
AND c.building_name = 'IICT')
) --SDO_ORDINATES: co-ordinates of the geomentry
)
);