Oracle 从圆心和半径创建SDO_几何多边形圆

我有圆心和半径的坐标，单位是米。 如何创建SDO_几何体类型的圆，因为它至少需要圆的三个点，如本例中所示

INSERT INTO cola_markets VALUES(
  4,
  'cola_d',
  SDO_GEOMETRY(
    2003,  -- two-dimensional polygon
    NULL,
    NULL,
    SDO_ELEM_INFO_ARRAY(1,1003,4), -- one circle
    SDO_ORDINATE_ARRAY(8,7, 10,9, 8,11)
  )
);

---

只有在投影数据的情况下，才能使用三个点表示圆。如果您的数据是大地测量的（即，您的中心位于经度/纬度），则表示圆的唯一方法是将其加密。您可以使用

SDO\u UTIL.CIRCLE\u POLYGON（）函数执行此操作

例如：

SQL> select sdo_util.circle_polygon (sdo_geometry(2001, 4326, sdo_point_type(-74.064962, 40.7113, null), null, null),500,1) from dual;

SDO_UTIL.CIRCLE_POLYGON(SDO_GEOMETRY(2001,4326,SDO_POINT_TYPE(-74.064962,40.711
-------------------------------------------------------------------------------
SDO_GEOMETRY(2003, 4326, NULL, SDO_ELEM_INFO_ARRAY(1, 1003, 1), SDO_ORDINATE_ARRAY(-74.064962, 40.7067975, -74.06422, 40.706833, -74.063491, 40.7069389, -74.062784, 40.7071136, -74.062112, 40.7073544, -74.061484, 40.7076573, -74.060912, 40.7080177, -74.060403, 40.7084299, -74.059966, 40.7088873, -74.059608, 40.7093828, -74.059335, 40.7099085, -74.05915, 40.7104562, -74.059057, 40.7110171, -74.059057, 40.7115826, -74.05915, 40.7121435, -74.059334, 40.7126912, -74.059608, 40.713217, -74.059966, 40.7137125, -74.060403, 40.7141699, -74.060911, 40.7145821, -74.061484, 40.7149426, -74.062111, 40.7152456, -74.062784, 40.7154863, -74.06349, 40.7156611, -74.06422, 40.715767, -74.064962, 40.7158025, -74.065704, 40.715767, -74.066434, 40.7156611, -74.06714, 40.7154863, -74.067813, 40.7152456, -74.06844, 40.7149426, -74.069013, 40.7145821, -74.069521, 40.7141699, -74.069958, 40.7137125, -74.070316, 40.713217, -74.07059, 40.7126912, -74.070774, 40.7121435, -74.070867, 40.7115826, -74.070867, 40.7110171, -74.070774, 40.7104562, -74.070589, 40.7099085, -74.070316, 40.7093828, -74.069958, 40.7088873, -74.069521, 40.7084299, -74.069012, 40.7080177, -74.06844, 40.7076573, -74.067812, 40.7073544, -74.06714, 40.7071136, -74.066433, 40.7069389, -74.065704, 40.706833, -74.064962, 40.7067975))

1 row selected. 

SQL> select circle (sdo_geometry(2001, 3857, sdo_point_type(-8244873.9, 4969851.29, null), null, null), 500) from dual;

CIRCLE(SDO_GEOMETRY(2001,3857,SDO_POINT_TYPE(-8244873.9,4969851.29,NULL),NULL,N
-------------------------------------------------------------------------------
SDO_GEOMETRY(2003, 3857, NULL, SDO_ELEM_INFO_ARRAY(1,1003, 4), SDO_ORDINATE_ARRAY(-8245373.9, 4969851.29, -8244873.9, 4970351.29, -8244373.9, 4969851.29))

如果数据是投影的，则使用以下函数生成一个3点圆：

create or replace function circle (
  center sdo_geometry,
  radius number
)
return sdo_geometry
is
  x number;
  y number;
begin
  x := center.sdo_point.x;
  y := center.sdo_point.y;
  return sdo_geometry (
     2003, center.sdo_srid, null,
     sdo_elem_info_array(1, 1003, 4),
     sdo_ordinate_array (
       x-radius, y,
       x, y+radius,
       x+radius, y
    )
  );
end;
/

例如：

SQL> select sdo_util.circle_polygon (sdo_geometry(2001, 4326, sdo_point_type(-74.064962, 40.7113, null), null, null),500,1) from dual;

SDO_UTIL.CIRCLE_POLYGON(SDO_GEOMETRY(2001,4326,SDO_POINT_TYPE(-74.064962,40.711
-------------------------------------------------------------------------------
SDO_GEOMETRY(2003, 4326, NULL, SDO_ELEM_INFO_ARRAY(1, 1003, 1), SDO_ORDINATE_ARRAY(-74.064962, 40.7067975, -74.06422, 40.706833, -74.063491, 40.7069389, -74.062784, 40.7071136, -74.062112, 40.7073544, -74.061484, 40.7076573, -74.060912, 40.7080177, -74.060403, 40.7084299, -74.059966, 40.7088873, -74.059608, 40.7093828, -74.059335, 40.7099085, -74.05915, 40.7104562, -74.059057, 40.7110171, -74.059057, 40.7115826, -74.05915, 40.7121435, -74.059334, 40.7126912, -74.059608, 40.713217, -74.059966, 40.7137125, -74.060403, 40.7141699, -74.060911, 40.7145821, -74.061484, 40.7149426, -74.062111, 40.7152456, -74.062784, 40.7154863, -74.06349, 40.7156611, -74.06422, 40.715767, -74.064962, 40.7158025, -74.065704, 40.715767, -74.066434, 40.7156611, -74.06714, 40.7154863, -74.067813, 40.7152456, -74.06844, 40.7149426, -74.069013, 40.7145821, -74.069521, 40.7141699, -74.069958, 40.7137125, -74.070316, 40.713217, -74.07059, 40.7126912, -74.070774, 40.7121435, -74.070867, 40.7115826, -74.070867, 40.7110171, -74.070774, 40.7104562, -74.070589, 40.7099085, -74.070316, 40.7093828, -74.069958, 40.7088873, -74.069521, 40.7084299, -74.069012, 40.7080177, -74.06844, 40.7076573, -74.067812, 40.7073544, -74.06714, 40.7071136, -74.066433, 40.7069389, -74.065704, 40.706833, -74.064962, 40.7067975))

1 row selected. 

SQL> select circle (sdo_geometry(2001, 3857, sdo_point_type(-8244873.9, 4969851.29, null), null, null), 500) from dual;

CIRCLE(SDO_GEOMETRY(2001,3857,SDO_POINT_TYPE(-8244873.9,4969851.29,NULL),NULL,N
-------------------------------------------------------------------------------
SDO_GEOMETRY(2003, 3857, NULL, SDO_ELEM_INFO_ARRAY(1,1003, 4), SDO_ORDINATE_ARRAY(-8245373.9, 4969851.29, -8244873.9, 4970351.29, -8244373.9, 4969851.29))

选择1行。
如果给定了中心（x，y）和半径r，则您可以简单地按如下方式形成3个点：
（x-r，y）、（x，y+r）、（x+r，y）
并在SDO坐标阵列中使用它们。在oracle文档中提到，它们需要3个非共线点，位于圆的圆周上。上面提到的点，给出这样的点。
使用这个问题的答案中给出的公式，我可以这样得到3点：点1=x-（180/pi）*（半径/6378137），y，点2=x，y+（180/pi）*（半径/6378137）/cos（x），点3=x+（180/pi）*（半径/6378137），y，就像我说的，三点圆和圆弧只允许用于投影坐标。不允许在大地坐标（即lat/long）中使用。如果您这样做，验证函数将拒绝它，并且如果您想使用它，将导致各种错误。此限制的一个应用是，当您在大地坐标系中围绕形状生成缓冲区时，所有圆边将自动加密。当您围绕一个点生成缓冲区时，也会发生这种情况：您得到一个加密的圆（就像
SDO\u UTIL.circle\u POLYGON（）
products）。