Question

给出PostgreSQL 9.4 jsonb列＆＃39; location＆＃39;内容如下：

string[] arr = new string[]
{
    "line1 text // random stuff",
    "line2 text // random stuff",
    "line3 text // random stuff",
    "line4 text // random stuff",
    "line5 text // random stuff"
};

arr = arr.Select(x => x.Split('/')[0].Trim()).ToArray();

要检索功能键＆＃39; geom＆＃39;我使用这个查询：

{"type": "FeatureCollection","features": [
  {
    "properties": {},
    "geom": "01030000000100000004000000000000000000244000000000000049400000000000002840000000000000494000000000000026400000000000004A4000000000000024400000000000004940",
    "type": "Feature",
    "geometry": {"type": "Polygon", "coordinates": [[[10,50],[12,50],[11,52],[10,50]]]}             
  },
  {..},
  {..}
]}

作品。但现在我想像这样做一个Postgis SELECT geom FROM ( SELECT jsonb_array_elements(t.location -> 'features') -> 'geom' AS geom FROM my_object t) AS g WHERE geom IS NOT NULL;查询：

ST_Intersects

不起作用，因为g.geom以SELECT ST_Intersects(g.geom, ST_GeomFromText('POINT(11 51)')) FROM ( SELECT t.iid, (jsonb_array_elements(t.location -> 'features') -> 'geom') AS geom FROM my_object t) AS g;传递：jsonb。我尝试转换为function st_intersects(jsonb, geometry) does not exist，然后错误为text。我如何处理function st_intersects(text, geometry) is not unique结果作为Postgis函数的输入？

Answer 1

首先，函数jsonb_array_elements()返回一个集合，因此你应该将它作为table function in a FROM clause。其次，您应该使用jsonb运算符获取->>字段的文本表示形式，然后将其转换为geometry，然后将其用于PostGIS函数。

检索非NULL的几何：

SELECT f.value ->> 'geom' AS geom
FROM my_object t
JOIN LATERAL jsonb_array_elements(t.location -> 'features') f ON true
WHERE f.value -> 'geom' IS NOT NULL;

做交叉：

SELECT t.iid, ST_Intersects((f.value->>'geom')::geometry, ST_GeomFromText('POINT(11 51)'))
FROM my_object t
JOIN LATERAL jsonb_array_elements(t.location -> 'features') f ON true;

您可能希望在选择列表中添加一些properties，以便区分表格中每行的多个几何图形。

Answer 2

要从geojson创建postgis几何体类型，您可以使用函数ST_GeomFromGeoJSON

例如：

SELECT ST_GeomFromGeoJSON('{"type":"Point","coordinates":[-48.23456,20.12345]}');

Postgis查询ST_Intersects与GeoJSON（jsonb）

2 个答案: