我访问了How to convert from PostgreSQL to GeoJSON format?中显示的问题
此 PostGIS SQL 将整个表转换为 GeoJSON 结果:
SELECT row_to_json(fc) AS geojson FROM
(SELECT 'FeatureCollection' As type, array_to_json(array_agg(f))
As features FROM
(SELECT
'Feature' As type,
ST_AsGeoJSON((lg.geometry),15,0)::json As geometry,
row_to_json((id, name)) As properties
FROM imposm3_restaurants As lg) As f ) As fc;
我发现在结果中,我们没有得到字段的名称。
我希望输出为 "properties":{"id":6323,"name":"锡纳亚餐厅"
但实际输出是 "properties":{"f1":6323,"f2":"锡纳亚餐厅"
看了row_to_json指令的规范,决定把最后一条row_to_json指令改掉
SELECT row_to_json(fc) AS geojson FROM
(SELECT 'FeatureCollection' As type, array_to_json(array_agg(f))
As features FROM
(SELECT
'Feature' As type,
ST_AsGeoJSON((lg.geometry),15,0)::json As geometry,
row_to_json((lg)) As properties
FROM imposm3_restaurants As lg) As f ) As fc;
但现在 geojson 还检索几何字段作为属性。
我的意思是,在结果中我可以看到以 geojson 格式和 PostGIS 格式格式化的几何图形(这第二个几何图形不是必需的,我可以浪费它)所以如果第一个结果是 1200Kb,第二个将是2300Kb。
我能做什么?任何替代品
row_to_json((id, name)) As properties
或
row_to_json((lg)) As properties
我也尝试过类似的东西
row_to_json(('id',lg.id ,'masa',lg.masa ,'parcela',lg.parcela)) As properties
和任何其他,但没有结果(只有 SQL 错误)
非常感谢
最佳答案
您需要做的是,首先选择您的列,然后选择 row_to_json。 根据您的值(value)观,这将给出以下示例:
SELECT
row_to_json(fc)
FROM (
SELECT
'FeatureCollection' AS type
, array_to_json(array_agg(f)) AS features
FROM (
SELECT
'feature' AS type
, ST_AsGeoJSON(geom)::json as geometry
, (
SELECT
row_to_json(t)
FROM (
SELECT
id
, name
) AS t
) AS properties
FROM imposm3_restaurants
) AS f
) AS fc
关于postgresql - 将 PostGIS 表 (postgresql) 转换为 GeoJSON,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56163349/