I访问了中显示的问题是如何从PostgreSql转换为Geojson格式的? this PostGIS SQL将整个桌子转换为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;
我已经在那里发现,在结果中,我们没有获得字段的名称。 我期望输出为 “ properties”:{“ id”:6323,“名称”:“餐厅ninaia”
但实际输出是 “属性”:{“ f1”:6323,“ f2”:“餐厅ninaia”
我阅读了row_to_json指令的规范,所以我决定更改最后一个row_to_json指令
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;
但是现在,Geojson也将几何场作为属性进行了检索。 我的意思是,在结果中,我可以看到以geojson格式形成的几何形状,并再次以gis后格式形成(不需要第二个几何形状,我可以浪费它),因此,如果第一个结果是1200kb,则第二个几何形状,第二个几何形状大约为2300kb。
我该怎么办?任何替代方案
row_to_json((id, name)) As properties
or
row_to_json((lg)) As properties
我也尝试了
row_to_json(('id',lg.id ,'masa',lg.masa ,'parcela',lg.parcela)) As properties
和其他任何人,但没有结果(只有SQL错误)
谢谢你
您需要做什么,首先选择列,然后选择row_to_json此选择。 有了您的价值观,这将提供以下示例:
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
对于路人,您不需要子征服,而JSONB的性能稍好一些。话虽如此,这在将11K记录倾倒在宽阔的桌子上时剃光了约150毫秒。为了使这个规模肯定需要
LIMIT
OFFSET
或扔在
WHERE
中,以滤除不需要的东西
您可以在控制台中使用OGR2OGR执行此代码
cd C:\\OSGeo4W64\\bin && ogr2ogr -f "GeoJSON" C:\Users\Documents\nameFile.json "PG:dbname=nameBD schemas=NameSchema host='localhost' port='5432' user='postgres' password='**'" -sql "select * from public.tableName"