如何在 row_to_json 中不提及列名的情况下获取表数据答案

【问题标题】：How to get table data without mention columns name in row_to_json如何在 row_to_json 中不提及列名的情况下获取表数据
【发布时间】：2020-11-11 23:15:07
【问题描述】：

下面的脚本非常适合指定的列，我得到了我期望的结果。但我想从表中获取所有列。谢谢你的帮助。

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(lg.geom)::json As geometry,
      row_to_json((id, name)) As properties FROM coffee_shops As lg
    ) As f
) As fc

【问题讨论】：

你好。您还可以添加表结构、一些数据样本和确切的预期结果吗？
"type":"FeatureCollection","features":[{"type":"Feature","properties":{"id":1,"name":"1369 Coffee House" ,"address":"1369 Cambridge St","city":"Cambridge","state":"MA","zip":"02139"},"geometry":{"type":"Point","坐标":[-71.10044,42.373695]}} 预期结果
请添加您的表结构以获得精确的 quary

标签： javascript sql node.js postgresql postgis

【解决方案1】：

要包含所有字段（包括几何），您可以将表名或别名传递给 row_to_json

select 
row_to_json(lg) As properties FROM coffee_shops As lg

【讨论】：

但是geom文件不能在geojson中转换
如果您将 all 字段放入 json 中，则它必须包含几何图形，此代码将正确转换它。有些读者会很好地使用它。是否要排除某些字段由您决定...但您的问题指定您想要所有字段
对不起，我提到的，“type”：“FeatureCollection”，“features”：[{“type”：“Feature”，“properties”：{“id”：1，”name ":"1369 Coffee House","address":"1369 Cambridge St","city":"Cambridge","state":"MA","zip":"02139"},"geometry":{"type ":"Point","coordinates":[-71.10044,42.373695]}} 预期结果