如何 select 只为 sql 中的函数返回没有行 ID 和列名的结果值？

Question

我执行 pgRouting，我需要从我的 SELECT.

向 dijkstra 算法插入整数结果

SELECT ST_AsGeoJSON(ST_Transform(way, 4326)) AS geometry
  FROM pgr_dijkstra(
'SELECT osm_id AS id, source, target, st_length(way) as cost FROM planet_osm_roads',
34, 3000, false
  ) as di
  JOIN planet_osm_roads pt
  ON di.edge = pt.osm_id ;

这是可行的，但是如果我想用这个查询将 dijkstra 函数中的节点 34 替换为我所在街道的节点号：

SELECT pl.source::integer 
FROM planet_osm_roads pl 
WHERE pl.name LIKE ''street_name'' 
LIMIT 1

一起：

SELECT ST_AsGeoJSON(ST_Transform(way, 4326)) AS geometry
  FROM pgr_dijkstra(
'SELECT osm_id AS id, source, target, st_length(way) as cost FROM planet_osm_roads',
'SELECT pl.source::integer FROM planet_osm_roads pl WHERE pl.name LIKE ''street_name'' LIMIT 1',
 3000, false
  ) as di
  JOIN planet_osm_roads pt
  ON di.edge = pt.osm_id ;

它将失败并出现错误：

ERROR:  function pgr_dijkstra(unknown, unknown, integer, boolean) is not unique
LINE 93:   FROM pgr_dijkstra(
                ^
HINT:  Could not choose a best candidate function. You might need to add explicit type casts.

我认为，这是因为我的 select 查询返回 sql 结果以及行 ID 和列名。但也许还有另一个问题。

如何只输出一个整数？

Answer 1

不要将查询作为字符串传递：

SELECT ST_AsGeoJSON(ST_Transform(way, 4326)) AS geometry
FROM pgr_dijkstra(
  'SELECT osm_id AS id, source, target, st_length(way) as cost FROM planet_osm_roads',
  (SELECT pl.source::integer FROM planet_osm_roads pl WHERE pl.name LIKE 'street_name' LIMIT 1),
   3000, false
) as di
  JOIN planet_osm_roads pt
  ON di.edge = pt.osm_id;

或者使用派生的 table

SELECT ST_AsGeoJSON(ST_Transform(way, 4326)) AS geometry
FROM (
  SELECT pl.source::integer as source
  FROM planet_osm_roads pl 
  WHERE pl.name 
  LIKE 'street_name' LIMIT 1
) pl 
  join lateral pgr_dijkstra(
    'SELECT osm_id AS id, source, target, st_length(way) as cost FROM planet_osm_roads',
    pl.source,
    3000, false
  ) as di on true
  JOIN planet_osm_roads pt ON di.edge = pt.osm_id;

Answer 2

Vertex 是整数参数，您正试图将文本放在那里。只需将 ' 更改为 ()

SELECT ST_AsGeoJSON(ST_Transform(way, 4326)) AS geometry
  FROM pgr_dijkstra('SELECT osm_id AS id, source, target, st_length(way) as cost FROM planet_osm_roads',
(SELECT pl.source::integer FROM planet_osm_roads pl WHERE pl.name LIKE ''street_name'' LIMIT 1),
 3000, false
  ) as di
  JOIN planet_osm_roads pt
  ON di.edge = pt.osm_id ;

这是一种方法，另一种方法是编写函数，将街道名称作为参数并将您的 geojson 作为输出。

如何 select 只为 sql 中的函数返回没有行 ID 和列名的结果值？

How to select just result value without row id and column name for function in sql?

sql

postgresql

postgis

dijkstra

pgrouting