to_city_name

Question

我必须在pSQL中做一个递归函数，才能得到下面的查询。

• 我有一个表叫 tb_route 与 from_city 和 to_city
• 我有另一列，显示不同城市之间的距离，单位是公里。

这个表是递归建立的。我必须做一个递归CTE来显示两个城市之间的路线（例如，从 "Santiago de compostela "到 "Saint Jean Pied de Port"），显示路线的总公里数和经过的城市。

输出的结果必须是这样的。

这是我试过的结果

WITH RECURSIVE cities AS (
      SELECT *
      FROM textil.tb_route
      WHERE to_city_name = 'Santigo de Compostela'
  UNION ALL
      SELECT e.from_city, e.to_city, e.route, e.km
      FROM textil.tb_route e
  INNER JOIN cities tb_route ON tb_route.from_city_name = e.from_city
)

SELECT *
FROM cities;

我发现了一个错误，比如：

ERROR:  column e.from_city does not exist
LINE 8: ...JOIN cities tb_route ON tb_route.from_city_name = e.from_cit...

表：

CREATE TABLE textil.tb_route
(
    from_city_name            CHARACTER VARYING(120) NOT NULL ,
    to_city_name              CHARACTER VARYING(120) NOT NULL ,
    km_distance_num           NUMERIC(5,2) NOT NULL ,
    CONSTRAINT pk_route PRIMARY KEY (from_city_name, to_city_name)
);

数据：

INSERT INTO textil.tb_route VALUES
  ('Saint Jean Pied de Port','Roncesvalles',25.7),
  ('Somport','Jaca',30.5),
  ('Roncesvalles','Zubiri',21.5),
  ('Jaca','Arrés',25),
  ('Zubiri','Pamplona/Iruña',20.4),
  ('Arrés','Ruesta',28.7),
  ('Pamplona/Iruña','Puente la Reina/Gares',24),
  ('Ruesta','Sangüesa',21.8),
  ('Puente la Reina/Gares','Estella/Lizarra',22),
  ('Sangüesa','Monreal',27.25),
  ('Estella/Lizarra','Torres del Río',29),
  ('Monreal','Puente la Reina/Gares',31.1),
  ('Torres del Río','Logroño',20),
  ('Logroño','Nájera',29.6),
  ('Nájera','Santo Domingo de la Calzada',21),
  ('Santo Domingo de la Calzada','Belorado',22.7),
  ('Belorado','Agés',27.4),
  ('Agés','Burgos',23),
  ('Burgos','Hontanas',31.1),
  ('Hontanas','Boadilla del Camino',28.5),
  ('Boadilla del Camino','Carrión de los Condes',24.6),
  ('Carrión de los Condes','Terradillos de los Templarios',26.6),
  ('Terradillos de los Templarios','El Burgo Ranero',30.6),
  ('El Burgo Ranero','León',37.1),
  ('León','San Martín del Camino',25.9),
  ('San Martín del Camino','Astorga',24.2),
  ('Astorga','Foncebadón',25.9),
  ('Foncebadón','Ponferrada',27.3),
  ('Ponferrada','Villafranca del Bierzo',24.1),
  ('Villafranca del Bierzo','O Cebreiro',28.4),
  ('O Cebreiro','Triacastela',21.1),
  ('Triacastela','Sarria',18.3),
  ('Sarria','Portomarín',22.4),
  ('Portomarín','Palas de Rei',25),
  ('Palas de Rei','Arzúa',28.8),
  ('Arzúa','Pedrouzo',19.1),
  ('Pedrouzo','Santiago de Compostela',20),
  ('Bayona','Ustaritz',14.3),
  ('Ustaritz','Urdax',21.2),
  ('Urdax','Elizondo',18.8),
  ('Elizondo','Berroeta',9.7),
  ('Berroeta','Olagüe',20.4),
  ('Olagüe','Pamplona/Iruña',25),
  ('Irún','Hernani',26.6),
  ('Hernani','Tolosa',18.9),
  ('Tolosa','Zerain',33),
  ('Zerain','Salvatierra/Agurain',28),
  ('Salvatierra/Agurain','Vitoria/Gasteiz',27.4),
  ('Vitoria/Gasteiz','La Puebla de Arganzón',18.5),
  ('La Puebla de Arganzón','Haro',31),
  ('Haro','Santo Domingo de la Calzada',20),
  ('Bayona','Irún',33.8),
  ('Tolosa','Zegama',37.9),
  ('Zegama','Salvatierra/Agurain',20.1),
  ('La Puebla de Arganzón','Miranda del Ebro',22.3),
  ('Miranda del Ebro','Pancorbo',16.7),
  ('Pancorbo','Briviesca',23.4),
  ('Briviesca','Monasterio de Rodilla',19.8),
  ('Monasterio de Rodilla','Burgos',28.5);
```

Answer 1

我理解你的问题是一个走图问题。正如你的问题中所描述的那样，边是有方向的（也就是说，你可以从 from_city_name 到 to_city_name，但不是反过来）。)

下面是一个使用递归CTE的方法。它的想法是从一个给定的城市开始，然后按照所有可能的路线，同时跟踪arry中的整体旅行路径。当检测到一个圆圈，或者到达目标城市时，递归就会停止。然后，外部查询会对成功的路径进行过滤（可能没有，也可能有一条或几条）。

with recursive cte as (
    select 
        from_city_name, 
        to_city_name, 
        km_distance_num, 
        array[from_city_name::text, to_city_name::text] path
    from tb_route
    where from_city_name = 'Saint Jean Pied de Port'
    union all
    select 
        r.from_city_name, 
        r.to_city_name, 
        c.km_distance_num + r.km_distance_num,
        c.path || r.to_city_name::text
    from tb_route r
    inner join cte c on c.to_city_name = r.from_city_name
    where 
        not r.to_city_name = any(c.path) 
        and c.from_city_name <> 'Santiago de Compostela'
)
select 
    path[1] from_city_name, 
    to_city_name, 
    km_distance_num, 
    array_to_string(path, ' > ') path
from cte
where to_city_name = 'Santiago de Compostela';  

DB Fiddle上的演示:

from_city_name Iruña > Puente la ReinaGares > EstellaLizarra > Torres del Río > Logroño > Nájera > Santo Domingo de la Calzada > Belorado > Agés > Burgos > Hontanas > Boadilla del Camino > Carrión de los Condes > Terradillos de los Templarios > El Burgo Ranero &gt；León &gt；San Martín del Camino &gt；Astorga &gt；Foncebadón &gt；Ponferrada &gt；Villafranca del Bierzo &gt；O Cebreiro &gt；Triacastela &gt；Sarria &gt；Portomarín &gt；Palas de Rei &gt；Arzúa &gt；Pedrouzo &gt；Santiago de Compostela。

Answer 2

在这里，我留下我最终得到的解决方案。

with recursive caminos(from_city_name, to_city_name, path, total_distance, terminar, ultima_ciudad) as (
    -- Consulta base
    select  to_city_name
            ,'Saint Jean Pied de Port' -- Cambiar Destino 
            ,concat(to_city_name, concat(' -> ', from_city_name)) 
            ,cast(km_distance_num as numeric(8,2))
            ,0 --No terminar
            ,from_city_name
    from    textil.tb_route 
    where   to_city_name = 'Santiago de Compostela' -- Cambiar Origen 
    union all
    -- Consulta recursiva
    select  caminos.from_city_name
            ,caminos.to_city_name
            ,concat(caminos.path, concat( ' -> ', tr.from_city_name))
            ,cast(caminos.total_distance + tr.km_distance_num as numeric(8,2))
            ,case when tr.from_city_name = caminos.to_city_name then 1 else 0 end
            ,tr.from_city_name  
    from    caminos inner join textil.tb_route tr on tr.to_city_name = caminos.ultima_ciudad and caminos.terminar != 1  
)
select from_city_name, to_city_name, path, total_distance 
from caminos 
where 1 = 1
and from_city_name = 'Santiago de Compostela' --Cambiar Origen 
and ultima_ciudad = 'Saint Jean Pied de Port' -- Cambiar Destino
;