第一次结对作业——SQL语句实现递归

今天在尝试编写地铁换乘查询时，在b站上发现了一个sql语句实现地铁换乘查询的案例，于是开始学习本案例。

此案例的数据表结构为：

 

 在案例讲解中，只在sql语句就实现了换乘查询的功能

以一号线的王府井到二号线的积水潭为例

WITH RECURSIVE transfer (start_station, stop_station, stops, path) AS (
  SELECT station_name, next_station, 1, CAST(CONCAT(line_name,station_name , '->', line_name,next_station) AS CHAR(1000))
    FROM bj_subway WHERE station_name = '王府井'
   UNION ALL
  SELECT p.start_station, e.next_station, stops + 1, CONCAT(p.path, '->', e.line_name, e.next_station)
    FROM transfer p
    JOIN bj_subway e
      ON p.stop_station = e.station_name AND (INSTR(p.path, e.next_station) = 0)
)
SELECT * FROM transfer WHERE stop_station ='积水潭';

-- Oracle
WITH transfer (start_station, stop_station, stops, path) AS (
  SELECT station_name, next_station, 1, line_name||station_name||'->'||line_name||next_station
    FROM bj_subway WHERE station_name = '王府井'
   UNION ALL
  SELECT p.start_station, e.next_station, stops + 1, p.path||'->'||e.line_name||e.next_station
    FROM transfer p
    JOIN bj_subway e
      ON p.stop_station = e.station_name AND (INSTR(p.path, e.next_station) = 0)
)
SELECT * FROM transfer WHERE stop_station ='积水潭';

-- SQL Server
WITH transfer(start_station, stop_station, stops, paths) AS (
SELECT station_name, next_station, 1 stops, 
       CAST(concat(line_name,station_name,'->',line_name,next_station) AS varchar(1000)) AS paths
FROM bj_subway
WHERE station_name = '王府井'
UNION ALL 
SELECT t.start_station, s.next_station, stops+1, CAST(concat(paths, '->', s.line_name, s.next_station)  AS varchar(1000))
FROM transfer t
JOIN bj_subway s 
ON (t.stop_station = s.station_name AND charindex(s.next_station, paths)=0)
)
SELECT *
FROM transfer
WHERE stop_station = '积水潭';

-- PostgreSQL
WITH RECURSIVE transfer (start_station, stop_station, stops, path) AS (
  SELECT station_name, next_station, 1, ARRAY[station_name::text, next_station::text]
    FROM bj_subway WHERE station_name = '王府井'
   UNION ALL
  SELECT p.start_station, e.next_station, stops + 1, p.path || ARRAY[e.next_station::text]
    FROM transfer p
    JOIN bj_subway e
      ON p.stop_station = e.station_name AND NOT e.next_station = ANY(p.path)
)
SELECT * FROM transfer WHERE stop_station ='积水潭';

-- SQLite
WITH RECURSIVE transfer(start_station, stop_station, stops, paths) AS (
SELECT station_name, next_station, 1 stops, 
       line_name||station_name||'->'||line_name||next_station AS paths
FROM bj_subway
WHERE station_name = '王府井'
UNION ALL 
SELECT t.start_station, s.next_station, stops+1, paths||'->'||s.line_name||s.next_station
FROM transfer t
JOIN bj_subway s 
ON (t.stop_station = s.station_name AND instr(paths, s.next_station)=0)
)
SELECT *
FROM transfer
WHERE stop_station = '积水潭';
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 在学习本案例时，我有些看不明白，并不理解为啥能够在SQL语句里面实现递归，但是通过讲解，明白了实质为深度查询，其实就是在利用算法来实现本功能，实际应用比较困难，写此博客来提供一种思路。