Question

我有下表，代表公共汽车及其停止的点。

bus_table

bus stop_no station_id 
  1       1          1 
  1       2          2
  1       3          3
  2       1          7
  2       2          8
  2       3          9
  3       1          3
  3       2          4
  3       3          5
  3       4          6
  3       5          7

我想从第1站到第9站。

我可以为此撰写什么查询？

Answer 1

举例说明。显然，这个解决方案不是递归的 - 但是你可以继续遍历总线路由，直到你从A到B。

DROP TABLE IF EXISTS bus_routes;


CREATE TABLE bus_routes 
(bus_no INT NOT NULL
,stop_no INT NOT NULL
,station_id INT NOT NULL
,PRIMARY KEY(bus_no,stop_no)
);

INSERT INTO bus_routes VALUES
(1,1,1),
(1,2,2),
(1,3,3),
(2,1,7),
(2,2,8),
(2,3,9),
(3,1,3),
(3,2,4),
(3,3,5),
(3,4,6),
(3,5,7);

SELECT a_from.bus_no
     , a_from.station_id start_from
     , b_from.station_id first_change_at
     , b_from.bus_no to_bus
     , c_from.station_id then_change_at
     , c_from.bus_no to_bus
     , c_to.station_id alighting_at
  FROM bus_routes a_from  
  JOIN bus_routes a_to 
    ON a_to.bus_no = a_from.bus_no 
   AND a_to.stop_no > a_from.stop_no
  JOIN bus_routes b_from
    ON b_from.station_id = a_to.station_id
  JOIN bus_routes b_to
    ON b_to.bus_no = b_from.bus_no
   AND b_to.stop_no > b_from.stop_no
  JOIN bus_routes c_from
    ON c_from.station_id = b_to.station_id
  JOIN bus_routes c_to
    ON c_to.bus_no = c_from.bus_no
   AND c_to.stop_no > c_from.stop_no
 WHERE a_from.station_id = 1
   AND c_to.station_id = 9;

+--------+------------+-----------------+--------+----------------+--------+--------------+
| bus_no | start_from | first_change_at | to_bus | then_change_at | to_bus | alighting_at |
+--------+------------+-----------------+--------+----------------+--------+--------------+
|      1 |          1 |               3 |      3 |              7 |      2 |            9 |
+--------+------------+-----------------+--------+----------------+--------+--------------+

Answer 2

公交车站图的递归遍历（在MSSQL 2016上测试）：

DECLARE @from NCHAR(4) = 'ort1';
DECLARE @to NCHAR(4) = 'ort9';

WITH [route] ([path], [lastBus], [current])
AS
    (SELECT CAST(N'' AS NVARCHAR(1000)) AS [path], CAST(N'' AS NCHAR(2)) AS [lastBus], @from AS [current]
    UNION ALL
    SELECT CAST( cur.[path] + N',' + busesAvailable.name + N' to ' + [next].stat AS NVARCHAR(1000)) AS [path],
        busesAvailable.name AS [lastBus],
        [next].stat AS [current]
    FROM [route] cur
    INNER JOIN dbo.BusStat busesAvailable ON busesAvailable.stat = cur.[current] AND busesAvailable.name <> cur.[lastBus]
    INNER JOIN dbo.BusStat [next]
        ON [next].stat <> cur.[current]
        AND [next].name = busesAvailable.name)
SELECT TOP 1 [route].[path]
FROM [route]
WHERE [route].[current] = @to;

考虑到该表是使用：

创建的

CREATE TABLE [dbo].[BusStat]
(
    [name] NCHAR (2) NOT NULL,
    [stat] NCHAR (4) NOT NULL
);
INSERT INTO [dbo].[BusStat] ([name], [stat])
SELECT 'b1',    'ort1' UNION ALL
SELECT 'b1',    'ort2' UNION ALL
SELECT 'b1',    'ort3' UNION ALL
SELECT 'b2',    'ort7' UNION ALL
SELECT 'b2',    'ort8' UNION ALL
SELECT 'b2',    'ort9' UNION ALL
SELECT 'b3',    'ort3' UNION ALL
SELECT 'b3',    'ort4' UNION ALL
SELECT 'b3',    'ort5' UNION ALL
SELECT 'b3',    'ort6' UNION ALL
SELECT 'b3',    'ort7';

在MySQL中进行子选择

2 个答案: