0

我在 SQL Server 中有一个问题。我有下表：

ROUTES = the route ID
STATIONS = the station ID
ORDER = the order the train pass from the stations
STOPS? = if the train stops at this station then is equal to 1 otherwise 0

样本数据：

-----------------------------------------------------
ROUTES      STATIONS    ORDER       STOPS?
-----------------------------------------------------
R1          S1          1           1
R1          S2          2           1
R1          S3          3           1
R1          S4          4           1
R1          S5          5           1
R2          S2          1           1
R2          S3          2           1
R2          S4          3           1
R3          S1          1           1
R3          S2          2           1
R3          S7          3           1
R3          S4          4           1
R3          S5          5           1
R3          S6          6           1
R4          S1          1           1
R4          S2          2           1
R4          S3          3           0
R4          S4          4           1
R5          S2          1           1
R5          S3          2           0
R5          S4          3           1
R6          S3          1           1
R6          S4          2           0
R6          S5          3           0
R6          S6          4           1
R7          S2          1           1
R7          S3          2           0
R7          S4          3           0
R7          S5          4           1

所以得出结论，我们有以下路线：

R1: S1-S2-S3-S4-S5
R2: S2-S3-S4
R3: S1-S2-S7-S4-S5-S6
R4: S1-S2-S3-S4
R5: S2-S3-S4
R6: S3-S4-S5-S6
R7: S2-S3-S4-S5

假设 S2 和 S4 是连接站

这意味着如果一条路线上的火车停在那里（STOPS = 1），乘客可以下车并从另一条路线乘坐另一辆火车

所以我们有一张表提到中转站

conn_stations
--------------
S2
S4

我的问题是如何获得所有可能的路线组合，例如从 S1 站出发并到达 S5 站。乘客可以根据上面的数据改变路线，我们应该得到以下结果（路线）：

R1:     S1-S2-S3-S4-S5
R3:     S1-S2-S7-S4-S5
temp1:  S1-S2(from R1)-S7-S4-S5(from R3)
temp2:  S1-S2(from R3)-S3-S4(from R1)-S5(from R3)
temp3:  S1-S2-S3-S4(from R1)-S5(from R3)
e.t.c

我希望你明白我在问什么。

如果有帮助，我有一张表，上面写着两个车站之间的距离，这也表明了哪些车站是连接的

Station A   Station B   Distance
-------------------------------------
S1          S2          5
S2          S3          1
S2          S7          8
S3          S4          15
S4          S5          16
S5          S6          25
S7          S4          10
4

2 回答 2

0

我认为，如果您不必包括连接站和终点站之间的所有中间站，那会更容易，但这是我为直达路线所做的：

select sroute, sstation, sorder, eroute, estation, eorder
from
 (select route as sroute, station as sstation, order as sorder
  from path
  where station = "s1" and stops = 1) pstart inner join
 (select route as eroute, station as estation, order as eorder
  from path
  where station = "s5" and stops = 1) pend on
 pstart.sroute = pend.eroute) direct

并通过一个连接站的路线：

SELECT sroute, sstation, sorder, 
        croute, cstation, corder,
        oroute, ostation, oorder,
        p.route , p.station, p.ORDER
FROM
    (SELECT sroute, sstation, sorder, 
                     croute, cstation, corder, 
                     p.route AS oroute, p.station AS ostation, p.ORDER AS oorder
    FROM
         (SELECT sroute, sstation, sorder, 
                 p.route AS croute, p.station AS cstation, p.ORDER AS corder
          FROM
              (SELECT route AS sroute, station AS sstation, ORDER AS sorder
               FROM path
               WHERE station = "s1" AND stops = 1
               ) pstart INNER JOIN
               path p ON 
               pstart.route = p.route INNER JOIN 
               conn_stations c ON 
               p.station = c.station
          WHERE p.stops = 1
          ) conn INNER JOIN 
          path p ON 
          conn.cstation = p.station
          WHERE p.stops = 1 AND p.route <> conn.route 
      ) conn_off INNER JOIN
        path p ON 
        conn_off.oroute = p.route
WHERE p.stops = 1 AND p.station = "s5"

但是连接站的数量会反映在SQL中，你最好使用编程工具。

高温高压

于 2012-11-20T19:22:29.847 回答
0

我认为在您的示例数据中有 39 条从S1to的路线S5（如果我的查询正确的话）。我不一定相信 SQL 是这项工作的正确工具（特别是如果根据评论，路由变成双向的）。

查询：

declare @StartStation char(2)
declare @EndStation char(2)
select @StartStation = 'S1'
select @EndStation = 'S5'

;With PartialRoutes as (
    select Route,Station,CONVERT(varchar(max),Station) as CurrentPath,RouteOrder,1 as HasTransferred,CONVERT(varchar(max),Route) as RoutesTaken,Stop
    from
        @Routes r
    where
        r.Station = @StartStation and r.Stop = 1
    union all
    --Continue on current route
    select pr.Route,r.Station,pr.CurrentPath + '-' + r.Station,r.RouteOrder,0,RoutesTaken,r.Stop
    from
        PartialRoutes pr
            inner join
        @Routes r
            on
                pr.Route = r.Route and
                pr.RouteOrder = r.RouteOrder - 1
    union all
    --Transfers
    select r.Route,r.Station,pr.CurrentPath,r.RouteOrder,1,RoutesTaken + '-' + r.Station + '(' + pr.Route + ',' + r.Route + ')',r.Stop
    from
        PartialRoutes pr
            inner join
        @Connections c
            on
                pr.Station = c.Station
            inner join
        @Routes r
            on
                pr.Route != r.Route and
                pr.Station = r.Station
    where
        pr.HasTransferred = 0 --Prevents us transferring multiple times in a single station
            and r.Stop = 1
)
select * from PartialRoutes where Station = @EndStation and Stop=1 option (MAXRECURSION 0)

结果：

Route Station CurrentPath         RouteOrder  HasTransferred RoutesTaken
----- ------- ------------------- ----------- -------------- --------------------------
R7    S5      S1-S2-S3-S4-S5      4           0              R4-S2(R4,R7)
R3    S5      S1-S2-S3-S4-S5      5           0              R4-S2(R4,R7)-S4(R7,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R4-S2(R4,R7)-S4(R7,R1)
R3    S5      S1-S2-S3-S4-S5      5           0              R4-S2(R4,R5)-S4(R5,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R4-S2(R4,R5)-S4(R5,R1)
R3    S5      S1-S2-S7-S4-S5      5           0              R4-S2(R4,R3)
R1    S5      S1-S2-S7-S4-S5      5           0              R4-S2(R4,R3)-S4(R3,R1)
R3    S5      S1-S2-S3-S4-S5      5           0              R4-S2(R4,R2)-S4(R2,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R4-S2(R4,R2)-S4(R2,R1)
R1    S5      S1-S2-S3-S4-S5      5           0              R4-S2(R4,R1)
R3    S5      S1-S2-S3-S4-S5      5           0              R4-S2(R4,R1)-S4(R1,R3)
R3    S5      S1-S2-S3-S4-S5      5           0              R4-S4(R4,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R4-S4(R4,R1)
R7    S5      S1-S2-S3-S4-S5      4           0              R3-S2(R3,R7)
R3    S5      S1-S2-S3-S4-S5      5           0              R3-S2(R3,R7)-S4(R7,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R3-S2(R3,R7)-S4(R7,R1)
R3    S5      S1-S2-S3-S4-S5      5           0              R3-S2(R3,R5)-S4(R5,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R3-S2(R3,R5)-S4(R5,R1)
R3    S5      S1-S2-S3-S4-S5      5           0              R3-S2(R3,R4)-S4(R4,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R3-S2(R3,R4)-S4(R4,R1)
R3    S5      S1-S2-S3-S4-S5      5           0              R3-S2(R3,R2)-S4(R2,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R3-S2(R3,R2)-S4(R2,R1)
R1    S5      S1-S2-S3-S4-S5      5           0              R3-S2(R3,R1)
R3    S5      S1-S2-S3-S4-S5      5           0              R3-S2(R3,R1)-S4(R1,R3)
R3    S5      S1-S2-S7-S4-S5      5           0              R3
R1    S5      S1-S2-S7-S4-S5      5           0              R3-S4(R3,R1)
R7    S5      S1-S2-S3-S4-S5      4           0              R1-S2(R1,R7)
R3    S5      S1-S2-S3-S4-S5      5           0              R1-S2(R1,R7)-S4(R7,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R1-S2(R1,R7)-S4(R7,R1)
R3    S5      S1-S2-S3-S4-S5      5           0              R1-S2(R1,R5)-S4(R5,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R1-S2(R1,R5)-S4(R5,R1)
R3    S5      S1-S2-S3-S4-S5      5           0              R1-S2(R1,R4)-S4(R4,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R1-S2(R1,R4)-S4(R4,R1)
R3    S5      S1-S2-S7-S4-S5      5           0              R1-S2(R1,R3)
R1    S5      S1-S2-S7-S4-S5      5           0              R1-S2(R1,R3)-S4(R3,R1)
R3    S5      S1-S2-S3-S4-S5      5           0              R1-S2(R1,R2)-S4(R2,R3)
R1    S5      S1-S2-S3-S4-S5      5           0              R1-S2(R1,R2)-S4(R2,R1)
R1    S5      S1-S2-S3-S4-S5      5           0              R1
R3    S5      S1-S2-S3-S4-S5      5           0              R1-S4(R1,R3)

我在末尾添加了一个列，显示了遵循的路线，例如R4-S2(R4,R7)-S4(R7,R6)，表示他们从 route 开始R4，然后在S2他们转移到 route R7，然后在S4他们转移到 route R6。

希望您可以使用此查询并根据需要对其进行修改（即，我仍然不知道您问题中最终表格的相关性）。

编辑我在火车需要到达Stop目的地车站的查询中添加了最后一个条件。我不确定这是否是一项要求，但无论如何它不会影响S1结果S5。取回来很容易。

基于此数据设置：

declare @Routes table (Route char(2) not null, Station char(2) not null,RouteOrder int not null,Stop bit not null)
insert into @Routes (Route,Station,RouteOrder,Stop) values
('R1','S1',1,1),('R1','S2',2,1),('R1','S3',3,1),('R1','S4',4,1),('R1','S5',5,1),
('R2','S2',1,1),('R2','S3',2,1),('R2','S4',3,1),
('R3','S1',1,1),('R3','S2',2,1),('R3','S7',3,1),('R3','S4',4,1),('R3','S5',5,1),('R3','S6',6,1),
('R4','S1',1,1),('R4','S2',2,1),('R4','S3',3,0),('R4','S4',4,1),
('R5','S2',1,1),('R5','S3',2,0),('R5','S4',3,1),
('R6','S3',1,1),('R6','S4',2,0),('R6','S5',3,0),('R6','S6',4,1),
('R7','S2',1,1),('R7','S3',2,0),('R7','S4',3,0),('R7','S5',4,1)

declare @Connections table (Station char(2) not null)
insert into @Connections (Station) values ('S2'),('S4')
于 2012-11-21T07:35:39.087 回答