Question

我有以下表格：

ROUTE（id，train_number，station_id，arrival_time，departure_time）
SERVICE（train_number，train_name）
STATION（id，name，city）

ROUTE表中的一行将包含列车编号，它的到达和离开时间（如果该站是第一个/最后一个，则为这两个中的一个）和station_id。

我想找出每个train_number，它走哪条路径。我的困难在于确定每条路线的到达和离开站名称。

例如，列车1337可能具有以下路径：

stationA - ＆gt; stationB：
- 路线（1,1137，stationA，null，1800）
- 路线（2,1337，stationB，1900,2000）
stationB - ＆gt; stationC：
- 路线（3,1337，stationC，2020,2100）
stationC - ＆gt; stationD：
- 路线（4,1337，stationD，2120,2200）

我想弄清楚这些路径，即stationA - ＆gt; stationB - ＆gt; stationC等。

如何实现（如果重要的话，我正在使用Oracle SQL）？

对于stationA（获取路线的开始）我可以执行以下操作：

SELECT name FROM STATION
WHERE station.id = ROUTE.station_id AND ROUTE.arrival IS NULL

Answer 1

设定：

CREATE TABLE STATION (
  id      NUMBER PRIMARY KEY,
  name    VARCHAR2(20),
  city    VARCHAR2(20)
);

CREATE TABLE SERVICE (
  train_number  NUMBER PRIMARY KEY,
  train_name    VARCHAR2(20)
);

CREATE TABLE ROUTE (
  id              NUMBER PRIMARY KEY,
  train_number    NUMBER REFERENCES SERVICE( train_number ),
  station_id      NUMBER REFERENCES STATION( id ),
  arrival_time    NUMBER,
  departure_time  NUMBER
);

INSERT INTO STATION VALUES ( 1, 'stationA', 'city1' );
INSERT INTO STATION VALUES ( 2, 'stationB', 'city1' );
INSERT INTO STATION VALUES ( 3, 'stationC', 'city2' );
INSERT INTO STATION VALUES ( 4, 'stationD', 'city3' );

INSERT INTO SERVICE VALUES ( 1337, 'train1' );
INSERT INTO SERVICE VALUES ( 1338, 'train2' );
INSERT INTO SERVICE VALUES ( 1339, 'train3' );

INSERT INTO ROUTE VALUES (1, 1337, 1, null, 1800);
INSERT INTO ROUTE VALUES (2, 1337, 2, 1900, 2000);
INSERT INTO ROUTE VALUES (3, 1337, 3, 2020, 2100);
INSERT INTO ROUTE VALUES (4, 1337, 4, 2120, 2200);
INSERT INTO ROUTE VALUES (5, 1338, 1, null, 1800);
INSERT INTO ROUTE VALUES (6, 1338, 4, 1900, 2000);
INSERT INTO ROUTE VALUES (7, 1338, 3, 2020, 2100);
INSERT INTO ROUTE VALUES (8, 1338, 2, 2120, 2200);

查询：

WITH INDEXED_ROUTES AS (
  SELECT train_number,
         s.name,
         ROW_NUMBER() OVER( PARTITION BY train_number ORDER BY COALESCE( arrival_time, 0 ), COALESCE( departure_time, 2400 ) ) AS IDX
  FROM   ROUTE r
          INNER JOIN
          STATION s
          ON ( r.station_id = s.id )
)
SELECT  train_number,
        SUBSTR( SYS_CONNECT_BY_PATH( NAME, ' -> ' ), 5 ) AS route
FROM    indexed_routes
WHERE CONNECT_BY_ISLEAF = 1
START WITH IDX = 1
CONNECT BY PRIOR IDX + 1 = IDX
AND        PRIOR train_number = train_number;

或者：

SELECT train_number,
       LISTAGG( s.name, ' -> ' ) WITHIN GROUP ( ORDER BY arrival_time ASC NULLS FIRST ) AS route
FROM   ROUTE r
        INNER JOIN
        STATION s
        ON ( r.station_id = s.id )
GROUP BY train_number;

要么输出：

TRAIN_NUMBER ROUTE
------------ --------------------------------------------
        1337 stationA -> stationB -> stationC -> stationD  
        1338 stationA -> stationD -> stationC -> stationB

如果您只想要成对的电台，请使用LAG或LEAD分析函数：

SELECT train_number,
       LAG( s.name ) OVER( PARTITION BY train_number ORDER BY COALESCE( arrival_time, 0 ), COALESCE( departure_time, 2400 ) ) AS prev_station_name,
       s.name AS station_name
FROM   ROUTE r
        INNER JOIN
        STATION s
        ON ( r.station_id = s.id );

输出：

TRAIN_NUMBER PREV_STATION_NAME    STATION_NAME       
------------ -------------------- --------------------
        1337 (null)               stationA             
        1337 stationA             stationB             
        1337 stationB             stationC             
        1337 stationC             stationD             
        1338 (null)               stationA             
        1338 stationA             stationD             
        1338 stationD             stationC             
        1338 stationC             stationB

Answer 2

如果路径中没有分支，那么问题只是在到达或离开时间点订站：

SELECT s.station_id, name FROM ROUTE r INNER JOIN STATION s r.station_id = s.station_id ON WHERE train_number = 1 ORDER BY departure_time

Answer 3

在MySQL中，您可以使用按到达时间排序的group_concat，以逗号分隔列表的形式获取每列火车的路径。

例如：

select train_number, group_concat(name order by arrival_time asc)
from route join station
  on route.station_id=station.id
group by train_number
order by train_number;

输出如下：

1337    stationA,stationB,stationC,stationD

在Oracle（来自11g第2版）中，以下查询将产生相同的输出：

select train_number,  LISTAGG(name, ',') WITHIN GROUP (ORDER BY arrival_time ASC NULLS FIRST)
from route join station
  on route.station_id=station.id
group by train_number
order by train_number;

SQL查询 - 根据时间表获取到达和离开站点

3 个答案: