用一种更简单的方法，使用self join 可以帮你实现。
试试这个：

select t1.* from train t1 
right join train t2 on t1.train=t2.train
where t1.station != 'Station3' and t1.time>(t2.time)
and t2.station='Station3' limit 1 ;

尝试更改 station 两个表上的值 t1 以及 t2 你想要的价值是 Station3 在这种情况下。
现场sql小提琴演示在这里。

如果你只需要知道下一站，你可以使用一个子查询3号站和158次列车的最长时间

select * from 
  my_table 
  where time > ( 
  select max(time) max_time
  from my_table 
  where Train = '158' AND Station = 'Station3' )
  and train = '158'
  order by time limit 1

另一种方法是先找出站点的最大时间。
MySQL8.0+让lead能够很好地获取下一个记录数据。
查询

SELECT
   Table1.Train
 , Table1.next_station AS station
 , Table1.next_station_time AS time
FROM ( 

   SELECT 
     *
     , LEAD(Station) OVER (PARTITION BY Train ORDER BY Time) AS next_station
     , LEAD(Time) OVER (PARTITION BY Train ORDER BY Time) AS next_station_time
   FROM
    Table1
   WHERE
     Train = 158 
)
 AS Table1

WHERE
   Table1.station = 'station3'
 AND
   Table1.station <> Table1.next_station

请参见演示https://www.db-fiddle.com/f/9ld7xznmeuzndrc3dmysut/1
在较旧的mysql版本中，我们需要更具创造性，因为lead不受支持。
模拟lead的最佳方法是使用mysql的用户变量和一个移位的self-left连接
记住mysql用户变量在mysql 5.1中工作+
查询

SELECT 
   current_train AS train
 , next_station AS station
 , next_time AS time
FROM ( 

  SELECT 
      t1.Train AS current_train
    , t1.Station AS current_station
    , t1.Time AS `current_time`
    , t2.Train AS next_train
    , t2.Station AS next_station
    , t2.Time AS next_time
  FROM (
    SELECT 
      *
      , @rownum1 := @rownum1 + 1 AS rownum
    FROM 
      Table1
    CROSS JOIN ( SELECT @rownum1 := 0 ) AS i 
    WHERE
      Train = 158
    ORDER BY 
      Time ASC
  ) AS t1
  LEFT JOIN (
    SELECT 
      *
      , @rownum2 := @rownum2 + 1 AS rownum
    FROM 
      Table1
    CROSS JOIN ( SELECT @rownum2 := 0 ) AS i   
    WHERE
      Train = 158 
    ORDER BY 
      Time ASC 
  ) AS t2
  ON   
   t1.rownum + 1 = t2.rownum
 ) 
 AS records
WHERE
   records.current_station = 'station3'
 AND
   records.current_station <> records.next_station

请参见演示http://sqlfiddle.com/#!9/ff23fc/38号