Question

我有三张桌子

驱动

defparam

每位车手都有轮班

  driver_id | driver_name |driver_number
  ----------------------------------------
    1       | Driver 1   | 99999
    2       | Driver 2   | 88888

在每次轮班期间，驾驶员可能会或可能不会多次出行

  shift_id | start_time            | end_time            | driver_id 
   -----------------------------------------------------------------
    4        |2015-04-02 10:09:00    |(NULL)               | 1
    3        |2015-04-02 09:19:00    |(NULL)               | 2
    2        |2015-04-02 11:09:00    |2015-04-02 19:09:00  | 1
    1        |2015-04-02 10:09:00    |2015-04-02 20:09:00  | 2

我希望得到一个查询，其中我在开放班次（结束时间为null）中获得最近最多行程以及驱动程序详细信息。

我所尝试的一切都失败了。

我理解查询应该是这样的：

     trip_id | start_time            | end_time            | shift_id 
       -----------------------------------------------------------------
        12       |2015-04-02 10:09:00    |(NULL)               | 4
        11       |2015-04-02 09:19:00    |(NULL)               | 4
        10       |2015-04-02 11:09:00    |2015-04-02 19:09:00  | 3
        9        |2015-04-02 10:09:00    |2015-04-02 20:09:00  | 2
        8        |2015-04-02 10:09:00    |(NULL)               | 4
        7        |2015-04-02 09:19:00    |(NULL)               | 4
        6        |2015-04-02 11:09:00    |2015-04-02 19:09:00  | 3
        5        |2015-04-02 10:09:00    |2015-04-02 20:09:00  | 4
        4        |2015-04-02 10:09:00    |(NULL)               | 4
        3        |2015-04-02 09:19:00    |(NULL)               | 4
        2        |2015-04-02 11:09:00    |2015-04-02 19:09:00  | 2
        1        |2015-04-02 10:09:00    |2015-04-02 20:09:00  | 1

请帮我查询一下。预期结果如下：

select * from
drivers 
inner join shifts on shifts.driver_id = drivers.driver_id
left join (some inner query on trips table) as trip 
on shifts.shift_id = trip.trip_id
where shifts.end_time is null;

Answer 1

这应该做你正在寻找的事情。

SELECT s.*, d.*, t.*
FROM shifts AS s 
INNER JOIN drivers AS d
    ON s.driver_id = d.driver_id
INNER JOIN (
    SELECT MAX(t.trip_id) AS trip_id
    FROM trips AS t
    INNER JOIN shifts AS s
        ON t.shift_id = s.shift_id
        AND s.end_time IS NULL
    GROUP BY s.shift_id) AS mt
INNER JOIN trips AS t
    ON s.shift_id = t.shift_id
    AND t.trip_id = mt.trip_id;

SQL Fiddle已添加且列名已修复：http://sqlfiddle.com/#!9/faeca/5

Answer 2

您可以使用变量来确定每个班次分区中的最近一次旅行：

SELECT d.driver_id, d.driver_name, s.shift_id, s.start_time AS shift_start_time, 
       t.trip_id AS recent_trip_id, t.start_time AS recent_trip_start_time
FROM drivers AS d
INNER JOIN shifts AS s ON s.driver_id = d.driver_id
INNER JOIN (
    SELECT trip_id, start_time, 
           @row_number:= CASE WHEN @sid = shift_id THEN @row_number+1
                              ELSE 1
                         END AS row_number,
           @sid:=shift_id AS shift_id  
    FROM trips, (SELECT @sid:=0,@row_number:=0) as vars
    ORDER BY shift_id, trip_id DESC ) t ON t.shift_id = s.shift_id AND t.row_number = 1
WHERE s.end_time IS NULL

谓词t.row_number = 1有效地选择每班最近的行程。查询的其余部分只是简单的JOIN子句，它们将驱动程序和（打开）移位数据聚集在一起。

SQL Fiddle Demo

Mysql左连接一行

2 个答案: