Question

汽车表（汽车ID，Cus_id，Model，Delivery_Date，Reg_No）

      101   1   Yaris - Active  05-Jan-2016 aa-168
      102   1   Yaris           06-Jan-2017 ab-168
      103   2   Fortuner        07-Jan-2017 aa-1111
      104   3   Vios            08-Jan-2017 aa-1234
      105   3   Altis           09-Feb-2017 ab-5555
      106   4   Camry           10-Feb-2016 ab-9999
      107   5   Hilux Revo      11-Feb-2017 aa-6666

服务表（Car_id，Service_Date，Service_Amount，Service_Detail）

      101   05-Feb-2016 2565    1000km
      101   10-Aug-2016 3400    10000km
      101   20-Feb-2016 2100    20000km
      102   06-Feb-2017 2565    1000km
      103   07-May-2017 3400    10000km
      104   08-Feb-2017 2565    1000km
      105   09-May-2017 2565    1000km
      106   10-Mar-2016 2565    1000km
      106   15-Aug-2016 3400    10000km
      106   20-Dec-2016 2100    20000km

我的目标是从两个表中查询值，以便其显示

预期表（型号，Reg_No，Last_Service，Previous_Service）

              Yaris-Active aa-168 20-Feb-2016 10-Aug-2016
              Yaris        ab-168 06-Feb-2017 -
              .............................................
              Camry        aa-9999 20-Dec-2016 15-Aug-2016

您会看到Last_Service将包含汽车使用该服务的最后日期，而Previous_Service将包含汽车在使用该服务之前的日期。我的基本步骤是在轿厢和服务台之间倾斜时，向左倾斜。

我打算执行的步骤是查询Car_id和Count（Car_id）并按Car_id分组，因为它将指示该汽车的存在数量。之后，使用计数值指示该模型的最后一行和最后第二行，并将值指定给指定的列...但是，我必须将其转换为SQL。...

 SELECT Model, Reg_No, Service_Date As Last_Service, Service_Date As Previous_Date
  FROM Car LEFT JOIN Service ON Car.Car_id = Service.Car_id

............................................... ......................

   SELECT Car_id, Count(Car_id) As No.Times
   From Service
   Group BY Car_id

我需要帮助...

Answer 1

在您第一次选择之前将其放置：

int main()
{
    MyClass* array = function("some_file.txt");
    // do something with the array above..
    delete[] array;  //clean up..
}

您现在可以将WITH countedservices AS( SELECT *, ROW_NUMBER() OVER(PARTITION BY car_id ORDER BY service_date DESC) service_num FROM service )连接到汽车桌两次-一次使用countedservices，一次使用service_num = 1来访问当前和以前的服务数据< / p>

如果您使用的是左联接，则不要在WHERE子句中强加这些谓词（大概是因为某些汽车将无法维修），因为它将左联接转换为内部联接。将它们放在ON子句中：

service_num = 2

Answer 2

使用row_number()根据service_dates获取car_id的序列，然后加入下一个序列。

select t1.Car_id, t1.Model, t1.Reg_No, t2.Service_Date as Last_Service, t1.Service_Date as Previous_Date
from 
    (select Car_id, Model, Reg_No, row_number() over (partition by Car_id order by s.Service_Date asc) rn, s.Service_Date
        from Car c
        inner join Service s on s.Car_id = s.Car_id) as t1
left join
    (select Car_id, Model, Reg_No, row_number() over (partition by Car_id order by Service_Date asc) rn, s.Service_Date
        from Car c
        inner join Service s on s.Car_id = s.Car_id) as t2 on t2.Car_id = t1.Car_id and t2.rn + 1 = t1.rn

Answer 3

像这样尝试...

 SELECT    Model,
            Reg_No,
            Service_Date
  INTO      #Temp_Output_1
  FROM      Temp_Car        Car
  JOIN      Temp_Service    Servic
    ON  (Car.Car_id =   Servic.Car_id)

  SELECT    Model,
            Reg_No,
            Service_Date,
            (SELECT MAX(Service_Date) FROM #Temp_Output_1 T1 WHERE T1.Reg_No = T2.Reg_No AND T1.Service_Date < T2.Service_Date)  AS Prev_Service_Date
  FROM      #Temp_Output_1 T2
  WHERE     T2.Service_Date =   (
                                    SELECT  MAX(Service_Date)
                                    FROM    #Temp_Output_1 T3
                                    WHERE   T2.Reg_No   =   T3.Reg_No
                                )

SQL解决方案？

3 个答案: