Mysql时间差异在组中

Question

我有一张times表。我想创建一个视图，它将为我提供所有传感器的主键和所有时间差异。

传感器休闲：

 0.   start
 1-3. on track timers
 4.   end

示例将显示最佳：

Times table:
----------------------------------------------------------
| TimeID | RaceID | RacerID | Round | Sensor |    Time    |
-----------------------------------------------------------
|   1    |    1   |    1    |   1   |    0   |   100540   |
-----------------------------------------------------------
|   2    |    1   |    2    |   1   |    0   |   154297   |
-----------------------------------------------------------
|   3    |    1   |    1    |   1   |    1   |   170134   |
-----------------------------------------------------------
|   4    |    1   |    2    |   1   |    1   |   185915   |
-----------------------------------------------------------
|   5    |    1   |    1    |   1   |    2   |   200137   |
-----------------------------------------------------------
|   6    |    1   |    2    |   1   |    2   |   235103   |
-----------------------------------------------------------
|   7    |    1   |    1    |   1   |    4   |   380123   |
-----------------------------------------------------------
|   8    |    1   |    2    |   1   |    4   |   421127   |
-----------------------------------------------------------

Desired view
-----------------------------------------------------------------------
| TimeID | RaceID | RacerID | Round | Time1 | Time2 | Time3 |   Total |
-----------------------------------------------------------------------
|   1    |    1   |    1    |   1   | 69594 | 99597 |  null |  279583 |
-----------------------------------------------------------------------
|   1    |    1   |    2    |   1   | 31618 | 80806 |  null |  266830 |
-----------------------------------------------------------------------

http://sqlfiddle.com/#!9/10c5f/1

Answer 1

如果我理解正确，只需使用一系列left join s：

SELECT t0.timeID , t0.raceID, t0.racerID, t1.time - t0.time as 'Time 1',
t2.time - t1.time as 'Time 2', t3.time - t3.time as 'Time 3', t4.time - t0.time as 'Total'
FROM time t0 left join
time t1 on t0.racerID = t1.racerID and t1.sensor = 1 left join
time t2 on t0.racerID = t2.racerID and t2.sensor = 2 left join
time t3 on t0.racerID = t3.racerID and t3.sensor = 3 left join
time t4 on t0.racerID = t4.racerID and t4.sensor = 4 
where t0.sensor = 0

Answer 2

您无法动态添加列，因此您需要通过在SELECT子句中的聚合函数表达式中表示Time1来指定Time1，以及所有您想要的值，这可能无法完成你表达了与其他价值观的关系。 MIN或MAX可能接近，但“第二个”或“第二个”值无法表达，因为它意味着与单独聚合的其他值的关系。

你可以考虑GROUP_CONCAT，它可能最接近你想要的。