Question

我有三张桌子（跑步者，阶段和时间）

跑步者表：

+--+----+
|id|name|
+--+----+
|1 |Karl|
+--+----+
|2 |Lou |
+--+----+

舞台表：

+--+-----+-----+---+
|id|name |order|end|
+--+-----+-----+---+
|1 |start|  1  | 0 |
+--+-----+-----+---+
|2 |bike |  2  | 0 |
+--+-----+-----+---+
|3 |run  |  3  | 0 |
+--+-----+-----+---+
|4 |end  |  4  | 1 |
+--+-----+-----+---+

跑步者数据（时间）表：

+------+-----+-----+
|runner|stage|time |
+------+-----+-----+
|   1  |  1  |10:00|
+------+-----+-----+
|   1  |  2  |10:30|
+------+-----+-----+
|   1  |  3  |11:00|
+------+-----+-----+
|   2  |  1  |10:00|
+------+-----+-----+
|   2  |  2  |10:43|
+------+-----+-----+
|   2  |  3  |11:56|
+------+-----+-----+
|   1  |  4  |12:14|
+------+-----+-----+
|   2  |  4  |12:42|
+------+-----+-----+

嗯......那么我现在想要的是得到如下结果（按总时间排序）：

+------+-----+-----+-----+-----+----------+
|runner|start|bike |run  | end |   Total  |
+------+-----+-----+-----+-----+----------+
| Karl |10:00|10:30|11:00|12:14| 01:44:00 |   <--- FIRST( one hour)
+------+-----+-----+-----+-----+----------+
| Lou  |10:30|10:30|11:56|12:42| 02:12:00 |   <--- SECONDS( two hours )
+------+-----+-----+-----+-----+----------+

知道如何才能做到这一点吗？问候！

Answer 1

以下情况应该有效（时间以秒为单位，而不是HH：MM：SS）

select r.name, rd_start.time as start, rd_bike.time as bike, rd_run.time as run, rd_end.time as end,  from runner as r, rd_start.time+rd_bike.time+rd_run.time+rd_end.time as total
inner join runnerdata as rd_start on r.id=rd_start.runner and rd_start.stage=1
inner join runnerdata as rd_bike on r.id=rd_bike.runner and rd_start.stage=2
inner join runnerdata as rd_run on r.id=rd_run.runner and rd_start.stage=3
inner join runnerdata as rd_end on r.id=rd_end.runner and rd_start.stage=4
order by (rd_start.time+rd_bike.time+rd_run.time+rd_end.time)

（如果您发布创建表格或更好地使用此工具：http://sqlfiddle.com/这将使我们更容易测试我们的陈述）

Answer 2

这需要连接，然后是条件聚合。最后一列使用timediff()减去两次：

select r.name,
       max(case when rt.stage = 1 then rt.time end) as start,
       max(case when rt.stage = 2 then rt.time end) as walk,
       max(case when rt.stage = 3 then rt.time end) as bike,
       max(case when rt.stage = 4 then rt.time end) as end,
       timediff(max(case when rt.stage = 4 then rt.time end),
                max(case when rt.stage = 1 then rt.time end)
               ) as TotalTime
from RunnersTime rt join
     Runners r
     on rt.runner = r.id 
group by r.id
order by TotalTime;

请注意，列名是固定的，因此不使用stages表。使它们动态化将使查询更加复杂。

Answer 3

查询看起来像这样，但计算总数的方法取决于时间的数据类型。

select runners.name as runner, starttime.time as start, biketime.time as bike, runtime.time as run, endtime.time as end, endtime.time - starttime.time as Total

from runners

inner join time as starttime on runners.id = starttime.runner
inner join stages as startstages on starttime.stage = startstages.id and startstages.name = 'start'

inner join time as biketime on runners.id = biketime.runner
inner join stages as bikestages on biketime.stage = bikestages.id and bikestages.name = 'bike'

inner join time as runtime on runners.id = runtime.runner
inner join stages as runstages on runtime.stage = runstages.id and runstages.name = 'run'

inner join time as endtime on runners.id = endtime.runner
inner join stages as endstages on endtime.stage = endstages.id and endstages.name = 'end'

order by endtime.time - starttime.time

Answer 4

你可能需要进行大量的内部加入，子查询，并比较这个时间与那个时间，如果你想要使用该模式，它真的不会很漂亮。或者，如果你的阶段是固定的，你可以简化为一个表，每列作为一个阶段。如果阶段的数量和名称需要变化（无论出于何种原因），那么我建议在赛跑者日期/时间表中存储开始时间和结束时间。

如果您的阶段已修复，那么直接从数据库中获取结果将很容易。如果阶段可能会有所不同（例如，取决于您的站点用户配置阶段），那么您将需要在PHP中交叉选择数据，或者如果您坚持在数据库中执行此操作，请查看this SO question（我d劝阻）。

MySQL - 按总时间排序结果

4 个答案: