Question

我正在进行流量分析，并且我对流量的平均旅行时间进行了汇总查看。我要做的是选择与给定时间段的最大平均旅行时间相关的时间。

以下是我正在寻找的代码和结果，没有时间：

SELECT dt.datekey, dt.peak_hours, dt.weekday_name, 
ROUND(MAX(avg_t_dur_traffic)/60,2) MaxATT
FROM cat.Analytics_AvgTimes att
INNER JOIN cat.Dim_Date dt on att.DateKey = dt.DateKey and att.timekey = dt.timekey
WHERE dt.DateKey = '20170523' AND dt.peak_hours <> 'off_peak'
GROUP BY dt.DateKey, dt.Peak_Hours, dt.Weekday_Name

datekey     peak_hours    weekday_name  MaxATT
2017-05-23  AM_Peak       Tuesday       28.93
2017-05-23  Midday_Peak   Tuesday       14.05
2017-05-23  PM_Peak       Tuesday       29.95

编辑：我没有专门针对这三个MaxATT时间，我只是寻求有关如何查询信息的帮助...

我在周二的AM高峰时段看到最大平均行程时间（MaxATT）。 29分钟当我加入时，我很难过。我得到以下结果：

请注意我已从结果中删除了几行以节省空间

SELECT dt.datekey, dt.TimeKey, dt.peak_hours, dt.weekday_name, 
ROUND(MAX(avg_t_dur_traffic)/60,2) MaxATT
FROM cat.Analytics_AvgTimes att
INNER JOIN cat.Dim_Date dt on att.DateKey = dt.DateKey and att.timekey = dt.timekey
WHERE dt.DateKey = '20170523' and dt.peak_hours <> 'off_peak'
GROUP BY dt.DateKey, dt.Peak_Hours, dt.Weekday_Name, dt.TimeKey

datekey     TimeKey     peak_hours  weekday_name    MaxATT
2017-05-23  05:15:00    AM_Peak     Tuesday         10.43
2017-05-23  05:30:00    AM_Peak     Tuesday         10.45
2017-05-23  07:15:00    AM_Peak     Tuesday         12.53
2017-05-23  07:30:00    AM_Peak     Tuesday         18.27
2017-05-23  07:45:00    AM_Peak     Tuesday         22.85
2017-05-23  08:00:00    AM_Peak     Tuesday         28.93

在AM峰值期间，我们可以看到与MaxATT = 28.93相关的时间是08:00。但我想要选择的是每个峰值期间的每个不同时间， 所以我要找的是

datekey     TimeKey     peak_hours    weekday_name  MaxATT
2017-05-23  08:00:00    AM_Peak       Tuesday       28.93
2017-05-23  13:45:00    Midday_Peak   Tuesday       14.05
2017-05-23  17:30:00    PM_Peak       Tuesday       29.95

这是尝试使用RANK（），但我失败了，因为MaxATT与正确的值不匹配，它拉的记录多于RANK（）= 1

SELECT id, datekey, peak_hours, weekday_name, MaxAtt
FROM(
        SELECT id, dt.DateKey, dt.Peak_Hours, dt.Weekday_Name, 
        ROUND(MAX(avg_t_dur_traffic)/60,2) MaxAtt,
        RANK() OVER(PARTITION BY id ORDER BY MAX(avg_t_dur_traffic) DESC) AS 
        [rank]
        FROM cat.Analytics_AvgTimes att
        INNER JOIN cat.Dim_Date dt ON att.datekey = dt.DateKey AND att.TimeKey = dt.TimeKey
        GROUP BY id, dt.DateKey, dt.Peak_Hours, dt.Weekday_Name
     ) mt
WHERE mt.[rank] = 1 AND DateKey = '20170523' AND peak_hours <> 'off_peak'
GROUP BY id, datekey, peak_hours, weekday_name, MaxAtt

感谢您的帮助，如果您需要进一步说明，请随时询问！

Edit: Sample data as requested

Answer 1

使用子查询查找所需的maxatt行，并将它们连接到原始表。

select a.* 
from cat.Analytics_AvgTimes a
Inner JOIN (
          select datekey, peak_hours, weekday_name, max(maxatt) as maxatt
          from cat.Analytics_AvgTimes
          group by datekey, peak_hours, weekday_name) b
ON a.datekey = b.datekey and a.peak_hours = b.peak_hours and 
   a.weekday_name = b.weekday_name and a.maxatt = b.maxatt

小提琴示例http://sqlfiddle.com/#!9/dcacb9/1/0

审核数据后进行编辑：

select one.*
from (select a.*, b.Avg_T_Dur_Traffic as max
      from dim_date a, Analytics_AvgTimes b
      where a.datekeyid = b.id) one
INNER JOIN (
      select a.*, max(b.Avg_T_Dur_Traffic) as max
      from dim_date a, Analytics_AvgTimes b
      where a.datekeyid = b.id
      group by datekey, monthid, weekid, dayid, peakhoursid) two
      ON one.datekey = two.datekey and one.monthid = two.monthid and
         one.weekid = two.weekid and one.dayid = two.dayid and one.max = two.max

摆弄提供的数据：http://sqlfiddle.com/#!9/d7889a/9

根据不同列的最大值选择时间

1 个答案: