如何在SQL中按多個列將連續的行分組在一起

Question

我在查詢中有返回以下內容的行：

Date        User    Time    Location    Service     Count
1/1/2018    Nick    12:00   Location A  X           1
1/1/2018    Nick    12:01   Location A  Y           1
1/1/2018    John    12:02   Location B  Z           1
1/1/2018    Harry   12:03   Location A  X           1
1/1/2018    Harry   12:04   Location A  X           1
1/1/2018    Harry   12:05   Location B  Y           1
1/1/2018    Harry   12:06   Location B  X           1
1/1/2018    Nick    12:07   Location A  X           1
1/1/2018    Nick    12:08   Location A  Y           1

查詢返回用戶訪問過的位置以及該位置進行的選擇計數。 結果按用戶和時間遞增排序。 我需要將其分組到用戶和位置相同的CONSECUTIVE行與Count列的SUM和Service Column中以逗號分隔的唯一值列表分組的地方，最終結果返回如下內容：

Date        User    Start Time  End Time    Location    Service Count
1/1/2018    Nick    12:00       12:01       Location A  X,Y     2
1/1/2018    John    12:02       12:02       Location B  Z       1
1/1/2018    Harry   12:03       12:04       Location A  X       2
1/1/2018    Harry   12:05       12:06       Location B  X,Y     2
1/1/2018    Nick    12:07       12:08       Location A  X,Y     2

我不確定從哪里開始。 也許滯后或分區條款？ 希望SQL專家可以在這里提供幫助...

Answer 1

這是一個懸而未決的問題。 解決它的一種方法是使用row_number() ：

select Date, User, min(Time) as start_time, max(time) as end_time,
       Location,
       listagg(Service, ',') within group (order by service),     
       count(*) as cnt
from (select t.*,
             row_number() over (date order by time) as seqnum,
             row_number() over (partition by user, date, location order by time) as seqnum_2
      from t
     ) t
group by Date, User, Location, (seqnum - seqnum_2);

解釋它是如何工作的有點棘手。 我的建議是運行子查詢，您將看到行號的差異如何定義您要查找的組。

Answer 2

使用lag獲取上一行的用戶和位置值。 然后，只要用戶和位置發生變化，就使用一個總和生成一個新的組。 最后匯總分類的組，用戶，位置和日期。

select Date, User, min(Time) as start_time,max(time) as end_time, Location,
listagg(Service, ',') within group (order by Service),
count(*) as cnt
from (select Date, User, Time, Location, 
      sum(case when prev_location=location and prev_user=user then 0 else 1 end) over(order by date,time) as grp
      from (select Date, User, Time, Location,
            lag(Location) over(order by date,time) as prev_location,
            lag(User) over(order by date,time) as prev_user,
            from t
           ) t
      ) t
group by Date, User, Location, grp;