Question

我有下表：

unixtime    byte_sent   byte_received   session_id
---------------------------------------------------
1543569405  8           12              1
1543569404  8           12              1
1543569403  8           12              1
1543569402  8           12              1
1543569401  8           12              1
1543569410  15          25              2
1543569409  15          25              2
1543569408  15          25              2
1543569407  15          25              2
1543569411  10          30              3
1543569410  10          30              3
1543569409  10          30              3
1543569408  10          30              3
1543569407  10          30              3
1543569406  10          30              3
1543569405  10          30              3

然后我有以下SQL查询：

select
    session_id,
    max(unixtime) as session_end_time,
    sum(byte_sent) as byte_sent_tot,
    sum(byte_received) as byte_received_tot,
    count(*) as duration
from
    table
group by
    session_id
order by 
    session_end_time

结果如下：

session_id  session_end_time    byte_sent_tot   byte_received_tot   duration
----------------------------------------------------------------------------
1           1543569405          40              60                  5
2           1543569410          60              100                 4
3           1543569411          70              210                 7

我需要反向查询才能从结果返回到原始表。

假设持续时间的每一秒都应将session_end_time一次跨一，并且总字节数可以除以持续时间，这样我将获得字节/秒的平均值。

谢谢。

Answer 1

假定列的每个session_id的记录分布均匀我们可以执行以下操作

with data
  as (select session_id, /*This is the Input query which the user has obtained.*/
             max(unixtime) as session_end_time,
             sum(byte_sent) as byte_sent_tot,
             sum(byte_received) as byte_received_tot,
             count(*) as duration
        from t
     group by session_id
     order by session_end_time
     )
select a.session_id
      ,a.session_end_time - b.col  + 1 as unixtime
      ,b.col  
      ,a.byte_sent_tot/a.duration as byte_sent /*Distribute evenly based on the count*/
      ,a.byte_received_tot/a.duration as byte_received
  from data a
   cross join generate_series(1,a.duration) as b(col)

SQL反向求和运算

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