Oracle SQL：不規則時間戳之間的計數

Question

很抱歉，如果以前在這里曾問過這個問題，但我似乎找不到。 我一直在尋找每小時的總和，但是我的問題是關於另一列中定義的時間戳之間的SUM和COUNT。

我有一個表叫incoming_orders：它顯示了預期的目標，並傳入訂單的時間戳。

我還有第二張表Scheduled_output ：它顯示每個目的地的每個計划輸出時刻。

我有第三個表，名為outgoing_orders ：它顯示了實際的目的地以及發出訂單的時間戳。

因此，數據可能是：

--Incoming_orders:
Destination  Timestamp
ROUTE B      14/03/2018 7:48:00 
ROUTE A      14/03/2018 7:58:00
ROUTE A      14/03/2018 12:48:00
ROUTE C      14/03/2018 13:28:00

--Scheduled_Output
ROUTE A      14/03/2018 8:00:00
ROUTE A      14/03/2018 11:00:00
ROUTE A      14/03/2018 12:00:00
ROUTE A      14/03/2018 17:00:00    
ROUTE B      14/03/2018 8:00:00
ROUTE B      14/03/2018 10:00:00
ROUTE B      14/03/2018 12:00:00
ROUTE C      14/03/2018 07:00:00 
ROUTE C      14/03/2018 14:00:00 
ROUTE C      14/03/2018 17:00:00 

--Which would lead to the following outgoing_orders:
ROUTE A      14/03/2018 8:00:00
ROUTE B      14/03/2018 8:00:00 
ROUTE C      14/03/2018 14:00:00
ROUTE A      14/03/2018 17:00:00

現在，我想檢查路由A的07:58傳入順序是否確實使它進入了路由A的08:00輸出周期。我正在考慮創建一個像這樣的表來顯示它：

Destination output moment   expected_output actual_output   diff
Route A     8:00            1               1               0
Route A     11:00           0               0               0
Route A     12:00           0               0               0
Route A     17:00           1               1               0

但是問題是：如何計算Expected_output列？ 如何將12:48的路線A的接收訂單分組到12：00-17：00組？ 它應該計算計划的輸出時刻之間的所有訂單，但是我不確定如何實現。

我能否將CEIL，FLOOR或ROUND設置為最接近schedule_output的值？ 還是可以通過行計數以某種方式在第n行和第n + 1行之間進行操作？ 還是有另一種更簡單的方法？

Answer 1

我認為以這種方式或多或少地確定計划輸出的先前時間，獲得時間間隔是最簡單的：

SELECT destination,
       time_stamp,
       ( SELECT max( time_stamp ) 
         FROM SCHEDULED_OUTPUT t1
         WHERE t.destination = t1.destination
           AND t1.time_stamp < t.time_stamp
        ) as previous_time_stamp
FROM SCHEDULED_OUTPUT t
order by 1,2

或更緊湊的形式使用分析功能：

SELECT destination,
       time_stamp,
       lag( time_stamp ) over (partition by destination order by time_stamp )
       as previous_time_stamp
FROM SCHEDULED_OUTPUT t
order by 1,2

演示： http : //sqlfiddle.com/#!4/c7bc9/1

| DESTINATION |            TIME_STAMP |   PREVIOUS_TIME_STAMP |
|-------------|-----------------------|-----------------------|
|     ROUTE A | 2018-03-14 08:00:00.0 |                (null) |
|     ROUTE A | 2018-03-14 11:00:00.0 | 2018-03-14 08:00:00.0 |
|     ROUTE A | 2018-03-14 12:00:00.0 | 2018-03-14 11:00:00.0 |
|     ROUTE A | 2018-03-14 17:00:00.0 | 2018-03-14 12:00:00.0 |
|     ROUTE B | 2018-03-14 08:00:00.0 |                (null) |
|     ROUTE B | 2018-03-14 10:00:00.0 | 2018-03-14 08:00:00.0 |
|     ROUTE B | 2018-03-14 12:00:00.0 | 2018-03-14 10:00:00.0 |
|     ROUTE C | 2018-03-14 07:00:00.0 |                (null) |
|     ROUTE C | 2018-03-14 14:00:00.0 | 2018-03-14 07:00:00.0 |
|     ROUTE C | 2018-03-14 17:00:00.0 | 2018-03-14 14:00:00.0 |

---

接下來，可以將上述結果集加入INCOMING_ORDERS以便計算計數：

SELECT x.destination, x.time_stamp as output_moment,
       count( y.DESTINATION ) as expected_output 
FROM (
   SELECT destination,
          time_stamp,
          lag( time_stamp ) over (partition by destination order by time_stamp )
          as previous_time_stamp
   FROM SCHEDULED_OUTPUT t
) x
LEFT JOIN INCOMING_ORDERS y
ON x.DESTINATION =  y.DESTINATION
AND y.TIME_STAMP <= x.TIME_STAMP
AND ( y.TIME_STAMP > x.previous_time_stamp OR x.previous_time_stamp IS NULL )
GROUP BY x.destination, x.time_stamp
ORDER BY 1,2

演示： http : //sqlfiddle.com/#!4/c3958/2

| DESTINATION |         OUTPUT_MOMENT | EXPECTED_OUTPUT |
|-------------|-----------------------|-----------------|
|     ROUTE A | 2018-03-14 08:00:00.0 |               1 |
|     ROUTE A | 2018-03-14 11:00:00.0 |               0 |
|     ROUTE A | 2018-03-14 12:00:00.0 |               0 |
|     ROUTE A | 2018-03-14 17:00:00.0 |               1 |
|     ROUTE B | 2018-03-14 08:00:00.0 |               1 |
|     ROUTE B | 2018-03-14 10:00:00.0 |               0 |
|     ROUTE B | 2018-03-14 12:00:00.0 |               0 |
|     ROUTE C | 2018-03-14 07:00:00.0 |               0 |
|     ROUTE C | 2018-03-14 14:00:00.0 |               1 |
|     ROUTE C | 2018-03-14 17:00:00.0 |               0 |

---

這種情況：

AND y.TIME_STAMP <= x.TIME_STAMP
AND ( y.TIME_STAMP > x.previous_time_stamp OR x.previous_time_stamp IS NULL )

告訴您，如果某個訂單在8:00:00下達，而該路線在8:00:00同時開始，則該訂單仍會分配給該“開始”路線。 如果這是不可能的（也就是說，當訂單在開始的確切時間下訂單時，必須將其分配給下一條路徑），然后將條件更改為：

AND y.TIME_STAMP < x.TIME_STAMP
AND ( y.TIME_STAMP >= x.previous_time_stamp OR x.previous_time_stamp IS NULL )