Question

select trunc(a.submitted_date,'HH') thehour
     , count(1) count_in_hour 
from your_order a 
LEFT JOIN my_order b ON a.order_id = b.order_id 
where a.state!='INCOMPLETE' 
  and b.substatus != 'DUMMY_IGNORE_INVENTORY' 
  and a.submitted_date>=trunc(to_date('20/01/2018 00:00:00','dd/mm/yyyy hh24:mi:ss')) 
  and a.submitted_date<=trunc(to_date('21/01/2018 00:00:00','dd/mm/yyyy hh24:mi:ss')) 
  group by trunc(a.submitted_date,'HH') 
  order by thehour asc;

我需要获取日期之间的订单计数，这会返回我

21-JAN-18 00:00:00 | 3
21-JAN-18 01:00:00 | 4
21-JAN-18 02:00:00 | 5
21-JAN-18 03:00:00 | 6
21-JAN-18 04:00:00 | 7
21-JAN-18 08:00:00 | 7
21-JAN-18 09:00:00 | 4
21-JAN-18 10:00:00 | 9

我没有得到5,6和7个小时的结果，因为这段时间之间没有下订单，但是即使此时未下订单，我也需要查询以将计数返回为0。

就像我需要24行一样，因为日期之间有24小时

Answer 1

我的方法：

进行基本查询，并期望它将以以下格式提供结果

小时计数_小时_
“ 21-JAN-18 00:00:00” 3
“ 21-JAN-18 01:00:00” 4

我正在编写一个子查询，该查询将在同一日期生成24小时，并对以上结果进行正确的联接。因此，即使您查询的整个24小时都没有获取数据，您也会从此子查询中获得一个条目。

子查询：

尝试以下查询。

 select 
  to_char(to_date('21-JAN-18 00:00:00', 'dd/mm/yyyy  HH24:MI:SS') + (level-1)/24, 'DD-MON-YY HH24:MI:SS') as col1 
  from dual
  connect by level <=24;

所有查询：

select 
  t2.col1,
  decode(t1.count_in_hour,null,0,t1.count_in_hour)  as count_in_hour
  from 
  (

 SELECT
    trunc(
        a.submitted_date,
        'HH'
    ) thehour,
    COUNT(1) count_in_hour
FROM
    your_order a
    LEFT JOIN my_order b ON a.order_id = b.order_id
WHERE
        a.state != 'INCOMPLETE'
    AND
        b.substatus != 'DUMMY_IGNORE_INVENTORY'
    AND
        a.submitted_date >= trunc(TO_DATE('20/01/2018 00:00:00','dd/mm/yyyy hh24:mi:ss') )
    AND
        a.submitted_date <= trunc(TO_DATE('21/01/2018 00:00:00','dd/mm/yyyy hh24:mi:ss') )
GROUP BY
    trunc(
        a.submitted_date,
        'HH'
    )) t1 right join 
    (select 
  to_char(to_date('21-JAN-18 00:00:00', 'dd/mm/yyyy  HH24:MI:SS') + (level-1)/24, 'DD-MON-YY HH24:MI:SS') as col1 
  from dual
  connect by level <=24 
  ) t2 on  t1.thehour = to_date(t2.col1,'DD-MON-YY HH24:MI:SS') order by t2.col1;

Answer 2

您需要有一个单独的小时表，该小时表可容纳24小时，并且需要与该表联接才能看到计数为0的小时。

CREATE TABLE OrderHours
(
OrderHour CHAR(2)
)

INSERT INTO OrderHours values('00'),('01')..,('24');


select h.OrderHour, count(*)
                count_in_hour from OrderHour as h LEFT JOIN your_order a on h.OrderHour = trunc(a.submitted_date,'HH')  LEFT JOIN my_order b ON a.order_id = b.order_id where a.order_id = b.order_id 
                and a.state!='INCOMPLETE' and b.substatus != 'DUMMY_IGNORE_INVENTORY' 
                and a.submitted_date>=trunc(to_date('20/01/2018 00:00:00','dd/mm/yyyy hh24:mi:ss')) 
                and a.submitted_date<=trunc(to_date('21/01/2018 00:00:00','dd/mm/yyyy hh24:mi:ss')) group by h.OrderHour order by thehour asc;

如果日期之间没有订单，则Oracle SQL返回0计数

2 个答案: