Sql 选择开始和结束日期跨越不同年份的每月小时数

我的情况：

我可以从以下视图检索数据：

|ID |      START_DATE      |        END_DATE      |
|80 | 09-JAN-2013 15:01:52 | 20-SEP-2014 15:01:52 |  
|82 | 09-SEP-2014 15:01:52 | 25-SEP-2014 15:01:52 | 

我想要的是这样的东西：

|TOTAL_TIME_IN_HOURS| MONTH| YEAR |
| 200               |   01 | 2013 |
| 250               |   02 | 2013 |
| etc.....          | etc. | etc..|
| 150               |   09 | 2014 |

一些补充资料： 我只能使用select语句，但我可以事先创建视图。 这是一个Oracle数据库，所以我不能使用诸如DATEDIFF等MYSQL函数

我已经做了以下工作：

SELECT 
ID, 
SUM(END_TIME - START_TIME) * 24 AS TOTAL_TIME_IN_HOURS, 
FROM TABLE_X
WHERE TO_CHAR(START_TIME, 'MM') IN (1,2,3,4,5,6,7,8,9,10,11,12) AND TO_CHAR(START_TIME, 'YYYY') BETWEEN 1965 AND 2050 AND 
TO_CHAR(END_TIME, 'MM') IN (1,2,3,4,5,6,7,8,9,10,11,12) AND TO_CHAR(END_TIME, 'YYYY') BETWEEN 1965 AND 2050
GROUP BY PROC_ID, TO_CHAR(START_TIME, 'YYYY'), TO_CHAR(START_TIME, 'MM') ORDER BY ID;

|ID| TOTAL_TIME_IN_HOURS |
|80|    5000             | 
|82|    300              |

这将返回以下内容：

SELECT 
ID, 
SUM(END_TIME - START_TIME) * 24 AS TOTAL_TIME_IN_HOURS, 
FROM TABLE_X
WHERE TO_CHAR(START_TIME, 'MM') IN (1,2,3,4,5,6,7,8,9,10,11,12) AND TO_CHAR(START_TIME, 'YYYY') BETWEEN 1965 AND 2050 AND 
TO_CHAR(END_TIME, 'MM') IN (1,2,3,4,5,6,7,8,9,10,11,12) AND TO_CHAR(END_TIME, 'YYYY') BETWEEN 1965 AND 2050
GROUP BY PROC_ID, TO_CHAR(START_TIME, 'YYYY'), TO_CHAR(START_TIME, 'MM') ORDER BY ID;

|ID| TOTAL_TIME_IN_HOURS |
|80|    5000             | 
|82|    300              |

我使用了虚构的结果，因为问题与实际结果无关

只要我只需要开始和结束日期之间的总小时数，这个逻辑就可以了。。然而，我需要的是开始日期和结束日期之间每月的总小时数

我想在我的视图中添加额外的列，比如开始月、结束月、开始年和结束年。然而，我在这些选项上遇到了新的问题，比如闰年

我的问题是：有可能达到我想要的结果吗？如果是这样，我应该用什么样的逻辑来达到这个结果？最好是动态查询，因此我不必输入数百行代码

with t(ID, START_DATE, END_DATE) as (
  select 80, to_date('09/01/2013 15:01:52', 'DD/MM/YYYY HH24:MI:SS'), to_date('20/09/2014 15:01:52', 'DD/MM/YYYY HH24:MI:SS') from dual union all
  select 82, to_date('09/09/2014 15:01:52', 'DD/MM/YYYY HH24:MI:SS'), to_date('25/09/2014 15:01:52', 'DD/MM/YYYY HH24:MI:SS') from dual
), t_mon(id, start_date, end_date, lvl, year, month) as (
   select id
        , start_date
        , least(trunc(add_months(start_date, 1), 'MONTH'), end_date)
        , 1
        , extract(year from start_date)
        , extract(month from start_date)
     from t
   union all
   select t.id
        , greatest(trunc(add_months(t.start_date, lvl), 'MONTH'), t.start_date)
        , least(trunc(add_months(t.start_date, lvl+1), 'MONTH'), t.end_date)
        , lvl + 1
        , extract(year from greatest(trunc(add_months(t.start_date, lvl), 'MONTH'), t.start_date))
        , extract(month from greatest(trunc(add_months(t.start_date, lvl), 'MONTH'), t.start_date))
     from t, t_mon
    where trunc(add_months(t.start_date, t_mon.lvl), 'MONTH') < t.end_date
), t_corr(id, start_date, end_date, year, month) as (
  select unique id, start_date, end_date, year, month
    from t_mon
)
select id, year, month, sum(end_date - start_date) * 24 hours
  from t_corr
 group by id, year, month
 order by id, year, month

       ID       YEAR      MONTH      HOURS
--------- ---------- ---------- ----------
       80       2013          1 536,968889
       80       2013          2        672
       80       2013          3        744
       80       2013          4        720
       80       2013          5        744
       80       2013          6        720
       80       2013          7        744
       80       2013          8        744
       80       2013          9        720
       80       2013         10        744
       80       2013         11        720
       80       2013         12        744
       80       2014          1        744
       80       2014          2        672
       80       2014          3        744
       80       2014          4        720
       80       2014          5        744
       80       2014          6        720
       80       2014          7        744
       80       2014          8        744
       80       2014          9 471,031111
       82       2014          9        384

---

另一个递归解决方案，至少需要Oracle 11gR2：

with t(id, start_date, end_date) as
  (select 80, to_date('09/01/2013 15:01:52', 'DD/MM/YYYY HH24:MI:SS'), to_date('20/09/2014 15:01:52', 'DD/MM/YYYY HH24:MI:SS') from dual
   union all
   select 82, to_date('09/09/2014 15:01:52', 'DD/MM/YYYY HH24:MI:SS'), to_date('25/09/2014 15:01:52', 'DD/MM/YYYY HH24:MI:SS') from dual
  )
, t_recur(id, start_date, end_date, month_start_date, month_end_date) as
  (select id
        , start_date
        , end_date
        , start_date
        , least(add_months(trunc(start_date, 'MM'), 1), end_date)
   from t
   union all
   select id
        , start_date
        , end_date
        , trunc(add_months(month_start_date, 1), 'MM')
        , least(add_months(trunc(month_start_date, 'MM'), 2), end_date)
   from t_recur
   where trunc(add_months(month_start_date, 1), 'MM') < end_date
  )
select id
     , extract(year from month_start_date) year
     , extract(month from month_start_date) month
     , (month_end_date - month_start_date) * 24 hours
from t_recur
order by id
       , year
       , month

---

使用层次结构查询的速度更快：

with w as
(
  select distinct id,
         greatest(start_date, trunc(add_months(start_date, level - 1), 'MON')) lim_low,
         least(trunc(add_months(start_date, level), 'MON'), end_date) lim_high
  from test t
  connect by add_months(start_date, level - 1) <= end_date
  order by 3, 1
)
select id, lim_low, (lim_high - lim_low) * 24 nb_hours
from w;

---

使用日历表左键连接，并按每月、每年分组。下面是生成一个日历表的方法，可能是输入错误，但视图中的行的开始日期晚于结束日期。@SylvainLeroux是的，确实是输入错误：谢谢您的帮助！在一般情况下，这不会更快，因为CONNECT BY查询会以指数方式增加行数，例如，如果源数据还包括记录812013年1月9日15:01:52、2014年9月20日15:01:52，则会生成600多万条记录，这些记录通过DISTINCT子句减少到43条。递归解决方案只生成43条记录。谢谢您的选择！