我有一个表UPCALL_HISTORY，其中有3列：SUBSCRIBER_ID，START_DATE和END_DATE。设唯一订阅者数为N。
我想创建一个包含3列的新表：
SUBSCRIBER_ID：所有唯一用户ID连续重复36次。MONTHLY_CALENDAR_ID：对于每个SUBSCRIBER_ID，此列的日期为2015年7月至2018年7月（36个月）。ACTIVE：此列将用作每个订阅者以及该月是否有订阅的标志。此订阅数据在名为UPCALL_HISTORY的表中。我对SQL还是很陌生，没有很多经验。我擅长Python，但是SQL似乎不像Python。
任何可以帮助我建立此表的查询想法吗？
让我的UPCALL_HISTORY表为：
+---------------+------------+------------+
| SUBSCRIBER_ID | START_DATE | END_DATE |
+---------------+------------+------------+
| 119 | 01/07/2015 | 01/08/2015 |
| 120 | 01/08/2015 | 01/09/2015 |
| 121 | 01/09/2015 | 01/10/2015 |
+---------------+------------+------------+
我想要一个看起来像这样的表
+---------------+------------+--------+
| SUBSCRIBER_ID | MON_CA | ACTIVE |
+---------------+------------+--------+
| 119 | 01/07/2015 | 1 |
| * | 01/08/2015 | 0 |
| * | 01/09/2015 | 0 |
| (36 times) | 01/10/2015 | 0 |
| * | * | 0 |
| 119 | 01/07/2018 | 0 |
+---------------+------------+--------+
持续120和121
答案 0 :(得分：3)
这就是我对这个问题的理解。
示例表和几行：
SQL> create table upcall_history
2 (subscriber_id number,
3 start_date date,
4 end_date date);
Table created.
SQL> insert into upcall_history
2 select 1, date '2015-12-25', date '2016-01-13' from dual union
3 select 1, date '2017-07-10', date '2017-07-11' from dual union
4 select 2, date '2018-01-01', date '2018-04-24' from dual;
3 rows created.
创建一个新表。对于不同的SUBSCRIBER_ID's，它会创建36个“月”行，固定行（如您所述）。
SQL> create table new_table as
2 select
3 x.subscriber_id,
4 add_months(date '2015-07-01', column_value - 1) monthly_calendar_id,
5 0 active
6 from (select distinct subscriber_id from upcall_history) x,
7 table(cast(multiset(select level from dual
8 connect by level <= 36
9 ) as sys.odcinumberlist));
Table created.
将ACTIVE表的MONTHLY_CALENDAR_ID和START_DATE中包含END_DATE的行的UPCALL_HISTORY列值更新为“ 1”。
SQL> merge into new_table n
2 using (select subscriber_id, start_date, end_date from upcall_history) x
3 on ( n.subscriber_id = x.subscriber_id
4 and n.monthly_calendar_id between trunc(x.start_date, 'mm')
5 and trunc(x.end_date, 'mm')
6 )
7 when matched then
8 update set n.active = 1;
7 rows merged.
SQL>
结果（仅ACTIVE = 1）：
SQL> select * from new_table
2 where active = 1
3 order by subscriber_id, monthly_calendar_id;
SUBSCRIBER_ID MONTHLY_CA ACTIVE
------------- ---------- ----------
1 2015-12-01 1
1 2016-01-01 1
1 2017-07-01 1
2 2018-01-01 1
2 2018-02-01 1
2 2018-03-01 1
2 2018-04-01 1
7 rows selected.
SQL>
答案 1 :(得分：2)
如果您使用的是12c，则可以对cross apply使用所有月份的内联视图，以获取具有所有ID的月份的组合：
select uh.subscriber_id, m.month,
case when trunc(uh.start_date, 'MM') <= m.month
and (uh.end_date is null or uh.end_date >= add_months(m.month, 1))
then 1 else 0 end as active
from upcall_history uh
cross apply (
select add_months(trunc(sysdate, 'MM'), - level) as month
from dual
connect by level <= 36
) m
order by uh.subscriber_id, m.month;
我已经将其设置为到当前月份为止的36个月的滚动窗口，但实际上您可能希望像问题中那样确定日期。
包含来自CTE的示例数据：
with upcall_history (subscriber_id, start_date, end_date) as (
select 1, date '2015-09-04', '2015-12-15' from dual
union all select 2, date '2017-12-04', '2018-05-15' from dual
)
生成72行：
SUBSCRIBER_ID MONTH ACTIVE
------------- ---------- ----------
1 2015-07-01 0
1 2015-08-01 0
1 2015-09-01 1
1 2015-10-01 1
1 2015-11-01 1
1 2015-12-01 0
1 2016-01-01 0
...
2 2017-11-01 0
2 2017-12-01 1
2 2018-01-01 1
2 2018-02-01 1
2 2018-03-01 1
2 2018-04-01 1
2 2018-05-01 0
2 2018-06-01 0
您可以使用它来创建新表或填充现有表；但是，如果您要做想要滚动窗口，则视图可能更合适。
如果您未使用12c，则cross apply不可用-您会收到“ ORA-00905：缺少关键字”。
您可以通过交叉连接两个CTE（一个获取所有月份，另一个获取所有ID）获得相同的结果，然后外部连接到您的实际数据：
with m (month) as (
select add_months(trunc(sysdate, 'MM'), - level)
from dual
connect by level <= 36
),
i (subscriber_id) as (
select distinct subscriber_id
from upcall_history
)
select i.subscriber_id, m.month,
case when uh.subscriber_id is null then 0 else 1 end as active
from m
cross join i
left join upcall_history uh
on uh.subscriber_id = i.subscriber_id
and trunc(uh.start_date, 'MM') <= m.month
and (uh.end_date is null or uh.end_date >= add_months(m.month, 1))
order by i.subscriber_id, m.month;
答案 2 :(得分：2)
您可以使用分区外部联接在11g中做到这一点，就像这样：
WITH upcall_history AS (SELECT 119 subscriber_id, to_date('01/07/2015', 'dd/mm/yyyy') start_date, to_date('01/08/2015', 'dd/mm/yyyy') end_date FROM dual UNION ALL
SELECT 120 subscriber_id, to_date('01/08/2015', 'dd/mm/yyyy') start_date, to_date('01/09/2015', 'dd/mm/yyyy') end_date FROM dual UNION ALL
SELECT 121 subscriber_id, to_date('01/09/2015', 'dd/mm/yyyy') start_date, to_date('01/10/2015', 'dd/mm/yyyy') end_date FROM dual),
mnths AS (SELECT add_months(TRUNC(SYSDATE, 'mm'), + 1 - LEVEL) mnth
FROM dual
CONNECT BY LEVEL <= 12 * 3 + 1)
SELECT uh.subscriber_id,
m.mnth,
CASE WHEN mnth BETWEEN start_date AND end_date - 1 THEN 1 ELSE 0 END active
FROM mnths m
LEFT OUTER JOIN upcall_history uh PARTITION BY (uh.subscriber_id) ON (1=1)
ORDER BY uh.subscriber_id,
m.mnth;
SUBSCRIBER_ID MNTH ACTIVE
------------- ----------- ----------
119 01/07/2015 1
119 01/08/2015 0
119 01/09/2015 0
119 01/10/2015 0
<snip>
119 01/06/2018 0
119 01/07/2018 0
--
120 01/07/2015 0
120 01/08/2015 1
120 01/09/2015 0
120 01/10/2015 0
<snip>
120 01/06/2018 0
120 01/07/2018 0
--
121 01/07/2015 0
121 01/08/2015 0
121 01/09/2015 1
121 01/10/2015 0
<snip>
121 01/06/2018 0
121 01/07/2018 0
我已经假设了一些有关您的开始/结束日期以及有效期的信息；希望您可以轻松地调整case语句以适合最适合您情况的逻辑。
答案 3 :(得分：0)
我也相信这可以作为CROSS JOIN的示例。我要做的就是创建一个包含所有日期的小表，然后将CROSS JOIN与订户表一起使用。