如何使用SQL为包含相同数据集的所有组生成标识符？

我有一个表格，包含公交线路时刻表，如下所示：

    SELECT
            NULL AS CALENDARDATE
           ,NULL AS BUS_ID
           ,NULL AS STOP_ID
           ,NULL AS TIME
      FROM DUAL UNION
    SELECT '20190101',  100, 'STOP_1000', '8:30' FROM DUAL UNION
    SELECT '20190101',  100, 'STOP_1000', '8:35' FROM DUAL UNION
    SELECT '20190101',  100, 'STOP_1001', '8:32' FROM DUAL UNION
    SELECT '20190101',  100, 'STOP_1001', '8:37' FROM DUAL UNION

    SELECT '20190102',  100, 'STOP_1000', '8:30' FROM DUAL UNION
    SELECT '20190102',  100, 'STOP_1000', '8:35' FROM DUAL UNION
    SELECT '20190102',  100, 'STOP_1001', '8:32' FROM DUAL UNION
    SELECT '20190102',  100, 'STOP_1001', '8:37' FROM DUAL UNION

    SELECT '20190103',  100, 'STOP_1000', '8:30' FROM DUAL UNION
    SELECT '20190103',  100, 'STOP_1000', '8:35' FROM DUAL UNION
    SELECT '20190103',  100, 'STOP_1001', '8:32' FROM DUAL UNION
    SELECT '20190103',  100, 'STOP_1001', '8:37' FROM DUAL UNION
    SELECT '20190103',  100, 'STOP_1003', '8:39' FROM DUAL UNION

    SELECT '20190104',  100, 'STOP_1000', '8:30' FROM DUAL UNION
    SELECT '20190104',  100, 'STOP_1000', '8:35' FROM DUAL UNION
    SELECT '20190104',  100, 'STOP_1001', '8:32' FROM DUAL UNION
    SELECT '20190104',  100, 'STOP_1003', '8:37' FROM DUAL UNION

    SELECT '20190105',  100, 'STOP_1000', '8:30' FROM DUAL UNION
    SELECT '20190105',  100, 'STOP_1000', '8:35' FROM DUAL UNION
    SELECT '20190105',  100, 'STOP_1001', '8:32' FROM DUAL UNION
    SELECT '20190105',  100, 'STOP_1001', '8:37' FROM DUAL UNION

    SELECT '20190101',  101, 'STOP_1002', '8:20' FROM DUAL UNION
    SELECT '20190101',  101, 'STOP_1002', '8:30' FROM DUAL UNION
    SELECT '20190101',  101, 'STOP_1002', '8:40' FROM DUAL UNION
    SELECT '20190101',  101, 'STOP_1002', '8:50' FROM DUAL UNION

    SELECT '20190102',  101, 'STOP_1002', '8:22' FROM DUAL UNION
    SELECT '20190102',  101, 'STOP_1002', '8:30' FROM DUAL UNION
    SELECT '20190102',  101, 'STOP_1002', '8:40' FROM DUAL UNION
    SELECT '20190102',  101, 'STOP_1002', '8:50' FROM DUAL UNION

    SELECT '20190103',  101, 'STOP_1002', '8:20' FROM DUAL UNION
    SELECT '20190103',  101, 'STOP_1002', '8:30' FROM DUAL UNION
    SELECT '20190103',  101, 'STOP_1002', '8:40' FROM DUAL UNION
    SELECT '20190103',  101, 'STOP_1002', '8:50' FROM DUAL ;

我们的目标是压缩数据，我们将为每天创建一个计划id，而不是冗余数据

也就是说，对于同一天，对于同一条公交线路，如果它的所有路线都在同一时间经过相同的站点，那么我们应该为这一天生成相同的计划id，并且每次生成相同的公交组合

我已经尝试通过两个循环来解决这个问题，外部循环每天都在进行，每个公交线路的内部循环，但是由于数据量巨大，需要花费太多的时间来完成。（每年约5亿美元）

我知道我可以在SQL Server中使用XML Path来实现这一点，但这在Oracle中不起作用，我确信Oracle中肯定还有其他elegent解决方案，但我对Oracle还是新手，所以我在这里寻求帮助

环境是Oracle 11G

预期结果：

    SELECT
            NULL AS SCHEDUL_ID
           ,NULL AS CalendarDate
           ,NULL AS BUS_ID
           ,NULL AS STOP_ID
           ,NULL AS TIME
        FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_01', '20190101',  100, 'STOP_1000', '8:30' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_01', '20190101',  100, 'STOP_1000', '8:35' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_01', '20190101',  100, 'STOP_1001', '8:32' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_01', '20190101',  100, 'STOP_1001', '8:37' FROM DUAL UNION

    SELECT 'SCHEDULE_2019_100_01', '20190102',  100, 'STOP_1000', '8:30' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_01', '20190102',  100, 'STOP_1000', '8:35' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_01', '20190102',  100, 'STOP_1001', '8:32' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_01', '20190102',  100, 'STOP_1001', '8:37' FROM DUAL UNION

    SELECT 'SCHEDULE_2019_100_02', '20190103',  100, 'STOP_1000', '8:30' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_02', '20190103',  100, 'STOP_1000', '8:35' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_02', '20190103',  100, 'STOP_1001', '8:32' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_02', '20190103',  100, 'STOP_1001', '8:37' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_02', '20190103',  100, 'STOP_1003', '8:39' FROM DUAL UNION

    SELECT 'SCHEDULE_2019_100_03', '20190104',  100, 'STOP_1000', '8:30' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_03', '20190104',  100, 'STOP_1000', '8:35' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_03', '20190104',  100, 'STOP_1001', '8:32' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_03', '20190104',  100, 'STOP_1003', '8:37' FROM DUAL UNION

    SELECT 'SCHEDULE_2019_100_01', '20190105',  100, 'STOP_1000', '8:30' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_01', '20190105',  100, 'STOP_1000', '8:35' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_01', '20190105',  100, 'STOP_1001', '8:32' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_100_01', '20190105',  100, 'STOP_1001', '8:37' FROM DUAL UNION

    SELECT 'SCHEDULE_2019_101_01', '20190101',  101, 'STOP_1002', '8:20' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_101_01', '20190101',  101, 'STOP_1002', '8:30' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_101_01', '20190101',  101, 'STOP_1002', '8:40' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_101_01', '20190101',  101, 'STOP_1002', '8:50' FROM DUAL UNION

    SELECT 'SCHEDULE_2019_101_02', '20190102',  101, 'STOP_1002', '8:22' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_101_02', '20190102',  101, 'STOP_1002', '8:30' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_101_02', '20190102',  101, 'STOP_1002', '8:40' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_101_02', '20190102',  101, 'STOP_1002', '8:50' FROM DUAL UNION

    SELECT 'SCHEDULE_2019_101_01', '20190103',  101, 'STOP_1002', '8:20' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_101_01', '20190103',  101, 'STOP_1002', '8:30' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_101_01', '20190103',  101, 'STOP_1002', '8:40' FROM DUAL UNION
    SELECT 'SCHEDULE_2019_101_01', '20190103',  101, 'STOP_1002', '8:50' FROM DUAL 

一旦得到了预期的结果，剩下的就很简单了，我可以将数据拆分为两个表： 第一个包含日期和计划id的映射，第二个包含计划id、停止和时间

另外，您可以忽略具有空值的行，它只是用于脚本中列名的可读性。

您可以使用

listag（）

和

dense\u rank（）

。首先使用

listag（）

进行分类：

然后指定一个值：

select b.*,
       rank() over (order by busid, stops) as scheule_number
from (select calendardate, busid,
             listagg(stopid || '-' || time, ',') within group (order by time) as stops
      from buslines
      group by calendardate, busid
     ) b;

---

请注意，这包括了排名中的

busid

，因此两个停靠相同站点的公交车具有不同的时刻表编号。

查询在test env上运行良好，但在prod env上它给出了错误：SQL错误：ORA-01489:字符串连接的结果太长，我想我必须创建一个用户定义的agg函数。@AdamYan。最长的母线有多长？stopid和时间应该大约为10个字符。我想100个车站对一辆公共汽车来说太多了。你也许可以过滤异常值。我可以达到50+站和100+次。这就是5000个委员会*5*2=每条总线每天大约50000个字符。@AdamYan。使用50个停止，您可能能够将停止和时间编码为少于15个字符。如果是这样，那么字符串可能不会超过内置限制。谢谢Gordon，我让它工作了，因为原始数据使用了重叠的日期范围，如date_from和date_to，所以我在date_from | date_to上使用了ListAGG函数，这绕过了01489错误。我还计划对listag结果使用ORA_散列，因此我将使用稳定的标识符，而不是非稳定的rank（）结果。

select b.*,
       rank() over (order by busid, stops) as scheule_number
from (select calendardate, busid,
             listagg(stopid || '-' || time, ',') within group (order by time) as stops
      from buslines
      group by calendardate, busid
     ) b;