count（distinct）over（partition by ...在Oracle SQL中不起作用

Question

我想计算过去30天的distinct day_number 。 但是，不同的功能不能与over一起使用

如果我删除distinct ，它会给我day_number的总数，但是day_number可以有很多重复。 所以这就是为什么我要添加distinct 。

select tr.*,
       count( distinct day_number) OVER (PARTITION BY ACCOUNT ORDER BY DAY_number range 29 PRECEDING) as result
from table tr;

任何人都可以告诉我如何计算over(partition by..)语句中的不同数字？ 提前致谢。

Answer 1

您可以通过首先创建一个列，该列仅列出每个id一次，然后对该列执行范围计数，例如：

WITH sd AS (SELECT 1 ID, 10 val FROM dual UNION ALL
            SELECT 1 ID, 20 val FROM dual UNION ALL
            SELECT 2 ID, 30 val FROM dual UNION ALL
            SELECT 2 ID, 40 val FROM dual UNION ALL
            SELECT 4 ID, 50 val FROM dual UNION ALL
            SELECT 4 ID, 60 val FROM dual UNION ALL
            SELECT 6 ID, 70 val FROM dual)
SELECT ID,
       val,
       COUNT(id_distinct) OVER (ORDER BY ID RANGE 3 PRECEDING) cnt_disinct_ids
FROM   (SELECT ID,
               val,
               CASE WHEN row_number() OVER (PARTITION BY ID ORDER BY val) = 1 THEN ID END id_distinct
        FROM   sd);

        ID        VAL CNT_DISINCT_IDS
---------- ---------- ---------------
         1         10               1
         1         20               1
         2         30               2
         2         40               2
         4         50               3
         4         60               3
         6         70               2

---

ETA：证明上述技术适用于您的数据：

WITH your_table AS (SELECT 'ABCDE' account_sk, 23 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 23 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 24 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 25 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 53 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 53 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 55 day_sk FROM dual UNION ALL
                    SELECT 'VWXYZ' account_sk, 10 day_sk FROM dual UNION ALL
                    SELECT 'VWXYZ' account_sk, 12 day_sk FROM dual UNION ALL
                    SELECT 'VWXYZ' account_sk, 40 day_sk FROM dual UNION ALL
                    SELECT 'VWXYZ' account_sk, 40 day_sk FROM dual)
SELECT account_sk,
       day_sk,
       COUNT(day_sk_distinct) OVER (PARTITION BY account_sk ORDER BY day_sk RANGE BETWEEN 29 PRECEDING AND CURRENT ROW) count_distinct_day_sk
FROM   (SELECT account_sk,
               day_sk,
               CASE WHEN row_number() OVER (PARTITION BY account_sk, day_sk ORDER BY day_sk) = 1 THEN day_sk END day_sk_distinct
        FROM   your_table);

ACCOUNT_SK     DAY_SK COUNT_DISTINCT_DAY_SK
---------- ---------- ---------------------
ABCDE              23                     1
ABCDE              23                     1
ABCDE              24                     2
ABCDE              25                     3
ABCDE              53                     3
ABCDE              53                     3
ABCDE              55                     2
VWXYZ              10                     1
VWXYZ              12                     2
VWXYZ              40                     2
VWXYZ              40                     2

Answer 2

count(distinct ...)适用于over子句，主要问题是order by。 你不能count (distinct ..) over (partition by ... order by ...)因为DISTINCT函数和RATIO_TO_REPORT不能有ORDER BY。 所以我这样做了：

select tr.*, count (distinct day_number) over (partition by account)
from (select t.*, row_number() over (partition by account order by day_number) row_number from table t) tr
where row_number < 30;

我已经在HR的员工计划中测试了它（你可以在任何地方找到免费oracle方案）我不确定它会在你的架构中有效，因为我没有它的副本，但是，如果没有，它应该给你一些思路：

 select count (distinct manager_id) over (partition by department_id), department_id, manager_id
from (select e.*, row_number() over (partition by department_id order by employee_id) row_number from employees e)
where row_number < 30;