[英]Oracle 12C - Using Width_Bucket with Over Partition By Clause
我有以下数据集(简化),由与该工作类型关联的“WORK_TYPE”和“TASKTIME”组成。
+-----------+----------+--------+
| WORK_TYPE | TASKTIME | OUTPUT |
+-----------+----------+--------+
| TYPE1 | 10 | 1 |
| TYPE1 | 20 | 1 |
| TYPE1 | 30 | 2 |
| TYPE1 | 30 | 2 |
| TYPE2 | 10 | 1 |
| TYPE2 | 10 | 1 |
| TYPE2 | 20 | 2 |
| TYPE2 | 20 | 2 |
+-----------+----------+--------+
我希望在这个数据集上使用 width_bucket function。 但是,我想按 work_types 对数据进行分区,因此每种类型都被分组,而与整个数据集无关。
SELECT
TASKTIME
,WORK_TYPE
,WIDTH_BUCKET(TASKTIME,0,100,30) AS TASKTIME_BUCKET
,WIDTH_BUCKET(TASKTIME,0,100,30) OVER (PARTITION BY WORK_TYPE) AS TASKTME_BUCKET_WT --This Errors
FROM TABLE1
第一个 width_bucket 有效,但是将值存储在整个数据集中。 我尝试在 width_bucket 之后使用OVER (PARITION BY WORK_TYPE)
,但这会导致以下错误: ORA-00923: FROM keyword not found where expected
有任何想法吗?
如果您想要每个组的宽度相等的桶,您可以为每个组计算单独的最小值和最大值:
SELECT TASKTIME, WORK_TYPE,
WIDTH_BUCKET(TASKTIME, 0, 100, 30) AS TASKTIME_BUCKET
WIDTH_BUCKET(TASKTIME, MIN_TASKTIME, MAX_TASKTIME, 30) AS TASKTME_BUCKET_WT
FROM (SELECT t1.*,
MIN(TASKTIME) OVER (PARTITION BY WORK_TYPE) as MIN_TASKTIME,
MAX(TASKTIME) OVER (PARTITION BY WORK_TYPE) as MAX_TASKTIME
FROM TABLE1 t1
) t1
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