基于数据的 Oracle Pivot 帮助
Posted
技术标签:
【中文标题】基于数据的 Oracle Pivot 帮助【英文标题】:Oracle Pivot Help based on Data 【发布时间】:2020-10-15 21:07:03 【问题描述】:我正在尝试使用 oracle 数据透视函数 以以下格式显示数据。我尝试使用我在 *** 中找到的示例,但我无法实现我所期待的。
With t as
(
select 1335 as emp_id, 'ADD Insurance New' as suuid, sysdate- 10 as startdate, null as enddate from dual
union all
select 1335 as emp_id, 'HS' as suuid, sysdate- 30 as startdate, null as enddate from dual
union all
select 1335 as emp_id, 'ADD Ins' as suuid, sysdate- 30 as startdate, Sysdate - 10 as enddate from dual
)
select * from t
输出:
+--------+-------------------+-------------------+---------+-------------------+
| EMP_ID | SUUID_1 | SUUID_1_STARTDATE | SUUID_2 | SUUID_2_STARTDATE |
+--------+-------------------+-------------------+---------+-------------------+
| 1335 | ADD Insurance New | 10/5/2020 15:52 | HS | 9/15/2020 15:52 |
+--------+-------------------+-------------------+---------+-------------------+
谁能建议如何使用 SQL Pivot 来获取这种格式?
【问题讨论】:
【参考方案1】:您可以使用条件聚合。理解您的问题的方法不止一种,但适用于您的示例数据的一种方法是:
select emp_id,
max(case when rn = 1 then suuid end) suuid_1,
max(case when rn = 1 then startdate end) suid_1_startdate,
max(case when rn = 2 then suuid end) suuid_2,
max(case when rn = 2 then startdate end) suid_2_startdate
from (
select t.*, row_number() over(partition by emp_id order by startdate desc) rn
from t
where enddate is null
) t
group by emp_id
Demo on DB Fiddle:
EMP_ID | SUUID_1 | SUID_1_STARTDATE | SUUID_2 | SUID_2_STARTDATE -----: | :---------------- | :--------------- | :-------- | :--------------- 第1335章添加保险 新 | 05-OCT-20 | HS | 20 年 9 月 15 日【讨论】:
【参考方案2】:您可以使用PIVOT
:
With t ( emp_id, suuid, startdate, enddate ) as
(
select 1335, 'ADD Insurance New', sysdate- 10, null from dual union all
select 1335, 'HS', sysdate- 30, null from dual union all
select 1335, 'ADD Ins', sysdate- 30, Sysdate - 10 from dual
)
SELECT emp_id,
"1_SUUID" AS suuid1,
"1_STARTDATE" AS suuid_startdate1,
"2_SUUID" AS suuid2,
"2_STARTDATE" AS suuid_startdate2
FROM (
SELECT t.*,
ROW_NUMBER() OVER ( ORDER BY startdate DESC, enddate DESC NULLS FIRST )
AS rn
FROM t
)
PIVOT (
MAX( suuid ) AS suuid,
MAX( startdate ) AS startdate,
MAX( enddate ) AS enddate
FOR rn IN ( 1, 2 )
)
输出:
EMP_ID | SUUID1 | SUUID_STARTDATE1 | SUUID2 | SUUID_STARTDATE2 -----: | :---------------- | :--------------- | :----- | :--------------- 第1335章添加保险 新 | 05-OCT-20 | HS | 20 年 9 月 15 日
db小提琴here
【讨论】:
以上是关于基于数据的 Oracle Pivot 帮助的主要内容,如果未能解决你的问题,请参考以下文章