基于Oracle的PIVOT具有多个列组 [英] Oracle based PIVOT with multiple columns group
问题描述
使用以下表格
生产力：
PRODUCTIVITYID PDATE EMPLOYEEID ROOMID ROOMS_SOLD SCR
81 03/26/2016 7499 21 56 43
82 03/26/2016 7566 42 - -
102 03/26/2016 7499 22 434 22
101 03/26/2016 7566 21 43 53
效率D：
PRODUCTIVITYID WORKHRS MEALPANELTY DESCRIPTION
2 50 4 -
21 6.4 1 -
102 6 - -
81 1.32 - -
101 3.6 - -
客房：
ID ROOM PROPERTCODE
22 102 6325
41 103 6325
42 104 6325
43 105 6325
EMP：
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
7566 JONES MANAGER 7839 04/02/1981 2975 - 20
7788 SCOTT ANALYST 7566 12/09/1982 3000 - 20
7902 FORD ANALYST 7566 12/03/1981 3000 - 20
7369 SMITH CLERK 7902 12/17/1980 800 - 20
7499 ALLEN SALESMAN 7698 02/20/1981 1600 300 30
以下查询生成以下输出，但我需要分组 employees 和< c> workhrs ，然后转动 RM_ROOM 和 RM_SCR
The following query is generating below output but I need to group employees and sum workhrs and then pivot RM_ROOM and RM_SCR
WITH pivot_data AS (
SELECT eNAME,workhrs,room, 'RM' as RM,SCR from PRODUCTIVITY p,PRODUCTIVITYd d, emp e, ROOMS R
where p.PRODUCTIVITYID=d.PRODUCTIVITYID and e.empno=p.employeeid
AND R.ID=P.ROOMID
)
SELECT *
FROM pivot_data
PIVOT (
MIN(room) as room,min(scr) as SCR --<-- pivot_clause
FOR RM--<-- pivot_for_clause
IN ('RM') --<-- pivot_in_clause
)
电流输出： / strong>
Current Output:
ENAME WORKHRS 'RM'_ROOM 'RM'_SCR
JONES 3.6 101 53
ALLEN 6 102 22
ALLEN 1.32 101 43
希望的输出：
ENAME WORKHRS 'RM'_ROOM 'RM'_SCR 'RM'_ROOM 'RM'_SCR
JONES 3.6 101 53 - -
ALLEN 7.32 101 43 102 22
推荐答案
你正在以一个固定值，字符串字面值'RM'进行循环，所以你真的没有做任何有用的枢轴 - 输出是相同的，自己运行'pivot_data'查询：
You are pivoting on a fixed value, the string literal 'RM', so you're really not doing anything useful in the pivot - the output is the same as you'd get from running the 'pivot_data' query on its own:
SELECT eNAME,workhrs,room, SCR from PRODUCTIVITY p,PRODUCTIVITYd d, emp e, ROOMS R
where p.PRODUCTIVITYID=d.PRODUCTIVITYID and e.empno=p.employeeid
AND R.ID=P.ROOMID;
ENAME WORKHRS ROOM SCR
----- ---------- ---------- ----------
JONES 3.6 101 53
ALLEN 1.32 101 43
ALLEN 6 102 22
您希望每个员工的总额 workhrs 以及他们售出的房间的枢纽。如果您更改该查询以获得 workhrs 的分析和，以及房间/ scr值的排名（并使用现代连接语法），您将获得：
You want the aggregate workhrs for each employee, and a pivot of the rooms they sold. If you change that query to get the analytic sum of workhrs and a ranking of the room/scr values (and using modern join syntax) you get:
select e.ename, r.room, p.scr,
sum(d.workhrs) over (partition by e.ename) as wrkhrs,
rank() over (partition by e.ename order by r.room, p.scr) as rnk
from productivity p
join productivityd d on d.productivityid = p.productivityid
join emp e on e.empno=p.employeeid
join rooms r on r.id = p.roomid;
ENAME ROOM SCR WRKHRS RNK
----- ---------- ---------- ---------- ----------
ALLEN 101 43 7.32 1
ALLEN 102 22 7.32 2
JONES 101 53 3.6 1
然后，您可以关注生成的 rnk 号码：
You can then pivot on that generated rnk number:
with pivot_data as (
select e.ename, r.room, p.scr,
sum(d.workhrs) over (partition by e.ename) as wrkhrs,
rank() over (partition by e.ename order by r.room, p.scr) as rnk
from productivity p
join productivityd d on d.productivityid = p.productivityid
join emp e on e.empno=p.employeeid
join rooms r on r.id = p.roomid
)
select *
from pivot_data
pivot (
min(room) as room, min(scr) as scr --<-- pivot_clause
for rnk --<-- pivot_for_clause
in (1, 2, 3) --<-- pivot_in_clause
);
ENAME WRKHRS 1_ROOM 1_SCR 2_ROOM 2_SCR 3_ROOM 3_SCR
----- ---------- ---------- ---------- ---------- ---------- ---------- ----------
ALLEN 7.32 101 43 102 22
JONES 3.6 101 53
您需要知道任何员工可能拥有的房间 - 即最高的 rnk 可能是 - 并且包括 in 这意味着您可能最终会出现空列，如本示例中没有 3_room 或 3_scr 。你不能避免这种情况，除非你得到一个XML结果或动态生成查询。
You need to know the maximum number of rooms any employee may have - i.e. the highest rnk could ever be - and include all of those in the in clause. Which means you're likely to end up with empty columns, as in this example where there is no data for 3_room or 3_scr. You can't avoid that though, unless you get an XML result or generate the query dynamically.
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